!" " " " # # $ # # % $ # # # & % # "# # ' # ( # (# (# ($ # % ( %#(""(" " " !" ! ' % # ##)* +% ! ,' '# ) ! (( ) - # ! ! " # $ %$ & '( %)* ! " # $ !% ! & $ ' ' ( $ ' # % % ) %* %' $ ' + " % & ' ! # $, ( $ - $$$ . /"/"#012" )*3- +) * !4 !& 5!6%55787& % 2 9:;567<=8>!<7>%2 ?9:;567<=8>!<7>; 0 @ ! * A & B $!+,-./010/2/20-3420$ &5)!* C 2 ! "C ( % 2 % % % % $ $ $C C CC - $$$ /"/"#012" )*3- +) * !4 !& 5!6%55787& %) C 9:;567<=8>!<7>%D?9:;567<=8>!<7>; 0 @ ! E & " E + CA B $!+,-./010/2/20-3420- C F8>78C /"/"#012" )*3- +) " & 8>78 About the Author Dr. Onofre Sumindo Corpuz is native in Matalam, Cotabato. He finished his Doctor of Philosophy in Forestry in 2008 at the University of the Philippines, Los Baños Laguna; Master of Science in Agriculture and Master of Arts in Educational Administration (Acad. Requirements) and BS Forestry studies at the Cotabato Foundation College of Science and Technology, Doroluman Arakan, Cotabato in 2002 and 1993 respectively. He also earned units in Research and Development Management at the University of the Philippines Open University in 1997. The author is an Associate Professor of the Cotabato Foundation College of Science and Technology, Doroluman, Arakan, Cotabato PHILIPPINES. At present, he was designated Dep. Director of the Research and Extension at the same time Chairman of the MS Degree Program of the same School. He has presented various scientific papers in National and International conferences and seminars. He has published four books entitled ROOT GROWTH POTENTIALS AND HERITABILITY OF GMELINA ARBOREA, CARBON BUDGET AND STEM CUT PROPAGATION TECHNOLOGIES FOR RUBBER TREES, FORESTRY STATISTICS AND RESEARCH METHODS, and PROCEDURES IN FORESTRY AND AGRICULTURAL RESEARCH. He also published scientific papers in an ISI index Journal such as the International ASIA Life Science Journal, and the Philippine Journal of Crop Science. Published two articles in the Book of Abstract of the 2nd World Agroforestry Congress held at Nairobi, Kenya in 2008. The author has various Professional and Scientific affiliations like; Society of Filipino Foresters, Philippine- ASIAN Japan Association, Forests and Natural Resources Research Society of the Philippines, Philippine Society for the Study of Nature, CSDi Community Development (www. Developmentcommunity.csd-i.org.). Member of the following group: Forest and Trees, Subsistence Farming, i O.S. Corpuz 2012 UNDERSTANDING STATISTICS Feature your Project, Adapting to Climate Change, Environmental Education, Conservation and Restoration, LinkedIn Group Member (www.linkedin.com): Society for Conservation Biology, Precision Forestry, Asia Life Science Academy (ALSA) Alumni and Network, Climate Eval: Evaluation of Climate Change, Esri Forestry Group, Freelance Natural Resources Professionals, GIS & Geospatial Technology, GIS Mapping and Geo Technology Professionals, GISuser-Mobil Mapping and Field GIS, GRASS GIS, Natural Resources Management Professionals, Professional in Environmental Risk Assessment, SEARCAL, Society for Conservation Biology, Society for Conservation GIS, Arboriculture and Landscape, Environmental Compliance Consultants, Environmental Sustainability Professionals, International Society of Arboriculture, National Spatial Data Infrastructure, Trees, Sustainable Forestry Initiatives and Supporters, and National Biodiversity Network. ii ACKNOWLEDGEMENT The authors express their sincere thanks and gratitude to their brothers and sisters, kids, friends and relatives. To all those who help and encouraged them finished this humble work. They have been indebted with ideas, inspirations, moral and spiritual supports, and suggestions which contributed much to the realization of this book. Above all, the Almighty ALLAH for his daily blessing and guidance. iii O.S. Corpuz 2012 UNDERSTANDING STATISTICS FOREWORD This book is design to College students and other Professionals who would like to enhance knowledge on social and educational statistics. The book deals with uses and importance of statistics, collection and presentation of data, sampling technique, data organization, descriptive statistics such as frequency distribution, measures of central tendencies, position, dispersion/ variability, skewness, kurtosis, and normal distribution of data. Discussion and concrete illustrations on regression and correlation analysis, t- test, F-test, z-test, chi-square test, alternative nonparametric analysis of variance, one-way classification and two-way classification analysis of data. The methods of social research presented will enlighten young researcher on the appropriateness of statistical tools they will use in the analysis of data. The Authors iv Title Page Chapter 1. INTRODUCTION Meaning and Uses Classification of Statistics Descriptive Statistics Inferential Statistics Parametric Statistics Non-Parametric Statistics Concept of Population and Sample Parameters and Estimates Subscript and Summation Notations Chapter 2. COLLECTION AND PRESENTATION OF DATA 1. Types of Data a. Qualitative Data b. Quantitative Data 2. Sources of data 3. Types of Error in Data 4. Methods of Gathering Data 5. Sampling Technique a. Probability/Scientific Sampling Random Sampling Systematic Sampling Stratified Sampling Cluster Sampling Multi Stage b. Non-Probability/non-scientific Sampling Purposive Sampling Quota Sampling Convenience Sampling 1 1 1 1 2 2 3 3 4 5 3 8 8 8 8 8 9 11 12 12 12 13 14 15 18 18 18 18 19 v O.S. Corpuz 2012 UNDERSTANDING STATISTICS Incidental sampling Chapter 3. DATA ORGANIZATION Tabular Presentation Graphs and Diagram Chapter 4. FREQUENCY DISTRIBUTION Relative frequency distribution Cumulative frequency distribution Cumulative percent Chapter 5. MEASURES OF CENTRAL TENDENCIES The Mean The Median The Mode Chapter 6. MEASURES OF POSITION The Quartiles The Deciles The percentiles Chapter 7. MEASURES OF DISPERSION/VARIABILITY The Range The Inter quartile Range The Quartile Deviation The Average Deviation The Standard Deviation Chapter 8. MEASURES OF SKEWNESS AND KURTOSIS Positive Skewness vi 19 20 20 21 24 25 27 30 32 32 36 39 42 42 46 53 61 61 62 63 65 67 71 71 Negative Skewness Moment Coefficient of Skewness Measures of Kurtosis Leptokurtic Mesokurtic Platykurtic Moment Coefficient of Kurtosis Chapter 9. NORMAL DISTRIBUTION The Normal Curve Standard Normal Scores Areas Under the Normal Curve Chapter 10. T-TESTS Test for Dependent or Correlated Samples T-test for Independent Samples Test for Equality of Variance Testing for the Difference in Means of the Two Independent Samples 71 72 76 76 76 76 80 80 81 82 84 84 94 96 97 Chapter 11. SIMPLE LINEAR REGRISSION ANALYSIS 110 Evaluation of the Simple Linear Regression Equation 112 Chapter 12. CORELATION ANALYSIS Estimation of the Correlation Coefficient Pearson Product-Moment Coefficient of Correlation Spearman Rank Correlation Chapter 13. ALTERNATIVE NON-PARAMETRIC TECHNIQUES McNemar Change Test Sign Test Wilcoxon Signed-Ranks Test 117 117 122 126 130 130 134 vii O.S. Corpuz 2012 UNDERSTANDING STATISTICS For Two Dependent Samples Independent Sample (two-sample case) Wilcoxon-Mann-Whitney Test Friedman Two-Way Analysis of Variance by Ranks Multiple Comparisons Between Groups of Conditions From the Result of the Friedman Two-Way ANOVA Chi-square Test for r x k Tables Kruskal – Wallis Test 141 143 146 149 153 155 159 Chapter 14. ANALYSIS OF VARIANCE One-Way Classification Analysis Two-Way Classification Analysis 162 162 174 APPENDICES 182 REFERENCES 193 viii Introduction Meaning and Uses of Statistics Statistics is a tools or methods use in data analysis. A scientific methods for collecting, organizing, analyzing, and interpreting quantitative data, as well as drawing valid conclusions and making reasonable decisions on the basis of such analysis. It is an essential tool in almost all fields of knowledge. Collecting data refers to the process of obtaining qualitative information and quantitative or numerical measurements needed in the study. Organizing is the tabulation/or presentation of data into tables, graphs or chart to formulate logical and statistical conclusions from the collected measurements. Analysis of data pertains to the process of extracting relevant information from the given data to formulate numerical descriptions. Interpretation of data on the other hand refers to the task of drawing conclusions from the analyzed statistical data. It also involves the formulation of forecast or predictions about larger populations based on the data collected from sample populations. Classification of Statistics Statistics can be classified into two: 1. Descriptive Statistics – describes and analyze a given group without drawing any conclusions and inferences about a larger group. It is also known as “Deductive 1 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Statistics”. Its concern includes the gathering, classification, and presentation of data such as in frequency distribution, cumulative frequency, percentage frequency, measures of central tendencies, measures of position, dispersion or variation, Skewness and Kurtosis. 2. Inferential Statistics – drawing important conclusions about the population inferred from the analysis of the sample. It is also known as “Inductive Statistics”. Testing of hypothesis through t-test, z-test, f-test, chi-square, Analysis of Variance (ANOVA) using parametric and non-parametric variables, regression and correlation analysis. Types: 1.1 Parametric Statistics. Are inferential techniques which make the following assumptions regarding the nature of the population from which the observations are drawn. - the observation must be independent. This means that in choosing any element from the population to be included in the sample must not affect the chances of other elements for inclusion. - the population must be drawn from normally distributed populations. The crude way of knowing that the distribution is normal is when the measures of central tendencies are all the same (mean=median=mode). Bell-shape curve will be produced when drawing a curve for this purpose. - the populations must have the same variance (homoscedastic populations) 2 - the variables must be measured in the interval or ratio scale, so that the result can be interpreted. 2.2. Non-Parametric Statistics. It makes fewer and weaker assumptions such as: - the observations must be independent and the variable has the underlying continuity. - the observations are measured in either the nominal or ordinal scales. To have better understanding on when to use the parametric and non-parametric statistics, please refer to the table below: INFERENTIAL STATISTICS Parametric DISTRIBUTION MEASUREMENT Normal Interval or ratio Non-Parametric Unknown Distribution Nominal or ordinal Concept of Population and Sample The central notion in any sampling problem is the existence of a population. A population is an aggregate of unit values, where the unit is the object upon which the observation is made, and the value is the property observed on that object. For example, in a class in which the unit being observed in the individual student is age. The population is the aggregate of all ages of the students in the class. The GPA of the same student would be another population. Often times, it is impossible or impractical to observe the entire group or aggregate especially if it is large. Hence, one may examine a small part of the entire group called sample. Parameters and Estimates To characterize the population as a whole, constants that are called parameters are often used. Examples are mean value of 3 O.S. Corpuz 2012 UNDERSTANDING STATISTICS height of student per class in a population of 3 sections and variability among these measured unit values. Often, but not always, one may estimate the population mean or total. The value of the parameter as estimated from a sample is referred to as the sample estimate or simply the estimate. Exercise No. 1 Test Yourself: (Follow instruction carefully) Get ½ sheet of yellow paper and answer the following by writing the letter of choice after the item number in your paper. COLUMN A COLUMN B __1.The process of extracting relevant information from a set of data a. Analysis of data __2.Refers to the drawing of conclusions or generalizations from the analyzed data b. Parametric Statistics __3.A science of collection, presentation, analysis and interpretation of data c. Data Interpretation __4.This refers to the collection of facts where the researcher get information d. Data Presentation __5.This refers to the organization of data into tables, graphs and charts to get a clear picture of relationship e. Organization of Data f. Descriptive Statistics __6.Each and every element of a large group of data g. Variable __7.Concern with summarizing values h. Inferential Statistics 4 to describe group characteristics of data __8.Statistics which make fewer and weaker assumptions using nominal and ordinal measurement i. Sample __9.An inferential technique, which assumes normality of distributions of population from which data are taken k. Non-parametric Statistics __10.Statistics that needs critical thinking and judgment using complex mathematical procedures l. Statistics j. Population Subscript and Summation Notations A subscript is a number or a letter representing several numbers placed at the lower right side of a variable. It is used to specify the item referred to. For example, if we have five trees of different diameters, and let X represent the tree, we will let X1 stand for the stem diameter of the first tree, X2 stand for the stem diameter of the second tree and X3 for the third tree. Sometimes we would like to summarize in just one term, the idea that there are five trees with their corresponding stem diameter. Here, instead of a numerical value, we may use a lettersubscript. We would then write the symbol as Xi where i stands for the numbers 1,2,3..n. in our particular example, i would stand for the numbers 1 to 5 because there are five trees with different stem diameters. Hence, Xi stands for X1, X2, X3,…,Xn The summation symbol is used to denote that the subscripted variables are to be added. n i=1 Is read as the summation of Xi where i is 1 to n this indicates that we get X1 + X2 + X3 +… Xn. 5 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The Summation Sign: denotes summation Consider these values: X1, X2, X3, X4 n 4 Xi i-1 i=1 = X1 + X2 + X3 + X4 Dot Notation: Another way of representing a sum of observations is to use a system called the DOT NOTATION system. Consider this set of data: Observation Group 1 Group 2 Group 3 Group 4 . Total 1 X11 - 3 X21 - 6 X31 - 1 X41 - 2 X.1 = 12 2 X12 - 5 X22 - 3 X32 - 5 X42 - 3 X.2 = 16 3 X13 - 4 X23 - 1 X33 - 3 X43 - 4 X.3 = 12 4 X14 - 2 X24 - 5 X34 - 2 X44 - 2 X.4 = 11 5 X15 - 1 X25 - 2 X35 - 4 X45 - 5 X.5 = 12 Total X1. = 15 X2. = 17 X3. = 15 X4. = 16 X.. = 63 Xij = stands for an ijth trees where i represents groups and j represents the number of observation. X1. = X11 + X12 + X13 + X14 + X15 = X2 . = X2j X3 . = X3j X4 . = X4j 6 5 i=1 5 i=1 5 i=1 5 Xi i=1 X.. 4 = X1. + X2. + X3. + X4.= Xi. = 63 i=1 Example: Variety I II III A 20 15 10 B 22 15 13 C 25 17 15 Total X1. = 67 Total X.3 = 57 X23 = 17 X32 = 13 X11 = 20 X.1 = 45 X2. = 47 X33 =15 X.2 = 50 X.. = 15 7 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Collection and Presentation of Data 1. Kinds of Data a. Qualitative data. are those that are not amenable to numerical measurements. Examples: b. yes or no responses sex, civil status ratings such as poor, fair, satisfactory, good or excellent In agriculture: climatic region, geographical locations , soil type, slope, land-used, kind of insecticides and fertilizers Quantitative data. are those that are amenable to numerical measurements and statistical analysis. These are either counts or measures, hence real nos. Examples: diameter, heights, test score, weight, distance, number of class, size of leaves, basal area, volumes, amount of currency, flying height, sea level, yield, leaf area index, panicle, number of branches, carbon density, biomass density 2. Sources of Data. Data can be obtained from principal sources such as: a. Direct or primary data. This are data arise from original investigations such as observations, interview, 8 questionnaires, experiments, measurements and the like b. Secondary data. Contained vague, general and at times chopped description of phenomena. This are more likely subjected to typographic errors. 3. Types of Error in Data a. Sampling error – is the difference between the estimated and the true values. That is, SE = Yi - μ (associated with the ith sample) Where: Yi = estimated mean of the ith sample μ = true population mean when Yi - μ = 0, this implies that there is no sampling error. Consider the set of data: Y1 = 1 Y2 = 2 Y3 = 3 To get the mean: sum all values divided by the number of values Y1 + Y2 + Y3 = 1 + 2 + 3 = 6 3 3 3 μ = 6/3, the population mean 9 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Now, consider the sampling experiment of drawing 2 values from this set of data without replacement. The number of possible samples is (3/2) = 3!/1!2! = 3. As follows: (N/ μ) = N!/(N-n)! n! Sample No. 1 (Y1,Y2) : (1, 2) Sample Mean Y1 = 3/3 2 (Y1,Y3) : (1, 3) Y2 = 4/3 3 (Y2,Y3) : (2, 3) Y3 = 5/3 Therefore, the sampling error for the ith sample is: 1. 3/3 – 6/3 = -1.00 2. 4/3 – 6/3 = -0.67 3. 5/3 – 6/3 = -0.33 Note: it is necessary to know the sampling error, so that we will be able to reduce, assess and investigate the nature and extent of SE. Sampling error can be reduced by increasing the size of the samples. a. Non-Sampling Error – this is usually committed in: - recording data - vague definition of terms - inaccurate measuring devices/techniques - inability of respondents to give accurate results - inconsistencies between experimental method and method of analysis used - computational mistakes Non-sampling error is unaffected by sample size 10 4. Methods of Gathering Data a. The direct or interview method. This is a method of personto-person exchange between the interviewer and the interviewee. The interview method provides consistent and more precise information since clarification may be given by the interviewee. Questions may be repeated or modified to suit each interviewee’s level of understanding. This method is time consuming, expensive and has limited field coverage. Types: a. b. c. d. Focus Group Discussion Personal interview House to house interview Phone interview b.The indirect or questionnaire method. Written responses are given to prepared questions. Prepared questionnaires will be constructed and submitted to the clients to answer each written questions. This method is inexpensive and can cover a wide area in a shorter span of time. Clients may feel a greater sense of freedom to express views and opinions because their anonymity is maintained. c.The registration method. Gathering of information is enforced by certain laws. Examples are the registration of births, death, motor vehicles, marriages, and licenses. The advantage of this method is that, information is kept systematically and made available to public because of the requirement of the law. d.The observation method. The investigator observes the behavior of the persons or organizations and their 11 O.S. Corpuz 2012 UNDERSTANDING STATISTICS outcomes. It is usually used when the subjects cannot talk or write. The method makes possible the recording of behavior at the appropriate time and situation. e.The experiment method. This method is used when the objective is to determine the cause and effect relationship of certain phenomena under controlled conditions. Scientific researchers usually use the experiment method in testing hypothesis. 5. Sampling Technique It is not necessary for a researcher to examine every member of a population to get data or information about the whole population. Cost and time constraints will prohibit one from undertaking a study of the entire population. At any rate, drawing sample units systematically or at random is appropriate. If sample is done in this way, we can validly infer conclusions about the entire population from the sample taken. There are two methods of drawing samples from given population. These are: 5.1. Probability/Scientific Sampling. Every element of the population has equal chance of being included in the sample and the probability that any specified unit of population is included in the sample is governed by this known chance. The types of probability sampling are the following: a. Random Sampling - is the method of selecting a sample size (n) from a population (P) such that each member of the population has an equal chance of being included in the sample and all possible combinations of 12 size have an equal chance of being selected as a sample. A prerequisite for the randomization is a complete listing of the population. Ways of drawing sample units at random. - Lottery/draw lot sampling - Table of random numbers - Drawing cards There are two ways of using the Table of random Numbers to wit: - Direct Selection method. Applicable when there are few samples to be selected. The sample unit can be directly from the table - Remainder Method. This method is actually a combination with direct selection. The method is used when the number obtained from the Table of Random Numbers exceeds the digit in the sampling frame c. Systematic Sampling – this method involves selecting every nth element of a series representing the population. A complete listing is also required in this method. Under this system, the sample units may be picked in the following manner: For instance, P = 100 n = 10 The value of n may be obtained by dividing the total number of elements in the population by the desired sample size. Thus, P/n = 100/10 = 10th 13 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The 10 sample units would therefore be the persons holding the following numbers: 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100 sample units. Types of Systematic Sampling: - Stratified Sampling. The population is divided into groups based on homogeneity in order to avoid the possibility of drawing samples whose members come from one stratum. The stratified random sampling may be used when it is known in advance that a special segment of the population would not have enough persons/objects in the sample when simple random sample were drawn. We allocate n either equally or proportionately to each stratum a. Equal Allocation. This methods involves taking the same number of units from each stratum to make up the desired sample size n, thus, k = n/ns where: k = number of sample elements per stratum n = desired sample size ns = number of strata Example: If a researcher decided the sample size n=200 taken from 5 strata. What will be the number of sample elements per strata? Solution: n = 200 ns = 5 14 k = n/ns k = 200/5 = 40 Hence, 40 units will be taken from each stratum to constitute the 200 sample elements decided by the researcher b. Proportional Allocation. Stratified sampling using proportionate allocation is used to guarantee a more representative sample from each stratum. It is expected that the more population in a stratum, the more sample units will be taken. Suppose the researcher decided to take 80 forestry students as sample which is proportionate in the four department of the College of Education presented below: Department Biological Science English Mathematics Pilipino Total Population (P) 150 140 200 75 565 Proportion wi = P/Tp 150/565 140/565 200/565 75/565 k = nswi 21 20 28 11 80 P = population wi = proportion of the population (P) with total population (Tp) k = number of elements to be taken per stratum ns = decided sample size Cluster Sampling. May refers to an area sample because it is frequently applied on a geographical basis. Districts or brgys of a municipality or city are selected. The districts or brgys constitute a cluster. It does not consider the homogeneity of a population instead; heterogeneity of the sample is more precise. 15 UNDERSTANDING STATISTICS O.S. Corpuz 2012 Determination of sample size in sampling Determining a sample size is basically anchored to the decision of researchers. Some researchers have no idea on how to determine sample size in a given population scientifically and arbitrarily. Major criterion to them is to select 50% + 1 is sufficient for their study. These ideas are not scientific in nature. Take note that sampling is advisable only if the population is equal or more than 100. Complete enumeration of population is advisable when the total population under study is less than 100. Calmorin 1995 suggested the equation to determine sample size scientifically: Ss NV + [Se2 (1 – p)] NSe + [V2 x p(1 – p)] = Where: Ss N V = sample size = total number of population = the standard value (2.58) or 1% level of probability with 0.99 reliability = sampling error (0.01) = the largest possible proportion (50%) Se p Example: The Forestry students want to determine the sample size of 1000 trees to measure diameter Solution: Ss = NV + [Se2 (1 – p)] NSe + [V2 x p(1 – p)] = 1000(2.58) + (0.01)2 x (1 – 0.50) 1000(0.01) + (2.58)2 x 0.50(1 – 0.50) 16 = 2,580 + 0.0001 x 0.50 10 + 6.6564 x 0.50 x 0.50 = 2,580 + 0.00005 10 + 1.6641 Ss = 2,580.00005 11.6641 = 221 Another method of determining sample is: n = P/1 +Pe2 where: n = sample size P = population e = sampling error (usually 1% and 5%) Example: P = 1000 e = 5% Solution: n = = = = = n = P/1 +Pe2 1000/1 + 1000(0.05)2 1000/1 + 1000(0.0025) 1000/(1 + 2.5) 1000/3.5 286 - Multi-stage Sampling. Uses several stages or phases in getting the sample from the general population but selection of sample is still done at random. This technique is useful in conducting nation-wide survey involving a large population. Example: A Country survey on climate change mitigation and adaptation practices. The researcher decided to conduct the survey in 5 Regions in the Philippines with 3 Provinces per Region, 4 municipalities 17 O.S. Corpuz 2012 UNDERSTANDING STATISTICS per Province and 4 barangays per municipality. Thus the total sample barangay will be: 5 x 3 x 4 x 4 = 240 barangays. Number of household will be another stage to be considered. Say, 10 household per barangay x 240. The total sample household then will be 2,400. 5.2. Non-Probability/non-scientific Sampling. is a technique wherein the manner of selecting units of the population depends on some inclusion rule as specified by the researcher. Types: 5.2.1. Purposive Sampling. This technique is based on the criteria or qualification given by the researcher. This sampling technique is used based on the knowledge of respondents on the given situation or questionnaires. Samples are taken if the researcher thinks that the respondents could supply/give the information needed by the researcher. 5.2.2. Quota Sampling. This sampling technique is quick and cheap method to employ. The researcher is given a definite instruction and quota about the subject population he will be working on, but the final choice of the actual respondent is left to his own preference, and is not predetermined by some carefully operated randomization plan. So that, not all population are given chance to be included. 5.2.3. Convenience Sampling. This uses some instruments or equipment that provides convenience like telephone, e-mail or hand set to pick his sample units. If the researcher calls at random or email at random, people without telephone or not opening an email will not be given chance to be a respondents. 18 5.2.4. Incidental Sampling. This sampling technique is applied to those samples which are taken because they are the most available (Guilford and Fruchter 1973). The researcher will just take individuals as sample until sample size is completed. Exercise No. 2 Test yourself: Answer the following questions: 1. If the Education students are stratified according to department where they belong (Biological Science, Mathematics, English, Pilipino, Physical Education), how many sample shall we get from each Department of we want to get 200 sample students allocated equally among each of the departments? Show your solution. 2. Suppose you would like to allocate proportionately the sample size of 200 among the Forestry Department with population given in the table below, how many sampling units would you allocate per stratum? DEPARTMENT Biological Science Mathematics English Pilipino Physical Education TOTAL P Proportion wi = P/Tp k = nswi 180 200 280 210 130 1000 19 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Data Organization Data that are collected in any form can be organized through tables, graphs, diagrams/or plots 1. Tabular Presentation There are basically two types of tables: the general or reference table and the summary or text table. The general table is used mainly as a repository of information. It is the table of the raw data. The summary table on the other hand is usually small in size and design to guide the reader in analyzing the data. It is usually accompanies a text discussion. Example 1: General reference table. Score of 4 First Year High School Students in two subjects Class Section Math Science A B C D 60 45 50 30 75 27 29 50 Example 2: Summary table. Mean score of 1st year high school students in Math and Science 20 Class Section Score A B C D Weighted Mean 67.5 36.0 39.5 40.0 45.75 2. Graphs and Diagrams a. Bar graphs – one of the most common and widely used graphical devices. This consists of bars or heavy lines of equal widths, either all vertical or all horizontal. The length of bars represents the magnitude of the quantities being compared. The illustration below is an example of bar graph: d. Line Graph. It is a line connecting points or intersection of the x and y axis in the graph. Example: 21 UNDERSTANDING STATISTICS O.S. Corpuz 2012 e. Pie Graph. Is usually used for percentage distribution of data. Example: Percentage Distribution of Educational Attainment Scatter Plot. Used to plot relationship of two variables Daily Income Example: AGE Scatter plot between age and daily income 22 Exercise No. 3 Test yourself: Answer the following 1. Give and define the two kinds of data 2. What are the types of error in data? how it is committed? 3. Give the different methods of collecting data. discuss atleast 3 methods 4. What are the different methods of organizing collected or measured data? 23 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Frequency Distribution Frequency distribution is a common technique of describing a set of data. It is the listing of the collected/measured data. To organize said data into a frequency distribution, we need to pick convenient intervals and tabulate the number of each data that falls into a particular interval. Frequency distribution – is used when there are many statistical data recorded/collected. Usually, it is utilized when the number of data collected exceeds or equal to 30 (n 30). The following steps are observed in preparing frequency distribution: 1. Look for the lowest and the highest data recorded 2. Subtract the lowest data from the highest data plus 1. 3. Decide on the number of steps or class intervals. The maximum number of intervals is 20, minimum number is 5, and the ideal number is between 10 and 15 inclusively. 4. Determine the interval size by dividing step 2 by the desired number of intervals. Unless specified, it is advisable to use the ideal number of intervals. 5. Choose an appropriate lower limit for the first class interval. This number should be minus 1 of the lowest data or is exactly divisible by the interval size. 6. Write the lowest limit at the bottom and from it, develop the lower limits of the next higher intervals by adding the interval size to a preceding lower limit until the highest data is included. From the lowest limits develop also the corresponding upper limits. 7. Read each data in a set of collected data and record a tally opposite the class interval to which it belongs. 24 8. Count the number of tallies falling within each class to get the frequency of each class intervals. 9. Add the frequencies to get the total number of data or samples (n). Sample Frequency Distribution of the Test Scores of 55 Students in Math 12 C.I Tally Frequency 65 - 69 60 - 64 55 – 59 50 – 54 45 – 49 40 – 44 35 - 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5–9 1–4 / / / // /// /// //// //// //// //// ////-/// //// - //// //// - // /// 1 1 1 2 3 3 4 3 4 5 8 10 7 3 Class Boundary 64.5 – 69.5 59.5 – 64.5 54.5 – 59.5 49.5 – 54.5 44.5 – 49.5 39.5 – 44.5 34.5 – 39.5 29.5 – 34.5 24.5 – 29.5 19.5 – 24.5 14.5 – 19.5 9.5 – 14.5 4.5 – 9.5 0 – 4.5 Class Mark 67 62 57 52 47 42 37 32 27 22 17 12 7 2 Relative frequency distribution – is a tabular arrangement of data showing the proportion in percent of each frequency to the total frequency. The relative frequency for each class interval is obtained by dividing the class frequency by the total frequency expressed in percent. 25 UNDERSTANDING STATISTICS O.S. Corpuz 2012 Example: In our frequency distribution, if the class frequency is 3, the relative frequency is 3/55 or 5.45%. The rest of the answers are shown in the following table. C.I 65 - 69 60 - 64 55 – 59 50 – 54 45 – 49 40 – 44 35 - 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5–9 1–4 Frequency 1 1 1 2 3 3 4 3 4 5 8 10 7 3 rf (%) 1.82 1.82 1.82 3.64 5.45 5.45 7.27 5.45 7.27 9.1 14.55 18.18 12.73 5.45 The relative frequency distribution can be shown graphically through the use of the histogram or the relative frequency polygon as in the following representations: Bar Graph 20 Relative Frequency (%) 18 16 14 12 10 8 6 4 2 0 2 7 12 17 22 27 32 37 42 47 52 57 62 Class Mark 26 67 Line Graph 20 18 16 14 12 10 8 6 4 2 0 2 7 12 17 22 27 32 37 42 47 52 57 62 67 Class Mark Cumulative frequency distribution – is a tabular arrangement of data by class interval whose frequencies are cumulated. There are two kinds of cumulative frequency distributions. These are: 1. The “less than” cumulative frequency distribution whose sum of frequencies for each class interval is less than the upper class boundary of the interval they correspond to. It is graphically represented by a rising frequency polygon which we shall call “less than” ogive or < ogive. Each number in the < cf column is interpreted as follows: One item is less than 69.5; two items are less than 64.5; 3 items are less than 59.5; five items are less than 54.5; eight items are less than 49.5; eleven items are less than 44.5; and so forth. 27 O.S. Corpuz 2012 UNDERSTANDING STATISTICS C.I 65 - 69 60 - 64 55 – 59 50 – 54 45 – 49 40 – 44 35 - 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5–9 1–4 Frequency 1 1 1 2 3 3 4 3 4 5 8 10 7 3 <cf 55 54 53 52 50 47 44 40 37 33 28 20 10 3 Line graph of < cumulative frequency with class mark 60 Cumulative Freq. < 50 40 30 20 10 0 2 7 12 17 22 27 32 37 42 47 52 57 62 Class Mark 28 67 2. The “greater than” cumulative frequency distribution whose sum of frequencies for each class interval is greater than the lower class boundary of the interval they correspond to. It is graphically represented by a falling frequency polygon which we shall call the “greater than” ogive or the > ogive. C.I 65 - 69 60 - 64 55 – 59 50 – 54 45 – 49 40 – 44 35 - 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5–9 1–4 Frequency 1 1 1 2 3 3 4 3 4 5 8 10 7 3 >cf 1 2 3 5 8 11 15 18 22 27 35 45 52 55 Each number in the > cf column is interpreted as follows. Fiftyfive items are greater than 69.6; fifty-four items are greater than 64.5 and so on and so forth. 60 Cumulative Freq. 50 40 30 20 10 0 2 7 12 17 22 27 32 37 42 47 52 57 62 67 Class Mark 29 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Cumulative percent – refers to the tabular presentation of the percentage of cumulative frequency by class interval. This can be done by getting first the respective relative frequency of each class. The first rf is equals to the first cumulative percent and add each rf for the succeeding cumulative percent done as follows: C.I 65 - 69 60 - 64 55 – 59 50 – 54 45 – 49 40 – 44 35 - 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5–9 1–4 30 Frequency 1 1 1 2 3 3 4 3 4 5 8 10 7 3 rf 1.82 1.82 1.82 3.64 5.45 5.45 7.27 5.45 7.27 9.09 14.55 18.18 12.73 5.45 <cf 55 54 53 52 50 47 44 40 37 33 28 20 10 3 Cum. % 100 98.17 96.35 94.53 90.89 85.44 79.99 72.72 67.27 60.00 50.91 36.36 18.18 5.45 Exercise No. 4 The following are the scores obtained by a group of 60 College students in Math 12 examination: MATH 12 SCORES 88 84 42 96 83 54 44 72 82 86 81 63 72 73 98 73 74 62 69 85 79 86 89 39 45 59 65 78 77 73 90 59 78 88 88 91 68 78 86 89 78 89 72 77 69 74 80 68 79 49 82 76 81 43 40 66 81 70 50 75 Problem Solving and Statistical Analysis: a. Prepare a frequency distribution of the above scores, using a class interval of 10 b. Present the frequency distributions as two frequency polygons (Line and histogram). Plot both frequency on the same graphs. c. Compute the relative frequency, cumulative frequency, and cumulative percent 31 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Measures of Central Tendencies The measures of central tendency give concise information about the nature of the distribution of enumerated raw data. They serve as the representatives of the entire distribution of the raw data and present appropriate ways of how the data tends toward the center. There are three commonly used measures of central tendency: the mean, the median and the mode. The Mean The mean is the most frequently used measured of central tendency because it is simple. Since it is based on all sets of data, it summarizes a lot of information. It is also the most reliable measure. Moreover, the mean is required as a basis for the computation of other statistical measurements. Thus, it is widely used in statistics. Mean can be applied in two ways: a. when the number of raw data is small (n<30); and b. when the data set is large (n 30). When data are small (n < 30), the mean will be calculated as follows: X = X n where: X X n 32 = mean = symbol for “summation” = individual data = total number of population Example: Calculate the mean scores of 10 students in Math and Science subjects No. Math Science 66 50 1 78 55 2 89 60 3 88 87 4 97 90 5 78 59 6 59 88 7 85 89 8 84 92 9 79 95 10 Total 765 803 Mean 76.5 80.3 When the number of data is large (n 30), it is easy to compute the mean by grouping the set of data in terms of frequency distribution. The mean from a frequency distribution may be obtained in almost the same ways as the mean from raw data is computed. There are many ways in determining the mean when the number of data is large: a. by midpoint; b. by the class-deviation method, lower limit method and upper limit method The mean from a frequency distribution by midpoint method may be computed using the formula below: X = fiXi n Where: fi = frequency of the ith interval. Xi = class mark of the ith class interval n = number of population 33 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Example: Calculation of Mean from frequency Distribution of scores of 40 students in Math using mid- point method. C.I 70-74 65-69 60-64 55-59 50-54 45-49 40-44 35-39 30-34 25-29 20-24 Frequency (fi) 2 2 3 2 8 9 2 3 5 3 1 n = 40 X = fiXi = 1,875 n 40 Xi 72 67 62 57 52 47 42 37 32 27 22 fiXi 144 134 186 114 416 423 84 111 160 81 22 fiXi = 1,875 = 46.875 Calculation of Mean from frequency Distribution using classdeviation method C.I 70-74 65-69 60-64 55-59 50-54 45-49 40-44 35-39 30-34 25-29 20-24 Frequency (fi) 2 2 3 2 8 9 2 3 5 3 1 n = 40 di +5 +4 +3 +2 +1 0 -1 -2 -3 -4 -5 X = Xo + {fidi} i = 47 + {-1} 5 = 46.875 n 40 34 fidi 10 8 9 4 8 0 -2 -6 -15 -12 -5 fidi = -1 The steps in determining the mean from frequency distribution using class-deviation method are as follows: 1. Take the middle lower class limit as an assumed mean 2. Assign zero (0) corresponding to the assumed mean and with positive integers above zero deviation and negative integers below it. This column is denoted by di. 3. Multiply the deviations by the corresponding frequencies to get values for column fidi and then sum up algebraically to get fidi. 4. Add the frequencies to get the total number of data (n). 5. Divide fidi by n and then multiply the quotient by the interval size to get the correction value. 6. Add the correction value obtained with the assumed mean. The result is the actual mean. Calculation of mean by Lower Class Limit from frequency of scores of 40 students in Math C.I 70-74 65-69 60-64 55-59 50-54 45-49 40-44 35-39 30-34 25-29 20-24 Frequency (fi) 2 2 3 2 8 9 2 3 5 3 1 n = 40 X = flc + i – 0. 5 n 2 flc 140 130 180 110 400 405 80 105 150 75 20 Flc=1,795 = 1,795875 + 5/2 – 0.5 = 46.875 40 Where: X = mean = summation notation 35 O.S. Corpuz 2012 UNDERSTANDING STATISTICS f = frequency lc = lower class limit i = class interval size n = number of observation uc= upper class limit Calculation of mean using the Upper Class Limit method C.I 70-74 65-69 60-64 55-59 50-54 45-49 40-44 35-39 30-34 25-29 20-24 Frequency (fi) 2 2 3 2 8 9 2 3 5 3 1 n = 40 X = fuc - i – 0. 5 n 2 fuc 148 138 192 118 432 441 88 117 170 87 24 fuc=1,955 = 1,95575 - 5/2 – 0.5 = 46.875 40 The Median Another measure of central tendency is the median. It is a point measure that divides the distribution of arranged data from highest to lowest or vice versa in half; thus, half of the data falls below the median and another half of the data falls above the median. It is the most stable measure of central tendency. The value of the median depends on the number of data, and not on its magnitude. If most of the data are high, the median is high, and if most of the data are low, the median is also low. In identifying median, data should be arranged in the order of magnitude, either in ascending or descending order, provided that n is small (n < 30). 36 The following are the steps for median determination from raw scores: 1. Arrange the data from highest to lowest or vice versa. 2. If n/2 is an integer, the median is taken to be the average of the two middlemost data. 3. If n/2 is not an integer, the median is taken to be the middlemost data. Example: Median from sample raw scores of 8 students in Stat 22 and 9 students in Math 12 STAT 22 MATH 12 17 15 17 19 26 20 28 24 28 30 30 30 32 31 32 37 40 Mdn = 28 + 30 = 29 Mdn = 28 2 When the data are large (n 30), using the less than cumulative frequency distribution, the median is computed by the formula: Mdn = Lm + {n/2 – lcf} i fm Where: Mdn = median Lm = lower class boundary of the median Class 37 UNDERSTANDING STATISTICS O.S. Corpuz 2012 lcf fm i n = less than cumulative frequency of the class immediately preceding the median class. = frequency of the median class = the size of the interval = the total number of population Example: Median from frequency distribution of scores of 40 students in Math using the Less Than Cumulative Frequency C.I fi <cf 40 2 70 -74 38 2 65 – 69 36 3 60 – 64 33 2 55 – 59 31 8 50 – 54 23 9 45 – 49 14 40 – 44 2 12 4 35 – 39 8 5 30 – 34 3 3 25 – 29 n = 40 Mdn = Lm + {n/2-lcf}i = 44.5 + {20-14}5 = 47.83 fm 9 Using the greater than cumulative frequency, the median is computed as follows: Mdn = Um – {n/2-gcf}i fm where: Um gcf fm i 38 = upper class boundary of the median class. = greater than cumulative frequency of the class immediately preceding the median class. = frequency of the median class = the size of the interval n = the total number of population Median from frequency distribution of scores of 40 students in Math using the Greater Than Cumulative Frequency C.I fi >cf 2 2 70 – 74 2 4 65 – 69 3 7 60 – 64 2 9 55 – 59 17 8 50 – 54 26 9 45 – 49 28 2 40 – 44 32 4 35 – 39 37 5 30 – 34 40 3 25 – 29 n = 40 Mdn = Um - {n/2-gcf}i = 49.5 - {20-17}5 = 47.83 fm 9 The Mode It is defined as the score that occurs frequently. The mode for a set of test scores need not be unique. Thus, it is possible to have two or more modes. Example: Calculation of Mode from scores of 8 Math students and 9 Science students Math Science 15 17 19 30 20 28 24 37 30 28 26 30 31 32 17 32 40 Mo = 17 and 30 Mo = 32 39 O.S. Corpuz 2012 UNDERSTANDING STATISTICS If the set of data is in the form of frequency distribution, the mode is calculated using the formula below: Mo = Lm + {fm-f1 } i 2fm-f1-f2 Where: Mo = mode Lm = lower class boundary of modal class fm = frequency of the modal class f1 = frequency of the class preceding the modal class f2 = frequency of the class following the modal class i = size of the interval Example: Calculation of mode from frequency distribution of the sample test scores of 40 students in Math . C.I 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 8 f2 9 fmo 2 f1 3 4 3 2 n = 40 Mo = Lm + {fm – f1 } i 2fmo-f1-f2 = 44.5 + {9 – 2} 5 2(9)-2-8 40 = 44.5 + ( 7 ) 5 8 = 44.5 + 35/8 = 44.5 + 4.375 Mo = 48.875 Problem Set No. 5 The following are the scores obtained by a group of 60 College students in Math 102 examination: Math 102 SCORES 88 84 42 96 83 54 44 72 82 86 81 63 72 73 98 73 74 62 69 85 79 86 89 39 45 59 65 78 77 73 90 59 78 88 88 91 68 78 86 89 78 89 72 77 69 74 80 68 79 49 82 76 81 43 40 66 81 70 50 75 Problem Solving and Statistical Analysis: a. Prepare a frequency distribution of the above marks, using a class interval of 15 b. Solve the mean, median and mode using the lower class method and upper class method to check your answer. 41 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Measures of Position These are the measures that are used to find the specific location of a point that determines percentage of test scores in the distribution. These measures are values below which specific fractions of the test scores in a given set would fall. The quartiles, deciles, and percentiles are the measures of position that are commonly used. The Quartiles Quartiles are points which divide the total number of test scores into four equal parts. Each set of test scores has three quartiles. 25% falls below the first quartile (Q1), 50% is below the second quartile (Q2), and 75% is below the 3rd quartile (Q3). The 1st and the 3rd quartiles are used in the computation of the interquartile range and quartile deviation. Quartiles are computed in the same way as the median is computed, since Q2 is the same as the median. The steps in finding the quartiles of raw scores can be summarized as follows: 1. Arrange the scores from highest to lowest or lowest to highest, 2. Determine Qk, where Qk is the kth quartile and k = 1, 2,3. - If nk/4 is an integer, Qk = (nk/4)th + (nk/4 + 1)th 2 If nk/4 is not an integer, Qk = ith score where i is the closest integer greater than nk/4. 42 To illustrate the procedure for calculating the quartiles from raw scores, consider the sample test scores given below: Calculation of quartiles from sample raw scores of eight students in Stat 22 and 9 students in Math102 Stat 22 17 17 26 28 30 30 31 37 Math 102 15 19 20 24 28 30 32 32 40 Q1 = (8)(1) = 2 --- Q1 = (2+3)th scores 4 2 Q1 = (9)(1) = 2.25 --- Q1 = 3rd score = 20 4 = 17 + 26 = 21.5 2 Q3 = (8)(3) = 6 --- Q3 = (6+7)th scores 4 2 = 30 + 31 = 30.5 2 Q3 = (9)(3) = 6.75 --- Q3 = 7th score = 32 4 Using the less than cumulative frequency distribution, the first quartile is computed as follows: Q1 = LQ1 + {n/4 – lcf} i fQ1 where: Q1 = the first quartile LQ1 = lower class boundary where Q1 lies lcf = less than cumulative frequency approaching or equal to but not exceeding n/4 fQ1 = the frequency where Q1 lies i = the size of the interval n = the total number of scores 43 O.S. Corpuz 2012 UNDERSTANDING STATISTICS For the third quartile, it is computed by the formula below: Q3 = LQ3 + {3n/4 – lcf} i fQ3 where: Q3 = the 3rd quartile LQ3 = lower class boundary where Q3 lies For the test scores in the form of a frequency distribution, the following are the steps in determining the quartiles: a). For the 1st Quartile 1. Take one fourth of the total number of scores 2. Get a less than cumulative frequency until one fourth of it is approached or equaled but not exceeded. 3. Subtract Step 2 from Step 1 4. Divide the difference in Step 3 by the frequency of the next higher step where Q1 lies. 5. Multiply the quotient in Step 4 by the size of the interval to get the correction value. 6. Add the correction value to the lower class boundary where Q1 lies. The result is the 1st quartile. b). For the 3rd Quartile 1. Take ¾ of the total number of scores 2. Get a less than cumulative frequency until 3/4 of it is approached or equaled but not exceeded. 3. Subtract Step 2 from Step 1 4. Divide the difference in Step 3 by the frequency of the next higher step where Q3 lies. 5. Multiply the quotient in Step 4 by the size of the interval to get the correction value. 6. Add the correction value to the lower class boundary where Q3 lies. The result is the 3rd quartile. 44 Example: Calculation of Quartiles from frequency distribution of test scores of 40 students in Math 102 Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 Q1 freq 2 2 3 2 8 9 2 3 4 3 2 n = 40 <cf 40 38 36 33 31 23 14 12 9 5 2 = nk/4 = 11(1) 4 = 2.75 or 3rd class interval Q1 = LQ1 + {n/4 – Fl} i = 29.5 + {10 – 9}5 = 30.75 fQ1 4 Q1 = nk/4 = 11(3) 4 = 8.25 or 9th class interval Q3 = LQ3 + {3n/4 – Fl} i fQ3 = 59.5 + {30 – 28}5 = 62 4 45 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The Deciles The deciles are points which divide the total number of test scores into ten equal parts. Each set of test scores has nine deciles. 10% falls below the 1st decile (D1); 20% falls below the 2nd decile (D2); 30% falls below the 3rd decile (D3); 40% falls below the 4th decile (D4); 50% falls below the 5th decile (D5); 60% falls below the 6th decile (D6); 70% falls below the 7th decile (D7); 80% falls below the 8th decile(D8); and 90% falls below the 9th decile (D9). The deciles are computed exactly in the same manner as the median is computed. Hence, the 5th decile (D5) is the same with the median. The steps in finding the deciles from raw scores can be summarized as follows: 1. Arrange the scores from highest to lowest or vice versa 2. Determine Dk, wher Dk is the kth decile and k = 1, 2, 3,…,9. - If nk/10 is an integer, Dk = (nk/10)th + (nk/10 +1)th 2 - If nk/10 is not an integer, Dk = ith score where i is the closest integer greater than nk/10. Example: Calculation of Deciles from raw score of 8 students in Statistics and 9 students in Math 12 Statistics Math 12 17 15 17 19 26 20 28 24 30 28 30 30 32 31 37 32 40 46 D1 = (8)(1) = 0.8 --- D1 = (1)st scores = 17 10 D6 = (9)(6) = 5.4 --- D6 = (6)th scores = 30 10 D2 = (8)(2) = 1.6 --- D2 = (2)nd scores = 17 10 D7 = (9)(7) = 6.3 --- D7 = (7)st scores = 32 10 D3 = (8)(3) = 2.4 --- D1 = (3)rd scores = 26 10 D8 = (9)(8) = 7.2 --- D8 = (8)th scores = 32 10 D4 = (8)(4) = 3.4 --- D1 = (4)th scores = 28 10 D9 = (9)(9) = 8.1 --- D9 = (9)th scores = 40 10 D5 = (8)(5) = 4 ---D5 = (4+5)th scores 10 = 28+30 = 29 2 D2 = (9)(2) = 1.8 --- D2 = (2)nd scores = 19 10 For the test scores in the form of frequency distribution, the following formulas are being employed: D1 = LD1 + {n/10 – lcf}i fD1 Where: D1 = the 1st decile LD1 = lower class boundary where D1 lies lcf = less than cumulative frequency approaching or equal to but exceeding n/10. fD1 i = the frequency where D1 lies = the size of the interval n = the total number of scores D2 = LD2 + {n/5 – lcf}i fD2 Where: D2 = the 2nd decile LD2 = lower class boundary where D2 lies fD2 = the frequency where D2 lies 47 O.S. Corpuz 2012 lcf UNDERSTANDING STATISTICS = less than cumulative frequency approaching or equal to but exceeding n/5. D3 = LD3 + {3n/10 – lcf}i fD3 Where: D3 LD3 fD3 lcf = the 3rd decile = lower class boundary where D3 lies = the frequency where D3 lies = less than cumulative freq. approaching or equal to but exceeding 3n/10. D4 = LD4 + {2n/5 – lcf}i fD4 Where: = the 4th decile = lower class boundary where D4 lies = the frequency where D4 lies = less than cumulative frequency approaching or equal to but exceeding 2n/5. D5 = LD5 + {n/2 – lcf}i fD5 D4 LD4 fD4 lcf Where: D5 = the 5th decile LD5 = lower class boundary where D5 lies fD5 = the frequency where D5 lies lcf = less than cumulative frequency approaching or equal to but exceeding n/2. D6 = LD6 + {3n/5 – lcf}i fD6 48 Where: D6 = the 6th decile LD6 = lower class boundary where D6 lies fD6 = the frequency where D6 lies lcf = less than cumulative frequency approaching or equal to but exceeding 3n/5. D7 = LD7+ {7n/10 – lcf}i fD7 Where: D7 LD7 fD7 lcf = the 7th decile = lower class boundary where D7 lies = the frequency where D7 lies = less than cumulative freq. approaching or equal to but exceeding 7n/10. D8 = LD8 + {4n/5 – lcf}i fD8 Where: D8 LD8 fD8 lcf D9 Where: D9 LD9 fD9 lcf = the 8th decile = lower class boundary where D8 lies = the frequency where D8 lies = less than cumulative freq. approaching or equal to but exceeding 4n/5. = LD9 + {9n/10 – lcf}i fD9 = the 9th decile = lower class boundary where D9 lies = the frequency where D9 lies = less than cumulative freq. approaching or equal to but not exceeding 9n/10. 49 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Example: Calculation of deciles from frequency distribution of test scores of 40 students in Math 102 Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 8 9 2 3 4 3 2 n = 40 For Decile 1 D1 = nk 10 = 11(1) 10 = 1.1 or 2nd class interval D1 = LD1 + {n/10 – lcf} i = 24.5 + {4 – 2}5 = 27.83 fD1 3 For Decile 2 D2 = nk 10 = 11(2) 10 = 2.2 or 3rd class interval 50 <cf 40 D9 38 D7 36 D8 33 31 D6 23 D5 14 D3; D4 12 9 D2 5 D1 2 D2 = LD2 + {n/5 – lcf} i fD2 = 29.5 + {8 – 5}5 = 33.25 4 For Decile 3 D3 = nk 10 = 11(3) 10 = 3.3 or 4th class interval D3 = LD3 + {3n/10 – lcf} i = 34.5 + {12 – 9}5 = 39.5 fD3 3 For Decile 4 D4 = nk 10 = 11(4) 10 = 4.4 or 5th class interval D4 = LD4 + {2n/5 – Fl} i fD4 = 39.5 + {16 – 14}5 = 39.5 2 For Decile 5 D5 = nk 10 = 11(5) 10 = 5.5 or 6th class interval D5 = LD5 + {n/2 – lcf} i fD5 = 44.5 + {20 – 14}5 = 47.83 9 51 O.S. Corpuz 2012 UNDERSTANDING STATISTICS For Decile 6 D6 = nk 10 = 11(6) 10 = 6.6 or 7th class interval D6 = LD6 + {3n/5 – lcf} i fD6 = 49.5 + {24 – 23}5 = 50.125 8 For Decile 7 D7 = nk 10 = 11(7) 10 = 7.7 or 8th class interval D7 = LD7 + {7n/10 – lcf} i = 54.5 + {28 – 23}5 = 67.00 fD7 2 For Decile 8 D8 = nk 10 = 11(8) 10 = 8.8 or 9th class interval D8 = LD8 + {4n/5 – lcf} i = 59.5 + {32 – 31}5 = 61.17 fD8 3 For Decile 9 D9 = nk 10 = 11(9) 10 52 = 9.9 or 10th class interval D9 = LD9+ {9n/10 – lcf} i = 64.5 + {36 – 33}5 = 72.00 fD9 2 The Percentiles The percentiles are the points that divide the total number of test scores or data into exactly one hundred equal parts. For each test score, it is understood that there are ninety-nine (99) percentiles which determine the points below which specific percentage of test score would fall. For instance, 9th percentiles (P9), would include 9% of the test scores in the distribution lie at below it and 91% lie at or above it. other percentiles would take similar meaning and interpretation. The percentiles are calculated exactly in the same manner as the computation of the median. In effect, the 50th percentile (P50) is the same with the median. The steps in finding the percentiles from raw scores can be summarized to wit: 1. Arrange the scores from highest to lowest or vice versa 2. Determine Pk, where Pk is the kth percentile and k = 1, 2,3, …, 99. a). If nk/100 is an integer, Pk=(nk/100)th+(nk/100+1)th scores 2 b). If nk/100 is not an integer, Pk = ith score where i is the closest integer greater than nk/100. To illustrate the procedure for calculating some percentiles from raw scores, consider the sample test scores in Stat 22 and Math 102 classes. 53 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Calculation of percentile from sample raw scores of 8 students in Stat 101 and 9 students in Math102 Stat 101 17 17 26 28 30 30 31 37 Math 102 15 19 20 24 28 30 32 32 40 P20 = (8)(20) = 1.6 --- P20 = (2)nd score = 17 100 P60 = (9)(60) = 5.4 --- P60 = (6)th score = 30 100 P25 = (8)(25) = 2 --- P25 = (2+3)th score 100 = 17 + 26 = 21.5 2 P70 = (9)(70) = 6.3 --- P70 = (7)th score = 32 100 P80 = (9)(80) = 7.2 --- P8 = (8)th score = 32 100 P97 = (8)(97) = 7.76--- P97 = (8)th score = 37 100 P90 = (9)(90) = 8.1 --- P90 = (9)th score = 40 100 P99 = (8)(99) = 7.92 --- P99= (8)th score = 37 100 P20 = (9)(20) = 1.8 --- P20 = (2)nd score = 19 100 For the test scores in the form of frequency distribution, the determining factor is (k/100)n, where k = 1, 2 ,3,…, 99, corresponding to the ith percentile. The (k/100)n is actually a value that serves as a reference point when determining a “less than” cumulative frequency (Fm-1) immediately preceding the class where the ith percentile lies. It is a value that should be approached to or equaled but not exceeded by Fm-1 when solving for certain percentiles. 54 The following equations will be utilized in determining some percentiles. P1 = LP1 + {n/100 – lcf} i fP1 Where: P1 = the 1st percentile LP1 = lower class boundary where P1 lies lcf = less than cumulative frequency approaching or equal to but exceeding n/100. fP1 = the frequency where P1 lies i = the size of the interval n = the total number of scores P2 = LP2 + {2n/100 – lcf}i fP2 Where: P2 LP2 fP2 lcf = the 2nd percentile = lower class boundary where P2 lies = the frequency where P2 lies = less than cumulative freq. approaching or equal to but exceeding 2n/100. P3 = LP3 + {3n/100 – lcf}i fP3 Where: P3 = the 3rd percentile LP3 = lower class boundary where P3 lies fP3 lcf = the frequency where P3 lies = less than cumulative freq. approaching or equal to but exceeding 3n/100. P4 = LP4 + {4n/100 – lcf}i fP4 55 O.S. Corpuz 2012 UNDERSTANDING STATISTICS = the 4th percentile = lower class boundary where P4 lies = the frequency where D4 lies = less than cumulative freq. approaching or equal to but exceeding 4n/100. P50 = LP5 + {n/2 – lcf}i fP50 Where: P4 LP4 fP4 lcf Where: P50 = the 50th percentile LP5 = lower class boundary where P50 lies fP5 = the frequency where P50 lies lcf = less than cumulative frequency approaching or equal to but exceeding n/2. P60 = LP60 + {3n/5 – lcf}i fP60 Where: P60 = the 60th percentile LP60 = lower class boundary where P60 lies fP60 = the frequency where P60 lies lcf = less than cumulative frequency approaching or equal to but exceeding 3n/5. P80 = LP80 + {4n/5 – lcf}i fP80 Where: P80 LP80 fP80 lcf 56 = the 80th percentile = lower class boundary where P80 lies = the frequency where P80 lies = less than cumulative frequency approaching or equal to but exceeding 4n/5. P90 = LP90 + {9n/10 – lcf}i fD9 Where: P90 = the 90th Percentile LP90 = lower class boundary where P90 lies fP90 = the frequency where P90 lies lcf = less than cumulative freq. approaching or equal to but exceeding 9n/10. To illustrate the procedure for calculating some percentiles from a frequency distribution, consider the scores of 40 students in Statistics Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 8 9 2 3 4 3 2 <cf 40 38 36 33 31 23 14 12 9 5 2 n = 40 For percentile 1 P1 = nk 100 = 11(1) 100 = 0.11 or 1st score P1 = LP1 + {n/100 – lcf} i = 19.5 + {0.4 – 0}5 = 20.167 fP1 3 57 O.S. Corpuz 2012 UNDERSTANDING STATISTICS For percentile 10 P10 = nk 100 = 11(10) 100 = 1.1 or 2nd score P10 = LP10 + {n/10 – lcf} i = 24.5 + {4 – 2}5 = 27.833 fP10 3 For percentile 20 P20 = nk 100 = 11(20) 100 = 2.2 or 3rd score P20 = LP20 + {n/5 – lcf} i fP20 = 29.5 + {8 – 5}5 = 33.25 4 For percentile 30 P30 = nk 100 = 11(30) 100 = 3.3 or 4th score P30 = LP30 + {3n/10 – lcf} i = 34.5 + {12 – 9}5 = 39.5 fP30 3 For percentile 40 P40 = nk 100 58 = 11(40) 100 = 4.4 or 5th score P40 = LP40 + {2n/5 – lcf} i fP40 For percentile 50 = 39.5 + {16 – 14}5 = 39.5 2 P50 = nk 100 = 11(50) 100 = 5.5 or 6th score P50 = LP50+ {n/2 – lcf} i fP50 = 44.5 + {20 – 14}5 = 47.83 9 For percentile 90 P90 = nk 100 = 11(90) 100 = 9.9 or 10th score P90 = LP90+ {9n/10 – Fl} i = 64.5 + {36 – 33}5 = 72.00 fP90 2 For percentile 99 P99 = nk 100 = 11(99) 100 = 10.89 or 11th score 59 O.S. Corpuz 2012 UNDERSTANDING STATISTICS P99 = LP9+ {99n/100 – Fl} i = 69.5 + {39.6 – 38}5 = 73.5 fP99 2 Exercise No.6 Calculate the quartiles (1 and 3), deciles (2,4,6,7 and 9), and Percentiles (70, 80, 85, 90, and 97) of the scores of 55 Education students in their examination in Stat 22 Prior to the computations fill-up the < cumulative column in the table. Class Interval 85 - 89 80 - 84 75 - 79 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 9 10 9 5 4 3 2 2 1 1 n = 55 60 <cf Measures of Dispersion/Variability It can be utilized in determining the size of the distribution of test score or a portion of it. They can be used to find the deviation of test scores from the mean scores. Measures of dispersion can also be used to establish the actual similarities or the differences of the distribution. In general, these measures are employed to further characterize the distribution of test scores. The most commonly used measures of dispersion are: the range; interquartile range; quartile deviation; average deviation and standard deviation. The Range It is the simplest and easiest measures of dispersion or variability. It simply measures how far the highest score is to the lowest score. It does not tell anything about the scores between these two extreme scores. Thus, it is considered as the least satisfactory measure of dispersion. The equations used are: a. For Raw Scores R=H–L Where: R = range H = highest score L = lowest score b. For Frequency Distributed Scores 61 O.S. Corpuz 2012 UNDERSTANDING STATISTICS R = (Hmpt – Lmpt) Hmpt = midpoint of the highest step Lmpt = midpoint of the lowest step The Inter quartile Range It refers to the range of score of specified parts of the total group usually the middle 50% of the cases lying between the 1st quartile and the 3rd quartile. The equation used is: I.Q.R = Q3 – Q1 Where: I.Q.R = inter quartile range Q3 = 3rd Quartile Q1 = 1st Quartile Example: a. Calculation of Interquartile Range from sample scores of 8 Stat 22 students DBH: 17 17 26 28 30 30 31 37 Recall: Q1 = 21.5 Q3 = 30.5 I.Q.R = Q3 – Q1 = 30.5 – 21.5 = 9.0 b. Calculation of Interquartile Range from frequency distribution of sample raw scores of 40 students in Stat 22 62 C.I 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 8 9 2 3 4 3 2 n = 40 Recall: Q1 = 30.75 Q3 = 62 I.Q.R = Q3 – Q1 = 62 – 30.75 = 31.25 The Quartile Deviation Is another measure of dispersion that divides the difference of the 3rd and the 1st quartiles into halves. It is the average distance from the median to the two quartiles, i.e., it tells how far the quartile points (Q1 and Q3) lie from the median, on the average. When quartile deviation (Q.D) is small, the set of test scores is more or less homogenous but when Q.D is large, the set of scores is more or less heterogeneous. This measure is used when there are big gaps between scores. It is also essentially used when the main concern is the concentration of the middle 50% of the scores around the median. Mathematically: 63 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Q.D = Q3 – Q1 2 Example: a. Calculation of Interquartile Range from sample raw scores of 8 Stat 22 students DBH: 17 17 26 28 Recall: Q1 = 21.5 Q3 = 30.5 IQR = Q3 – Q1 2 = 30.5 – 21.5 2 = 9/2 IQR = 4.5 30 30 31 37 b. Calculation of Interquartile Range from frequency distribution of raw scores of 40 Stat 22 students. Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 freq 2 2 3 2 8 9 2 3 4 3 2 n = 40 Recall: Q1 = 30.75 IQR = Q3 – Q1 2 64 Q3 = 62 = 62 – 30.75 2 = 31.25 2 IQR = 15.625 The Average Deviation Average deviation is a measure of absolute dispersion that is affected by every individual score. It is the mean of the absolute deviation of the individual score from the mean of all the scores. A large average deviation would mean that a set of scores is widely dispersed about the mean while a small average deviation would imply that a set of scores is closer to the mean. The formula in calculating average deviation is as follows: 1. Raw Score _ A.D = /X-X/ n–1 Where: A.D = average deviation X = Individual score _ X = mean of all scores n = total no. of W. lauan = summation symbol fi = frequency of the ith class interval Xi = midpoint of the ith Diameter class 2. Frequency Distributed Data _ A.D = fi/Xi-X/ n–1 65 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Example: a. Calculation of Average Deviation from sample raw scres of 8 Stat 22 students _ X X–X _ 17 -10 Recall: X = 27 17 -10 26 28 30 -1 1 3 _ A.D = /X-X/ n–1 30 3 = 42 31 4 7 37 10 42 A.D = 6.0 c. Calculation of Average Deviation from frequency distribution of raw scores of 40 Stat 22 students Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 _ Recall: X = 47.25 fi 2 2 3 2 8 9 2 4 5 3 n = 40 Xi 72 67 62 57 52 47 42 37 32 27 (X – X) 24.75 19.75 14.75 9.75 4.75 -0.25 -5.25 -10.25 -15.25 -20.25 _ A.D = fi/Xi-X/ n–1 = 381.50 39 66 fi(Xi – X) 49.50 39.50 44.25 19.50 38.00 -.2.25 -10.50 -41.00 -76.50 -60.75 381.50 A.D = 9.78 The Standard Deviation Is the measure of dispersion that involves all scores in the distribution rather than through extreme scores. It may be referred to as the root-mean square of the deviation from the mean. It is considered the most important measure of dispersion. Mathematically, it is equated as: 1. Raw Score S.D = ¥(X – X)2 n-1 2. Frequency Distribution a. Midpoint Method S.D = ¥fi(Xi – X)2 n-1 b. Class-deviation Method S.D = i¥fidi2 – (fidi)2 n-1 n(n-1) Where: S.D = standard deviation fi = frequency of the ith diameter class di = deviation of the ith diameter class di2 = square of the deviation of the ith diameter class Xi = midpoint of the diameter class X = mean of the DBH Example: a. Calculation of Standard Deviation from sample raw scores of 8 Stat 22 students 67 O.S. Corpuz 2012 UNDERSTANDING STATISTICS X 17 _ X–X - 10 _ (X – X)2 100 17 - 10 100 26 28 30 -1 1 3 1 1 9 30 31 37 3 4 10 9 16 100 336 Recall: _ X = 27 _ S.D = ¥(X-X)2 n–1 = ¥336 7 S.D = 6.93 b. Calculation of Standard Deviation from frequency distribution of raw scores of 40 students in Stat 22 using Midpoint Method. C.I fi Xi (X – X) (X – X)2 fi(Xi –X)2 1225.12 612.56 24.75 72 70 – 74 2 780.12 390.06 19.75 2 65 – 69 67 652.68 217.56 14.75 3 60 – 64 62 190.12 95.06 9.75 55 – 59 2 57 180.48 22.56 4.75 8 52 50 – 54 0.567 0.063 -0.25 45 – 49 9 47 55.12 27.56 -5.25 2 40 – 44 42 420.24 105.06 -10.25 4 37 35 – 39 1162.80 232.56 -15.25 5 32 30 – 34 1230.18 410.06 -20.25 3 27 25 – 29 n = 40 5,897.43 _ _ Recall: X = 47.25 S.D = ¥fi(Xi-X)2 n–1 = 5,897.43 39 S.D = 12.30 68 c. Calculation of Standard Deviation from frequency distribution of raw scores of 40 students in Stat 22 using Class-Deviation Method. C.I 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 fi 2 2 3 2 8 9 2 4 5 3 n = 40 di +5 +4 +3 +2 +1 0 -1 -2 -3 -4 fidi 10 8 9 4 8 0 -2 -8 -15 -12 2 fidi2 50 32 27 8 8 0 2 16 45 48 236 SD = i¥fidi2 – (fidi)2 n-1 n(n-1) SD = 5¥236 – (2)2 39 40(39) = 5¥6.05128 – 4/1560 = 5¥6.05128 – 0.0025641 = 5¥6.048716 = 5(2.4594) SD = 12.30 69 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Exercise No. 7 Test yourself: Given the frequency distributed data on the raw scores of students in Math Class Interval fi 75 - 79 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5-9 2 2 3 2 8 9 2 4 10 3 10 5 5 5 5 n = 75 Compute for: 1. Quartile Deviation 2. Inter-quartile deviation 3. Average Deviation and 4. Standard Deviation 70 Xi (X – X) (X – X) 2 fi(Xi –X) 2 Measures of Skewness and Kurtosis Skewness – the deviation from the symmetrical distribution. It is a degree of asymmetry of a distribution or departure from symmetry of a distribution. It indicates not only the amount of asymmetry but also the direction of the distribution of the raw data. Thus, a distribution is said to be skewed in the direction of the extreme values, or speaking in terms of the curve, in the direction of the excess tail. The greater the value of skewness departs from 0, the more skewed is the distribution and the nearer the value of skewness to 0, the nearer the distribution is to a normal distribution. Types of Skewness: 1. Positive Skewness or Skewed to the Right – refers to the distribution that tapers more to the right than to the left. In this kind of distribution, there are more small data than bigger data. In examination, this type of skewness may be obtained if the test is very difficult. If skewness has a positive value, then the distribution is skewed to the right (see illustration below). 2. Negative Skewness or Skewed to the Left – refers to the distribution that tapers more to the left than to the right. This means a longer tail is directed to the left. In this distribution, there are more high scores than low scores; 71 O.S. Corpuz 2012 UNDERSTANDING STATISTICS this type of skewness may happen if the test is very easy. If skewness has a negative value, then the distribution is skewed to the left (see graph below). Moment Coefficient of Skewness It is an essential method of describing the distribution of raw data. It helps determine whether the majority of the data in the distribution are below or above the mean. In examination, it tells whether the test itself is difficult or easy. In interpreting the distribution of data, the value of the moment coefficient of skewness (degree of asymmetry) and the direction of the distribution should be highly considered. When its value is positive, it indicates that the distribution is skewed to the right. This could mean that majority of the data are below than above the mean. In examination, it can be said that the test is difficult. On the other hand, if its value is negative, the distribution of the data can be interpreted as skewed to the left, this further implies that there are more data above than below the mean. The test is easy when talking to examination. In cases where the value, be it positive or negative, of the moment coefficient of skewness is approaching or near zero (0) indicates that the distribution of the data is approaching the normal distribution, the mean, the median, and the mode are approximately equal. The get the moment coefficient of skewness, the equations used are the following: a. Raw Scores _ 72 SK = (X – X)3 (n-1)(SD)3 Where: SK = skewness Xi = midpoint of the ith class interval SD = standard deviation b. Frequency Distribution _ SK = fi(X – X)3 (n-1)(SD)3 fi = frequency of the ith class interval n = total number of data = symbol of summation _ X = mean of all data Example: a. Calculation of Moment Coefficient of Skewness from sample raw scores in Stat 22 of 8 students in the previous examples. X 17 17 X–X - 10 - 10 (X – X)3 -1000 -1000 26 -1 -1 28 30 1 3 1 27 _ SK = (X – X)3 (n-1)(SD)3 30 31 37 3 4 10 42 27 64 1000 -882 = -882 7(6.93)3 SK = -882 7(332.813) _ Recall: X = 27 SD = 6.93 SK = -882 2,329.69 SK = - 0.38 Take note of the result which is negative. This implies that the distribution of the scores in Stat 22 of the 8 students is skewed to 73 O.S. Corpuz 2012 UNDERSTANDING STATISTICS the left; hence, majority of the scores are above the mean and that, the test is very easy. d. Calculation of Moment Coefficient of Skewness from frequency distribution of raw scores of 40 students in Stat 22 using Midpoint Method. Class Interval fi Xi (X – X) 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 2 2 3 2 8 9 2 4 5 3 n = 40 72 67 62 57 52 47 42 37 32 27 24.75 19.75 14.75 9.75 4.75 -0.25 -5.25 -10.25 -15.25 -20.25 Recall: X = 47.25 (X – X)3 fi(Xi –X)3 15160.92 30321.84 7703.73 15407.46 9627.15 3206.09 1853.72 926.86 107.17 857.36 -0.18 -0.02 -289.40 -144.70 -1076.89 -4307.56 -3546.58 -17732.90 -8303.76 -24911.28 10,826.21 S.D = 12.30 SK = fi(X – X)3 (n-1)(SD)3 SK = 10,826.21 (39)(12.30)3 = 10,826.21 (39)(1,860.867) = 10,826.21 72,573.813 SK = 0.15 The result of the moment coefficient of skewness is positive (0.15). This indicated that the distribution of the raw scores is skewed to the right; hence, most of the scores are below the mean. It can be said that the examination is difficult. 74 Exercise No. 8 Test yourself: Given the frequency distributed data on the scores of students in Mathematics, find the moment coefficient of skewness Diameter Class 75 - 79 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5-9 fi x (X – X) (X – X)3 fi(Xi –X)3 2 2 3 2 8 9 2 4 10 3 10 5 5 5 5 n = 75 75 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Measures of Kurtosis Curves of distributions having the same coefficient of skewness may still differ significantly. Symmetrical curves, for instance may vary in shape because they may not have the same peakedness, a property of curves which can be described by computing for a value called measure of kurtosis. Kurtosis – is a measure of the degree of peakedness or flatness of a distribution. Hence, the concern is on the height of the curve along the y-axis. Types: 1. Leptokurtic – refers to the distribution having a relatively high peak 2. Mesokurtic – refers to the distribution neither very peaked nor very flat-topped 3. Platykurtic – refers to the distribution having relatively flattop Leptokurtic Mesokurtic Platykurtic If KU < 3, the distribution is platykurtic or less peaked than the normal curve. If KU = 3, the distribution is mesokurtic, and if KU > 3, the distribution is leptokurtic or more peaked than the normal curve. Moment Coefficient of Kurtosis It is utilized when the degree of height (peakedness) and the degree of flatness of raw data relative to the normal is to be 76 determined. It helps to determine whether the data are far higher or far lower than the normal curve The equations used in determining the value of kurtosis are: 1. If the data is arrange in raw _ KU = (X – X)4 (n-1)(SD)4 2. If the data is arrange in Frequency Distributions _ KU = fi(Xi – X)4 (n-1)(SD)4 Where: KU = kurtosis Others as defined Example: a. Calculation of Moment Coefficient of Kurtosis from sample raw scores of 8 students in Stat 22 in the previous examples _ _ X X–X (X – X)4 _ 17 - 10 10,000 Recall: X = 27 17 - 10 10,000 SD = 6.93 26 -1 1 _ 28 1 1 KU = (X – X)4 30 3 81 (n-1)(SD)4 30 3 81 = 30,420 31 4 256 7(6.93)4 37 10 42 10,000 30,420 KU = 30,420 7(2,306.40) KU = 30,420 16,144.8 KU = 1.88 77 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The result is less than 3 (1.88). This implies that the distribution of the DBH of W. lauan is platykurtic; hence, the data are far below the normal curve and the distribution is said to be flattopped. b. Calculation of Moment Coefficient of Kurtosis from frequency distribution of sample test scores of 40 students in Stat 22 using Midpoint Method. Class Interval 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 fi 2 2 3 2 8 9 2 4 5 3 n = 40 Xi 72 67 62 57 52 47 42 37 32 27 _ Recall: X = 47.25 KU = fi(X – X)4 (n-1)(SD)4 KU = 2,039,462.07 (39)(12.30)4 = 2,039,462.07 (39)(22,888.664) = 2,039,462.07 892,657.9 KU = 2.28 78 (X – X) 24.75 19.75 14.75 9.75 4.75 -0.25 -5.25 -10.25 -15.25 -20.25 (X – X)4 375,232.82 152,148.75 47,333.44 9,036.88 509.07 0.004 759.69 11,038.13 54,085.32 168,151.25 S.D = 12.30 fi(Xi–X)4 750,465.64 304,297.50 142,000.32 18,073.76 4,072.56 0.04 1,519.38 44,152.52 270,426.60 504,453.75 2,039,462.07 The result is less than 3 (2.28). This implies that the distribution of the test scores is platykurtic; hence, the data are far below the normal curve and the distribution is said to be flat-topped. Exercise No. 9 Test yourself: Given the frequency distributed data on the test scores of 75 students in Arts, find the moment coefficient of kurtosis Diameter Class 75 - 79 70 – 74 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 5-9 fi x (X – X) (X – X)4 fi(Xi –X)4 2 2 3 2 8 9 2 4 10 3 10 5 5 5 5 n = 75 79 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The Normal Distribution It is one of the most important continuous distributions. It is regarded as the most significant probability in the entire scope of statistical inferences. The probability is expressed in terms of a number from 0 to 1. Zero probability implies that there is assurance that the event will not occur; however, when the probability value is 1, there is an assurance that the event will occur or happen. For the probability value of 0.5, one is half sure that a case or event will happen. The Normal Curve It is graphically represented by a symmetrical, bell-shaped curve (see graph below). A curve is symmetry if ½ is the exact shape of the other half when folded vertically at the middle. This curve is also described as mesokurtic. It is unimodal or has only one mode and is centered at its origin. The curve is asymptotic at the extremities of the horizontal line. De Moivre derived the mathematical equation of the normal curve in 1773. Normal distribution is called Gaussian distribution in honor of Gauss, who also derived the mathemathetical equation in the 19th century. ȝ Normal Curve The equation of the normal curve is: 80 f(x) = le ½(x-µ/į)2 ¥2ʌį2 Where: f(x) = the height of the curve above the x-axis x = raw score laid off along the x-axis ȝ = mean of the distribution of the test score į = standard deviation of the distribution of the test score e = 2.711828 ʌ = 3.1415 Characteristics of a normal curve: 1. The mean, median, and mode have the same value which is a point on the horizontal axis of which the curve is a maximum. 2. The curve is symmetrical and bell-shaped about a vertical axis through the mean. This means that the line at both sides fall off toward the opposite directions at exactly equal distances from the center. 2. The normal curve approaches the horizontal axis asymptotically as we proceed in either direction away from the mean. 4. The total area under the curve and above the horizontal axis is equal to 1. Standard Normal Scores These are the converted scores, which are needed and utilized when constructing areas of the normal probability curve. They are the scores having definite mean of 0 and standard deviation of 1. These scores are referred to as z-scores. 81 The Normal Distribution O.S. Corpuz 2012 UNDERSTANDING STATISTICS A z-score expresses the deviation of given raw scores in terms of standard deviation units from the mean. The formula in determining z-scores is: _ z =X–X s Where: z = standard normal score X = any given raw score s = standard deviation of the distribution of X scores X = mean of the distribution Areas Under the Normal Curve P(X1<X<X2) P(X<X1) P(X>X2) X2 X1 µ Examples: 1. Determine the area of the normal curve given the following: a. from z = 0 to z = 2.56 b. from z = -1.25 to z = 0 c. from z = 1.19 to z = 2.59 d. from z = - 2.5 to z = 1.45 82 Solutions a. First, draw the standard normal curve and indicate the required area by shading it, then read the corresponding area from the table. To locate the area, look for 2.5 along the leftmost column and then locate 0.06 at the topmost row. The area under the normal curve is the intersection of 2.5 and 0.06, which gives 0.4406 0 1 2 3 2.56 83 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The T- Tests Test for Dependent or Correlated Samples Note that types of data that consist of information obtained from matched pairs or repeated measures are classified as dependent or correlated samples. Examples of correlated data are effects of socio-economic characteristics with academic performance. Data on socio economic characteristics serves as dependent variables and their performance as independent variables. If the data were taken twice on the same criterion variable, this can be called repeated measure design. Testing the Significance of the Difference between Means when the samples are Dependent or Correlated. Procedure: 1. Ho: μ1=μ2 that the mean of population 1 is equal to the mean of population 2. The alternative hypothesis may be defined in one of the three hypotheses presented below: a. Ha: μ1>μ2 that the mean of population 1 is greater than the mean of population 2. b. Ha: μ1<μ2 that the mean of population 1 is less than the mean of population 2 c. Ha: μ1μ2 that the mean of population 1 is not equal to the mean of population 2 84 The alternative hypothesis a and b are directional hypotheses and so if one uses either of these two hypotheses, a onetailed test will be employed. However, a two tailed test is used when one indicated the alternative hypothesis c since this is a non-directional hypothesis. 2. Test statistics: Use t-test for dependent samples 3. Level of significance: Use Į level of significance 4. Decision Criterion: Note that the decision criterion would entirely depend on the formulated alternative hypothesis. As such, use the decision criterion that corresponds to the alternative hypothesis to wit: a. Reject Ho if tctĮ (n-1)df b. Reject Ho if tc -tĮ (n-1)df c. Reject Ho if /tc/-tĮ (n-1)df 2 5. Computations Individual 1 2 3 : : : n Total Experimental Group x11 x21 Control Treatment x12 x22 Di Di2 D1 D2 D12 D22 xi1 xi2 Di Di2 xn1 xn2 Dn Dn2 n Xi1 i=1 n Xi2 i=1 n Di i=1 n Dn2 i=1 85 O.S. Corpuz 2012 where: UNDERSTANDING STATISTICS Di = difference of the observed values in the first and second condition of the ith individual D = mean difference Solve for the mean difference using the equation Solve for the variance of the mean difference by the equation Finally, the test-statistic will be calculated using the equation 6. State your decision based on the decision criterion and the tcomputed. 7. State your conclusion. Illustration 1: In a study of the effect of vitamins on the weight increase of students, a group of 10 students were weighted before and after a two months of taking the vitamins. 86 The data were as follows: Seedling No. 1 2 3 4 5 6 7 8 9 10 Wt. Before Wt. after 196 171 170 207 177 162 199 173 231 140 200 178 169 212 180 165 201 179 243 144 Use the 0.05 level of significance to test if there significant difference among weight before and after two months taking of the vitamins. Solution: 1. That the mean of the weight before and after taking the vitamins do not differ significantly. That the mean weight before and after taking the vitamins differ significantly. 2. Test-Statistic: Use t-test for dependent samples. 3. Level of Significance: Use a a=0.05 level of significance. 4. Decision Criterion: Reject 87 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 5. Computation: Seedling No. 1 2 3 4 5 6 7 8 9 10 Total Wt. before 196 171 170 207 177 162 199 173 231 140 1826 Wt. after 2 Months 200 178 169 212 180 165 201 179 243 144 1871 Solving the mean difference: Solving for the variance of the mean difference: = 1.1833 88 4 7 -1 5 3 3 2 6 12 4 45 16 49 1 25 9 9 4 36 144 16 309 Solving for the t-statistics: 6. Decision: Since , reject Ho. 7. Conclusion: Based on the result, it can be concluded that the weight after two months of taking the vitamins significantly heavier than the mean weight before. The result further reveals that there exists a significant increase of weight as indicated in the test. Therefore, taking the vitamin in two months is effective for weight increase. Illustration 2: In a study of effectiveness of physical exercise in weight reduction, a group of 10 students engaged in a prescribed program of physical exercise for one showed the following results: INDIVIDUAL WEIGHT WEIGHT BEFORE AFTER (pounds) (pounds) 1 210 196 2 168 170 3 165 170 4 202 200 5 170 157 6 185 152 7 211 189 8 189 170 9 245 221 10 149 130 Use the 0.05 level of significance to test if the prescribed program of physical exercise is effective in reducing weight. 89 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Solution: 1. That the mean weights of persons before and after the physical exercise program do not differ significantly. That the mean weight of persons before the physical exercise program is greater that their mean weight after the exercises. 2. Test-Statistic: Use t-test for dependent samples. 3. Level of Significance: Use a a=0.05 level of significance. 4. Decision Criterion: Reject 5. Computation: INDIVIDUAL 1 2 3 4 5 6 7 8 9 10 Total WEIGHT BEFORE (pounds) 200 178 169 212 180 165 201 179 243 144 1871 Solving the mean difference: 90 WEIGHT AFTER (pounds) 196 171 170 207 177 162 199 173 231 140 1826 14 -2 -5 2 13 33 22 19 24 19 139 196 4 25 4 169 1089 484 361 576 361 3269 Solving for the variance of the mean difference: SD2 = 14.85 Solving for the t-statistics: tc 6. Decision: Since = 13.65 , reject Ho. 91 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 7. Conclusion: Based on the result, it can be concluded that the weight before the physical exercises is significantly heavier than the mean weight of the students after the physical exercises as the treatment. The result further reveals that there exists a significant reduction of weight as indicated in the test. Therefore, the prescribed program of physical exercise is effective in reducing weight. Exercise No. 10 1. A certain reducing weight program has produced the following weight changes (lb) in ten students: STUDENTS 1 2 3 4 5 6 7 8 9 10 Before 124 138 113 129 149 149 177 138 139 129 After 115 110 110 131 122 155 125 142 122 105 Use the 0.05 level of significance to test if the diet is effective in reducing weight. 2. Two samples of 10 students each has been matched on IQ before beginning an experiment in learning to calculate simple statistical problem. They were then allowed 30 minutes to study the problems, after which they were tested for the number of the problem answered correctly. Group A studied in pairs, one student reading the problems and the other Group B studied alone. The following table shows the results: NUMBER OF PROBLEMS ANSWERED 92 CORRECTLY Group A Group B 19 23 18 10 11 13 25 20 12 10 16 18 11 10 15 11 20 12 18 13 Is there a significant difference at 5% level in the number of statistical problems answered correctly between Group A and Group B? 3. A program is designed to enhance readers’ speed and comprehension. To evaluate the effectiveness of this program, a test given both before and after the program, and sample result follow. At the 0.05 significance level, test the claim that comprehension is higher after the program. 1 Before 102 After 109 2 3 4 5 6 7 8 9 111 117 139 151 169 185 208 189 105 127 102 133 129 127 116 115 10 125 128 4. An exercise is design to lower systolic blood pressure of 10 randomly selected faculty of statistics. The results are as follows: 2 3 4 5 6 7 8 9 10 Samples 1 Before 120 130 160 90 110 110 180 190 130 120 After 110 120 140 100 90 90 180 170 130 130 At 0.05 level of significance, test if systolic blood pressure of the faculty is not affected by the design exercise. That is, test the claim 93 O.S. Corpuz 2012 UNDERSTANDING STATISTICS that the before and after the design exercise program values are equal. T-test for Independent Samples Problems on two samples are most commonly associated with small population/samples and with equal but unknown variances. For this situation the small samples must be used to provide an estimate of the common but unknown variances . The sample variances and each provide an estimate of . However, a can be obtained by combining the two better estimate of estimates in a weighted average as follows: For equal sample sizes, this formula reduces to which is the simple average of the two estimates. Otherwise, each of the estimates receives values proportional to its respective degrees of freedom. In the presentation of the two-sample t-test the estimates and were pooled together to form as the estimate of the common variance However, this pooling is usually done under the assumption that the populations had equal variance; otherwise, the pooling is not justified and the t-test cannot be used. In addition, the populations must come from a normal distribution. The t-statistic is composed using the equation (case1): 94 Where: = computed t-value = sample size of group 1 = sample size of group 2 = pooled variance Note that the degrees of freedom (df) is equal to n1 + n2 – 2. With the assumption that the populations have an unequal variance (case2), the test-statistic is given by: Where: = computed t-value = sample size of group 1 = sample size of group 2 = sample variance of group 1 = sample variance of group 2 The degree of freedom (df) will be determined using the equation: Where: = sample size of group 1 = sample size of group 2 = = = sample variance of group 1 = sample variance of group 2 95 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Test for Equality of Variances The test for equality of variance involves the ratio of the two sample variances. A random variable that consists of the ratios of the two sample variances has an F distribution if the two samples are independent and from normal population. The following are the steps to be followed for testing the variances: a. State the null and alternative hypotheses. b. Test-Statistic: Use F-test at a level of significance. c. Decision Criterion: Reject if Note that Where: = sample size of the group with the larger variance = sample size of the group with smaller variance d. Computation of the Test-Statistics: e. State your decision based on the rejection criterion and the computed test-statistic. f. State your conclusion based on your decision. 96 Testing for the Difference in Means of the Two Independent Samples Case 1: 1. That the mean of population 1 is equal to the mean of population 2. The alternative hypothesis may be defined in one of the three hypotheses same as the of the t-test dependent samples presented below: , That the mean of population 1 is greater a. than the mean of population 2. b. , That the mean of population 1 is lesser than the mean of population 2. c. , That the mean of population 1 is not equal to the mean of population 2. The alternative hypotheses a and b are directional and so if one uses either of these two hypotheses, a one-tailed test will be employed. However, a two-tailed test is used when one indicated the alternative hypothesis c since this is a non-directional hypothesis. 2. Test-Statistics: Use t-test for independent samples (case 1) 3. Level of Significance: Use a level of significance. 4. Decision Criterion: Note that the decision criterion would entirely depend on the formulated alternative hypothesis. As such, use the following decision criterion that corresponds to the alternative hypothesis. a. Reject b. Reject c. Reject if if if , , 97 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 5. Computation Where: = computed t-value = sample size of group 1 = sample size of group 2 = pooled variance For equal sample sizes, this formula reduces to which is the simple average o the two estimates. Otherwise, each of the estimates receives values proportional to its respective degrees of freedom. 6. State your decision based on the rejection criterion and the computed test-statistic. 7. State your conclusion based on your findings. Illustration for t-test for independent samples (case 1): An experiment was undertaken to compare the diameter increase of Gmelina and mahogany seedlings after 2 months measured in centimeters. Do the following data present sufficient evidence to conclude that the stem diameter of the two species are significantly different at 5% level? 98 Example 1: TEST SCORES Math English 45 55 67 78 90 39 58 69 80 89 58 89 59 59 58 60 94 49 90 91 Solution: The first step to take is to test the quality of variances. a. a. Test-Statistic: Use F-test at α = 0.05 level of significance. b. Decision Criterion: Reject if Note that c. Computation of the Test-Statistic: 99 O.S. Corpuz 2012 UNDERSTANDING STATISTICS TEST SCORES Math English Sample size Total Mean Variance 45 55 67 78 90 39 58 69 80 89 10 670 67 311.1111 58 89 59 59 58 60 94 49 90 91 10 707 70.7 316.0111 Fc = 316.0111 311.1111 = 1.01575 a. Decision: Since Fc < F0.05(9.9) = 1.015, accept the b. Conclusion: . Comparison of the variances of the diameter increase of Benguet pine and mangiumy seedlings after 2 months measured in centimeters, the following is undertaken to solve the main problem 100 Example 2: SEEDLINGS DIAMETER (CM) Benguet pine Mangium 3.08 2.38 3.10 2.68 2.35 2.17 3.56 3.86 3.73 3.91 1.48 2.65 1.72 1.85 2.30 1.86 2.80 2.76 3.50 2.68 Total 29.27 25.15 Mean 2.93 2.52 Variance 0.4994 0.5291 1. , That the two species do not differ in their stem diameter after two months , That the Beguet pine seedlings differs significantly from mangium seedlings in terms of diameter in centimeter after two months. 2. Test-Statistics: Use t-test for independent samples (case 1). 3. Level of Significance: Use α = 0.05 level of significance. 4. Decision Criterion: Reject if . 5. Computation: 101 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Since the two groups tested have equal sample sizes. Therefore, = 1.278 6. Decision: Since , there is no sufficient evidence to reject . 7. Conclusion: The data indicate that the fruit-bearing capabilities of the two varieties of tomatoes do not differ at 5% level of significance. For Case 2: 1. That the mean of population 1 is equal to the mean of population 2. The alternative hypothesis may be defined in one of the three hypotheses same as the of the t-test dependent samples: a. , That the mean of population 1 is greater than the mean of population 2 b. , That the mean of population 1 is lesser that the mean of population 2. c. , That the mean of population 1 is not equal to the mean of population 2. The alternative hypotheses a and b are directional and so if one uses either of these two alternative hypotheses, a one-tailed test will be employed. However, a two-tailed test is used when one 102 indicated the alternative hypothesis c since this is a non-directional hypothesis. 2. Test-Statistics: Use t-test for independent samples (case 2) 3. Level of Significance: Use α level of significance. 4. Decision Criterion: Note that the decision criterion would entirely depend on the formulated alternative hypothesis. As such, use the following decision criterion and corresponds to the alternative hypothesis. a. Reject b. Reject c. Reject if if if , , 5. Computation: Where: = computed t-value = sample size of group 1 = sample size of group 2 = sample variance of group 1 = sample variance of group 2 Note that 103 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Where: = sample size of group 1 = sample size of group 2 = = = sample variance of group 1 = sample variance of group 2 6. State your decision based on the rejection criterion and the computer test-statistic. 7. State your conclusion based on your decision. Illustration for t-test for independent samples (case 2): A study was conducted to compare the scores of 4 Math students and 6 Science students. The results are as follows: Math Science 15 15 32 28 40 30 50 17 20 22 The researcher wants to determine if the mean scores is significantly higher for Math than Science. Use 1% level of significance. Solution: Math Science Total Mean Variance 15 32 40 50 137 34.25 218.9167 15 28 30 17 20 22 12 22.00 35.6000 First, we will test the equality of variance: a. 104 b. Test-Statistic: Use F-test at α = 0.05 level of significance. if c. Decision Criterion: Reject Note that Where: = sample size of the group with the larger variance = sample size of the group with smaller variance d. Computation of the Test-Statistic = 6.1493 e. Decision: Since f. Conclusion: , reject . With the result on the test for the quality of variances, we can proceed to the solution of the main problem. Thus, 1. , That the mean scores of 4 students in Math is equal to the mean scores of 6 science students. , That the mean scores of 4 students in Math is significantly higher with the mean scores of 6 science students 2. Test-Statistics: Use t-test for independent samples (case 2). 3. Level of Significance: Use α = 0.01 level of significance. 4. Decision Criterion: Reject if . =4 =6 105 O.S. Corpuz 2012 UNDERSTANDING STATISTICS = = 5. Computation of the test-statistic: 6. Decision: Since , we fail to reject . 7. Conclusion: The mean scores of 4 Math students is equal to the mean scores of 6 science students. 106 Exercise No. 11 T-TEST FOR INDEPENDENT SAMPLES 1. The weight of 2 groups of children (randomized samples) taken were found to be as follows: Group A B 22.5 14.1 24.4 20.6 26.4 24.1 Weight (kg) 25.5 24.9 22.5 24 23.7 31.2 26.5 21.6 23.3 Test the hypothesis, at 5% level of significance, that the two groups of students are equal against the alternative hypothesis that they are unequal. 2. Find out whether poly bags affects height growth (m) of White lauan (Shore contarta). Group A were planted on poly bags and the other Group B planted in a bed. Group A Group B 1.9 2.1 Plant height (m) 0.5 2.8 1.3 1.4 3.1 0.6 0.9 Test the hypothesis at 1% level of significance. 3. In a study designed to estimate the volumes of water discharge in two major creeks (gal/sec). Creek 1 Creek 2 1 15 20 2 20 24 No of Trials 3 4 12 10 21 18 5 25 28 6 14 Determine if Creek 1 significantly have higher water discharge (gal/sec) compared with Creek 2. 107 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 4. Determine whether extra-curricular activities have detrimental effects to the grades of education students, the following GPA were recorded over a period of 8 years. Group With Without YEAR 1 2 3 4 5 6 7 8 85 95 86 87 83 79 89 91 78 89 92 95 85 81 84 89 Assuming the population to be normally distributed, test at 5% level of significance whether to participate actively in extra-curricular activities is detrimental to the student grades (GPA). Exercise No. 12 1. A student want to buy battery for her flashlight. She has a choice of three brands of rechargeable batteries that vary in cost. She obtains the sample data in the following table. She randomly selects three batteries for each brand, and test them for operating time (in hours) before recharging is necessary. Are three brands have the same mean usable time before recharging is required? Brand A B C 108 24.7 28.4 27.5 OPERATING TIME (in hours) 27.9 20.9 29.6 23.7 26.7 25.0 28.2 29.5 24.6 2. The following data shows the mid-term grades obtained by the five students statistics, biology, math, and FOR 102: STUDENT SUBJECTS Statistics Biology Math FOR 102 88 80 79 90 1 86 91 94 83 2 81 83 88 85 3 76 84 80 78 4 87 85 87 83 5 Use a 0.05 level of significance to test the hypothesis that a. Students have equal ability b. Courses are just the same 3. A Forester examined the effect of maintaining the water table at three different heights on seminal length of three hardwood seedlings. Evaluate at 5% level of whether water table and type of cereal significantly affect root length. Tree Species Molave Ipil-ipil Dao Low 9.2 20.6 24.3 WATER TABLE Medium High 9.7 11.5 9.2 6.8 20.3 7.9 Yield (kgs/ha) 4. Four varieties of rice have been tested for yield production recorded as follows: A 4,500 5,500 8,000 4,000 7,735 Rice Varieties B C 7,000 5,000 4,125 8,000 9,000 5,000 5,235 9,900 3,699 6,342 D 5,325 7,985 6,689 7,321 6,390 Are the yields the same with respect to varieties? 109 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Simple Linear Regression Analysis Regression analysis is a statistical technique used to determining the functional form of relationship between two or more variables, where one variable is called the dependent or response variable and the rest are called independent or explanatory variables. The ultimate objective is usually to predict or estimate the value of the response variable given the values of the independent variables. The relationship between the variables X and Y is represented by a statistical model of the form Y = f(x) + ε Where: Y = response or dependent variables (measures an outcomes of the study) X = explanatory or independent variable (attempts to explain the observed outcomes) In the simple linear regression the model is given by: Where: = ith observed value of the variable Y = ith observed value of the variable X = regression constant. It is the true Y-intercept = regression coefficient. It measures the true increases in Y per unit increase in X 110 = random error associated with Yi and Xi. The general idea in regression analysis is to for the model mentioned. However the model is based on population values or parameters. So, estimate the model based on a simple random sample. The estimated model is denoted as, Ǔ = b0 + b1X where, b0 and b1 are estimates of β0 and β1, respectively, On the other hand, the Ordinary Least Squares (OLS) methods may estimate the population parameters by minimizing the error sum of squares. That is, minimize, The OLS estimators of β0 and β1 are, Note that The computed values of β0 and β1 determine the equation of the regression line given by Ǔ = b0 + b1X. This equation is used to predict/estimates the value of Y, denoted by Ǔ, given a value of X. Hence, it is also defined as the predicting equation. Note further that if the regression coefficient b1 is positive, the relationship between the X and Y is directly proportional. That is, when X increases, Y also increases and when X decreases, Y also decreases. However, 111 O.S. Corpuz 2012 UNDERSTANDING STATISTICS if the regression coefficient b1 is negative, the relationship between X and Y is inversely proportional. This denoted that when X increases, Y decreases and when X decreases, Y increases. Evaluation of the Simple Linear Regression Equation An overall measure of adequacy of the equation is provided by the coefficient of determination, R2. R2 gives the proposition of total variation in Y that is accounted for by the independent variable X. R2 ranges from 0 to 1 or 0 to 100 if expressed in %. The nearer its value to 1 or 100 the better is the fit of the regression line. Individual 1 2 x x1 x2 y y1 y2 xy x1 y1 x2 y2 x2 x12 x22 y2 y12 y22 i xi yi xi yi xi2 yi2 n Total xn Σx yn Σy xn yn Σxy xn2 Σx2 yn2 Σy2 The Test of Hypothesis for the Regression Coefficient (β1) 1. , That there is no significant linear relationship that exists between X and Y. , That there is a significant linear relationship between X and Y. 2. Test-Statistics: Use F-test at α level of significance if , 3. Rejection Criterion: Reject 4. Computation: 112 a. b. c. d. e. f. ANOVA Source of Variation Regression Error Total Degrees of Freedom 1 n–2 Sum of Squared SSReg SSError n–1 SSTotal Mean Squares MSReg MSError Computed F Fc 5. State your decision. 6. State your conclusion. Sample Illustration of Simple Regression Analysis Problem: Suppose that we are given the following sample data to study the relationship between the Math scores and Science scores of 7 students. Math Scores 98 99 118 94 109 116 97 Science Scores 75 70 95 72 88 85 94 113 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 85 70 83 100 99 114 a. Fit an equation of the form Ǔ = b0 + b1X to a given data. Interpret the values obtained in the equation. b. Determine if there is a significant linear relationship between tree heights and basal area using F-test. Solution for problem the is: SUM MEAN Math Scores (X) 98 99 118 94 109 116 97 100 99 114 1044 104.4 i) Where: Therefore: 114 Science Scores (Y) 75 70 95 72 88 85 94 85 70 83 827 82.7 x2 y2 xy 9604 9801 13924 8836 11881 13456 9409 10000 9801 12996 109708 5625 4900 9025 5184 7744 7225 8836 7225 4900 8649 69313 7350 6390 11210 6768 9592 9860 9118 8500 6930 10602 86860 ii) Where: Therefore: iii) The resulting regression equation is Ǔ = 6.5336 + 0.7296x The prediction equation has a y-intercept of 6.534 and a slope of a line equal to 0.7296. The y-intercept of 6.534 indicates that even if the Math scores is 0 (which is impossible), the Science scores predicted to be 6.534 or 7. The slope of 0.7296 indicates that the Science score increases to about 0.7296 for every unit increase in the Math scores. Solution for problem b: 1. , That there is no significant linear relationship that exists between Math and Science scores of the 7 students . 2. , That there is a significant linear relationship between the Math and Science scores . 3. Test-Statistics: Use F-test at α =0.05 level of significance if , 4. Rejection Criterion: Reject 115 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 5. Computation: a. b. c. = 920.10 – 380.2484 = 539.8516 d. e. f. The ANOVA Table Source of Degrees of Variation Freedom 1 Regression Error 8 Total 9 Sum of Squared 380.2484 539.8516 920.1000 Mean Squares 380.2484 67.4815 Computed F 5.6349* * = significant at 5% level 6. Decision: Since 7. Conclusion: The result indicated that there is a significant linear relationship between the Math and Science scores of 7 students. 116 Correlation Analysis Correlation Analysis is a statistical technique used to determine the strength or degree of linear relationship between two variables. A measure of the degree of linear relationship is called correlation coefficient, r. The more pronounced the linear relationship and the greater is the magnitude of the correlation coefficient. The value of r range from -1 to +1. The correlation coefficient, r is interpreted as follows. Values of r ±1 ± 0.81 to ± 0.99 ± 0.61 to ± 0.80 ± 0.41 to ± 0.60 ± 0.21 to ± 0.40 ± 0.01 to ± 0.20 0 Qualitative Interpretation Perfect linear relationship Very strong linear relationship Strong linear relationship Moderate linear relationship Weak linear relationship Very weak linear relationship No linear relationship Note: Researcher should be careful in interpreting the correlation coefficient when it is near zero. It is possible that variable X and Y are strongly correlated but not in linear form Estimation of the Correlation Coefficient, ρ Based on a Simple Random Sample (SRS) of size n, an estimator of ρ is the sample correlation coefficient, r, defined as 117 UNDERSTANDING STATISTICS O.S. Corpuz 2012 Where: sum of the cross produce of X and Y sum of squares of x sum of squares of x Test of Hypothesis about Correlation Coefficient, ρ: 1. , There is no correlation between X and Y. , There is a significant positive correlation between X and Y (one tailed test). b. , There is a significant negative correlation between X and Y (one tailed test). c. , There is a significant correlation between X and Y (two-tailed test) a. 2. Test-Statistics: Use t-test at a level of significance 3. Rejection Criterion: The following rejection is used based on the alternative hypothesis: a. Reject b. Reject c. Reject if if if , , 4. Computation for the test-statistic. 118 Where: Tc = computed t-statistic r = correlation coefficient n= sample size 5. Decision: State your decision based on the rejection criterion and computer t-statistic. 6. Conclusion: State your conclusion. Illustration: Problem: Compare and interpret the correlation coefficient for the following grades of ten education students selected at random Student 1 2 3 4 5 6 7 8 9 10 Total Mean where: Stat 101 (X) 78 90 79 80 88 90 78 82 90 70 819 81.9 Math 102 (y) 76 88 80 78 90 92 80 82 89 68 823 82.3 x2 y2 xy 5184 8100 6241 6400 7744 8100 6084 6724 8100 4900 67577 5776 7744 6400 6084 8100 8464 6400 6724 7921 4624 68237 5472 7920 6320 6240 7920 8280 6240 6724 8010 4760 67886 sum of the cross produce of X and Y 119 O.S. Corpuz 2012 UNDERSTANDING STATISTICS = 482.3 sum of squares of x = 500.9 sum of squares of x =504.10 Therefore: r = 0.960 Based on the qualitative interpretation of r, the result indicates that there is a direct very strong linear relationship between the grades of students in Stat 101 and Math 102. That is, the higher is the student’s Stat 101 grade, the higher is his Math 102 grades. Scatter plot presentation: Math Regression Equation Coefficient of Determination Regression Equation Coefficient of Determination 120 Interpretation: Regression (Y- intercept) The prediction equation has a y-intercept of -6.171 and a slope of a line equal to 1.0724. The y-intercept of -6.171 indicates that if the score in STAT 101 is 0, the Math 102 score is predicted to be -6. The slope of 1.0724 indicates that the STAT 101 score increases to about 1.07 for every unit increase in the Math 102 score. Coefficient of Determination (R2) The R2 value of 0.9456 implies that the independent variable STAT 101 (as pre-requisite course) accounts 94.56% on the scores in Math 102. Only about 5.44% accounts for other factors not included in the model. Testing the Degree of Relationship Between Stat 101 and MATH 102 Grades: 1. , There is no correlation between the Student’s Stat and MATH 102 grades. , There is a significant degree of relationship between the student’s grades in Stat 101 and MATH 102. 2. Test-Statistics: Use t-test at 0.05 level of significance 3. Rejection Criterion: Reject if 4. Computation for the Test-Statistics: = 9.697 5. Decision: Since reject 6. Conclusion: The result reveals that there is a significant degree of relationship between student’s grades in Stat 101 and MATH 102. 121 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Measures of Correlation + Relationship No Relationship - Relationship No Relationship Pearson Product-Moment Coefficient of Correlation Equation: r= nXY – (X)(Y)_____ ¥[nX – (X)2][nY2 – (Y)2 2 Where: r = coefficient of correlation X = the 1st set of test score Y = the 2nd set of scores N = total number of pairing Calculation of Pearson Product-Moment Coefficent of Correlation Between Heights and Diameter of Molave (Vitex parviflora) Seedlings 122 Heights (X) Diameter (Y) 0.22 76 0.34 55 0.34 78 0.21 89 0.19 47 0.22 57 0.35 57 0.43 66 0.47 79 604 2.77 ȡ = XY 16.72 18.70 26.52 18.69 8.93 12.54 19.95 28.38 37.13 187.56 X2 5,776 3,025 6,084 7,921 2,209 3,249 3,249 4,356 6,241 42,110 Y2 0.0484 0.1156 0.1156 0.0441 0.0361 0.0484 0.1225 0.1849 0.2209 0.9365 9(187.56) – (604)(2.77) ¥[9(42,110) – (604)2][(9)(0.9365)2 = 1,688.04 – 1,673.08 ¥(378,990 – 364,816)(8.4285)2 = 14.96 ¥(14,174)(71.04) = 14.96 ¥1,006,921 = 14.96 1,003.455 = 0.0000149 The result simply means that there is no linear relationship between DBH and height of the seedlings using Pearson Product-Moment Coefficient of Correlation. 123 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Scatter plot presentation The plot explained that tree height is not related to dbh. The Regression equation represented by y=0.0011x + 0.2369 explained that the y-intercept of 0.2369 indicates that even if tree height is 0, dbh remains 0.24. The slope of 0.0011 indicates that dbh increases to about 0.0011 for every unit increase of tree heights. Simple Linear Correlation Analysis It deals with the estimation and test of significance of the simple linear correlation coefficient r, which is a measure of the degree of linear relationship between two variables X and Y. this can be computed using the equation: r= xy ¥(x )(y)2 2 Where: x = deviate of data X y = deviate of data Y Example: 124 X 196 171 170 207 177 162 199 173 231 140 1826 182.6 r= Y 200 178 169 212 180 165 201 179 243 144 1871 187.1 x y x2 y2 xy 13.4 12.9 179.56 166.41 172.86 -11.6 -9.1 134.56 82.81 105.56 -12.6 -18.1 158.76 327.61 228.06 24.4 24.9 595.36 620.01 607.56 -5.6 -7.1 31.36 50.41 39.76 -20.6 -22.1 424.36 488.41 455.26 16.4 13.9 268.96 193.21 227.96 -9.6 -8.1 92.16 65.61 77.76 48.4 55.9 2342.56 3124.81 2705.56 -42.6 -43.1 1814.76 1857.61 1836.06 1643.4 1683.9 6042.4 2835519.21 2767321 13.4 12.9 x y ¥(x2 )(y)2 = 6492.859 ¥( 6042.4)( 6976.9) = 6492.859 ¥6492.859 = 0.995 Interpretation: Based on the qualitative interpretation of r, the result indicates that there is a direct very strong linear relationship between the X and Y variables. That is, the higher is the X variable, the higher is the Y value. 125 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Scatter plot presentation The plot explained that X variable is linearly related to Yvariable. The Regression equation represented by y=1.0685x – 8.011 explained that the y-intercept of -8.011 indicates that even if X variable is 0, Y remains -8. The slope of 1.0685 indicates that Y increases to about 1.07 for every unit increase of X. Spearman Rank Correlation (for Non-parametric data) ȇ = 1 – 6D2 n(n2-1) Example: Entry No. 1 2 3 4 5 6 7 8 9 126 1st Judge 9 2 7 4 5 8 6 3 1 2nd Judge 7 6 8 4 9 1 2 5 3 D 2 -4 -1 0 -4 7 4 -2 -2 D2 4 16 1 0 16 49 16 4 4 110 Computation: ȇ = 1 – 6D2 n(n2-1) = 1 – 6(110) 9(81-1) = 1- 660 720 = 1 – 0.9167 = 0.083 Based on the result, no relationship detected between judge no. 1 and judge no. 2. Scatter plot presentation The plot proved that there is no linear relation of the judging of the two judges. Judge No. 1 only accounts 0.007% with judge no. 2. The remaining large percentage of about 99.99% is due to other factor not included in the analysis. 127 UNDERSTANDING STATISTICS O.S. Corpuz 2012 Exercise No. 13 SIMPLE LINEAR REGRESSION AND CORRELATION ANALYSIS 1. The following data shows the score (X) of students during examination and the number of hours (Y) they studied for the examination: X Y 1 2 3 4 5 6 7 8 9 85 8 78 3 90 10 92 15 87 11 89 12 80 6 82 8 81 7 a. Estimate the regression equation. b. Predict the score of a student who studied 13 hours for the examination. c. Determine if there is a significant relationship between the number of hours the students studied for an examination and their scores. 2. The following data were obtained on a study of the relationship between the weight and volumes of wood samples: WEIGHT(g) 2,080 2,150 4,041 3,152 3,221 4,132 2,231 4,430 3,573 Wood Volumes (bdft) 2.5 3.5 3.4 3.4 4.5 6.5 3.0 6.0 5.5 At 5% level of significance, is there a significant correlation between the weight and volumes of wood samples? 128 3. The figures given below pertain to the monthly disposable income and consumption expenditure of five families: 1 FAMILY 2 3 4 5 Disposable Income (X) 800.5 100.5 104.5 200.6 522.0 Consumption Expenditure(Y) 97.6 53.2 62.8 116.9 218.6 a. Compute the correlation coefficient and test for the significance of correlation between disposable income and consumption expenditure at the 5 percent level of significance. b. Compute for the coefficient of determination and interpret the derived value. 4. The following are the data on soil pH and height (H) of seedlings in cm. Seedling Soil pH H 1 2 3 4 5 6 7 8 9 10 11 12 6.8 7.0 6.9 7.2 7.3 7.0 7.0 7.5 7.3 7.1 6.5 6.4 154 167 162 175 190 158 166 195 189 186 148 140 A research is interested in learning how strong the association and how well he can predict the effect soil pH on the height of seedlings. a. b. c. d. e. f. Compute a Person r and interpret the result. Test at 5% level of significance. Compute r2 and explain what it means. Calculate the regression equation for the data. Test at the 1% level of significance. What would be likely height growth of seedlings under soil pH of 6.3? 129 O.S. Corpuz 2012 UNDERSTANDING STATISTICS ALTERNATIVE NON-PARAMETRIC TECHNIQUES Non-Parametric technique is a test of significance appropriate when the data represent an ordinal or nominal scale or when assumptions required for parametric test cannot be met. Related Samples (two sample Case) McNemar Change Test Function: This test for the significant of changes is particularly applicable to those “before and after” designs in which each person or subject is used as his own control and in which measurement is in the strength of either nominal or ordinal scale. Method To test the significance of any observed changes by this test, one sets up a fourfold table of frequencies (see figure below) to represent the first and second sets of responses from the same individuals. Note that + and – are used to signify different responses. 130 After Before - + + A B - C D Note also that those cases which show changes between the 1st and 2nd response appear in cells A and D. An individual is tailled in cell A if her changes from + to -, tallied in cell D if he changed from – to +, and tallied in both cells B and C if no changes is observed. The formula for a McNemar Change test is equated as: The significance of any observed as computed from the equation given for McNemar Change test is determined by reference to the Table for the Critical values for If the observed value of is equal to or greater that that shown in the said table for a particular significance level at df = 1t, the implication is that a “significant” effect was demonstrated in the “before” and “after” responses. Sample Illustration by Tagaro and Tagaro: A sociologist is interested in 25 fish cage operators’ sources of information before (prior to engaging in fish cage operation) and after (period in operation). He has data that operators prior to engaging in fish cage culture got their information from co-fisher farmers. With increasing familiarity and experience in fishcage culture (period in operation) the sources of information are other than co-fisher farmers (e.g. extension workers, pamphlets, others). 131 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The data cast in the form show below. After Operation Co-Fisher Others farmers Co-Fisher farmers Others Total Prior Operation Total 14 4 18 3 17 4 8 7 25 The researcher wants to determine if increasing familiarity and experience, fish cage operators will increasingly change source of information from co-fisher farmers to other sources of information. Solution: 1. , For those operators who change, the probability that any operators will change his source of information from co-fisher farmers to other sources (PA) is equal to the probability that he will change his source of information from others to co-fisher farmers (PD). 2. 3. 4. 5. 6. 132 , For those operators who change, the probability that any operators will change his source of information from co-fisher farmers to other sources (PA) is greater than the probability that he will change his source of information from others to co-fisher farmers (PD). Test-Statistics: The McNemar change test for the significance of changes is chosen because the study uses two related samples, each of the before and after type uses normal (classificatory) measurement. Significance Level” Let α = 0.05, N=25 Rejection Criterion: Reject if . Computation: = 4.50 7. Decision: Since reject 8. Conclusion: It can be concluded that for those operators who change, the probability that any operators will change his source of information from co-fisher farmers to other sources (PA) is greater than the probability that he will change his source of information from others to co-fisher farmers (PD).That is, fish cage operators show a significant tendency to change their sources of information from co-fisher farmers to other source when they get familiarity and experience in fish culture (Tagaro and Tagaro). Exercise No. 15 McNemar Change Test 1. A scientist is interested in finding out whether the students in Political Science changed their perception of the government two years after the election. Immediately after the elections, he asked the opinion of the community toward the government. Two years later, he asked the same question to the same people. The result are as follows: Before Favorable(+) Unfavorable(-) After Operation Unfavorable(-) Favorable(+) 30 19 15 24 2. During the deliberation of the bill reproductive health, a group of people was asked about their opinions toward the bill. Three year after becoming a law, the same people were asked the same question. The intention is to find out whether the events of 133 O.S. Corpuz 2012 UNDERSTANDING STATISTICS the first year during the law was in effect made them change their minds, The data are summarized as: One Year After Against Favor Favor 25 12 3 Years Before Against 16 28 3. A group of faculty were asked about their degree of satisfaction with the university before and after a change in administration. The test uses significance at 0.05 level of the change in opinion of the faculty regarding the school administration. The data are as follows: Before Satisfied Not Satisfied After Not Satisfied Satisfied 40 56 22 18 Sign Test Functions The sign test gets its name from the fact that it uses + and signs rather than quantitative measures in its data. It is particularly useful for research in which quantitative measurement is impossible, but possible to rank with respect to each other the two members of each pair. The sign test is applicable to the case of low related samples when the researcher wishes to establish that two conditions are different. 134 Method The null hypothesis tested by the sign tests that Where is the judgment or score under one condition (or before is the judgment or score under the other the treatment) and condition (after the treatment). That is are the two “scores” is that the median for a matched pair. Another way of stating difference of X and Y is zero. In applying the sign test, we focus on the direction of the difference , nothing where sign of the difference is + or -. between every is true, we would expect the number of pairs which have When to be equal to the number of pairs which have . That is if the null hypothesis is true, we would expect about half of the difference to be a negative and half to be positive. is rejected if too few differences of one sign occur. Hypothesis Testing Procedure of Sign Test: 1. 2. Test – Statistic: Use (Sign Test) at α level of significance. . 3. Rejection Criterion: 4. Computation: a. Determine the sign of the difference between the two numbers of each pair. b. By counting, determine the value o n, the number of pairs whose difference shown a sign. Let a be the number of negative signs, and b be the number of positive signs. c. For small samples, n≤ 35 use the formula. 135 O.S. Corpuz 2012 UNDERSTANDING STATISTICS d. For large samples, use the formula: Where: x is the number of fewer signs. Here we use +1 when x<n/2 and -1 when x>n/2. The significance of an obtained x may be determined by reference to Appendix Table A (Probabilities associated with the upper tail of the normal distribution). 5. State your decision. 6. State your conclusion. Sample Illustration No.1 (for small samples) Suppose we wish to determine whether increase of wages would increase the daily output of employees. Let X be the daily output in units before the increase in wages and Y be the daily output after the increase of wages. Sample of 15 employees yield the following data: EMPLOYEES 1 2 3 4 5 6 7 8 9 10 11 12 13 136 X 91 88 70 79 85 86 90 66 72 60 75 84 80 Y 88 87 67 69 83 81 93 67 76 55 74 86 72 di 3 1 3 10 2 5 -3 -1 -4 5 1 -2 8 14 80 90 -10 15 70 75 -5 Using the sign test, it is reasonable to say that the data presented sufficient evidence to conclude that the daily output after the wage increase is higher than before the increase. Solution: 1. The daily output after the wage increase is the same as the daily output before the increase. The daily output after the increase is higher than the daily output before the wage increase. (Sign Test) at α = 0.05 level of 2. Test – Statistic: Use significance. 3. Rejection Criterion: 4. Computation: EMPLOYEES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X 91 88 70 79 85 86 90 66 72 60 75 84 80 80 70 Y 88 87 67 69 83 81 93 67 76 55 74 86 72 90 75 di + + + + + + + + + - a. Determine the sign of the difference between the two numbers of each pair. 137 O.S. Corpuz 2012 UNDERSTANDING STATISTICS b. By counting, the value of n = 15, the number of pairs whose difference shows a sign. In the data given, a = 6 (the number of negative signs), and b = 9 (the number of positive sign. c. For small samples, n≤ 35 use the formula. 5. Decision: Since , there is no sufficient evidence to reject . 6. Conclusion: The hourly output after the raise of salary does not significantly differ from the hourly output before the raise. Illustrative Example No. 2 (For large Samples, n > 35) A company claims that its hiring practices are fair, it does not discriminate on the basis of gender, and the fact that 40 of the last 50 new employees are men is just a fluke (an unexpected random event). The company acknowledges that applicants are about half men and women. Test the null hypothesis that men and are equal in their ability to be employed by this company. Use the significance level of 0.05. Solution: 1. The proportion of men and women in the company are equal. The proportion of men and women in the company are not equal. 2. Test – Statistic and Significant Level: Use Sign Test at 5% level of significance. 3. Rejection Criterion: 138 4. Computation: If we denote players by + and non-players by - , we have 10 positive signs and 40 negative signs. The test statistic X is the smaller of 10 and 40, so X=10. We note that the value of n = 50 is above 35, so the test-statistic X is converted to the test-statistic Z. Here, we use +1 since X < n/2 or X = 10<50/2=25. 5. Decision: Since reject 6. Conclusion: There is a sufficient evidence to warrant rejection of the claim that the hiring practices is fair. 139 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Exercise No. 12 Sign Test 1. Two different firms design their own IQ test and psychologist administer both test to randomly selected students with the results given below. At the 0.05 level of significance, test the hypothesis that there is no significant difference between the two tests. Test I II A B C D E F G H I J 99 115 95 103 115 113 102 98 108 112 105 106 92 97 88 97 101 107 99 103 2. A test of running ability is given to a random sample of 8 runners before and after they completed a formal athletic lesson course in track and field. The results follow. At the 0.05 significance level, test the hypothesis that the course does not affect running scores. Runners 1 2 3 4 5 6 7 8 Before After 97 115 99 103 110 113 112 98 106 112 105 109 92 97 98 95 140 Wilcoxon Signed-Ranks Test For Two Dependent Samples The Wilcoxon signed-ranks test is very useful test for the behavioral scientist. With behavioral data, it is not uncommon that the researchers (1) tell which member of a pair is “greater than”, i.e. tell the sign of the difference between any pair, and (2) rank the difference in order of absolute size. That is, the researcher can make the judgment of “greater than” between any pair’s two values as well as between any two different scores arising from any two pairs. Procedures: 1. For each pair of data, find the difference d by subtracting the second score from the first. Retain signs, but discard any pairs for which d = 0. 2. Ignoring the signs of those differences, rank them from lowest to highest. When differences have the same numerical value, assign to them the mean of the ranks involved in the tie. 3. Assign to each rank the sign of the differences from which it can. 4. Find the sum of the absolute values of the negative ranks. 5. Let T be the smaller of the two sums found in step 4. 6. Let N be the number of pairs of data for which the difference d is not zero. 7. If n ≤ 30, use Table A-8 to find the critical value of T. Reject H0 if the sample data yield a value of T less than or equal to value in Table A-8. Otherwise, fail to reject the null hypothesis. If n > 30, compute the test statistics z by using the formula: 141 O.S. Corpuz 2012 UNDERSTANDING STATISTICS When eq. z is used, the critical z values are found from Appendix Table A. Again, reject the null hypothesis if the test statistics z is greater than or equal the critical value(s) of z. Otherwise, fall to reject the null hypothesis. Consider the example given by Tagaro and Tagaro. Thirteen students were tested for logical thinking. They are then given a tranquilizer and retested. Let us use the Wilcoxon signed-ranks test to test the hypothesis that the tranquilizer has no effect, so that there is no significant difference before and after testing scores. We will assume a 0.05 level of significance. a. H0: The tranquilizer has no effect on logical thinking. Ha: The tranquilizer has an effect on logical thinking. b. Test Statistics: Use Wilcoxon signed-ranks test. c. Significance Level: Use 0.05 level of significance. d. Rejection Criterion: Reject H0 if T is less or equal to the critical value in Table A-8 since < 30 in the example. e. Computation: T = min (Ȉ (+), Ȉ (-)) COURSE BEFORE AFTER DIFFERENCE RANK OF DIFFERENCE SIGNED RANKS A B C D E F G H I J K L M 67 78 81 72 75 92 84 83 77 65 71 79 80 68 81 85 60 75 81 73 78 84 56 61 64 63 -1 -3 -4 +12 0 +11 +11 +5 -7 +9 +10 +15 +17 1 2 3 10 8.5 8.5 4 5 6 7 11 12 -1 -2 -3 +10 +8.5 +8.5 +4 -5 +6 +7 +11 +12 Sum of the absolute value of negative ranks = 1+2+3+5=11 142 Sum of the absolute value of the positive ranks = 10+8.5+8.5+4+6+7+11+12=67. T – 11 (smaller of the two sums) f. Decision: since T = 11 is less than the critical value = 14 in Table A-8, reject H0. g. Conclusion: It appears that the drug affects scores. Independent Sample (two-sample case) Chi-square Test for Two Independent Samples Function When the data of research consist of frequencies in discrete categories, x2 test may be used to determine the significance of difference between two independent groups. The measurement involved may be as weak as nominal scaling. The hypothesis under the test is usually that the two groups differ with respect to some characteristics and therefore with respect to the relative frequency with which group members fall in several categories. Method The null hypothesis may be tested by Where: = the observed number of cases in the ijth category 143 O.S. Corpuz 2012 UNDERSTANDING STATISTICS = the expected number of cases in the ijth category when the null hypothesis is true k = the number of categories Note that the most common of all uses of the x2 test is the test whether an observed breakdown of frequencies in a 2 x 2 contingency table could have occurred under H0. When applying a Chi-square test where both r and k equal 2, the following formula should be used: 2 x 2 Contingency Table GROUP I A II C Total A+C + B D B+D Total A+B C+D N with df =1 To test the significance of x2 –computed, the researcher has to refer to the table of critical values x2 (Appendix Table C). If the x2 computed is greater than or equal to the critical value of x2 , then reject H0. Sample Illustration by Tagaro and Tagaro: A researcher studied the relation of feeding management in tilapia raising with efficiency. Efficiency is measured in terms of total weights (in kg) at harvest time. The greater the weight means that the feeding management is more efficient that other assuming all other factors of production constant. A purposive sampling of 60 fishponds operators using feeding a day (A) and 65 fishpond operators using 3 times in a day (B) were interviewed. The amounts of feed for both groups are equal. 144 The response of respondents is summarized in a 2x2 table below: FEEDING MGT. A B Total Efficient Not Efficient Total 40 30 70 20 35 55 60 65 125 Solution: 1. H0: There is no difference in efficiency between the two feeding management. Ha: Management has a significant effect on the tilapia raising.. 2. Test Statistics: Use The x2 test for two independent samples is chosen because the two groups (A and B) are independent and because the “score” under study are frequencies in discrete categories (efficient and not efficient). 3. Level of Significance: Use 5% as the level of significance, N = 125 (number of fishpond operators). 4. Rejection Criterion: Reject H0 if 5. Computation: = 4.53 , reject H0. 6. Decision: since 7. Conclusion: We conclude that the management has a significant effect on the tilapia raising. 145 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Wilcoxon-Mann-Whitney Test Function The Wilcoxon-Mann-Whitney test is an examination of equality of two population distributions. The test is most useful in testing for equality of two population means. As such, the test is an alternative to the two-sample t-test and is used when the assumption of normal population distribution is not met. The test is slightly weaker than the t-test. The only assumptions required for the test are random samples from the two populations of interest and that they are also drawn independently of each other. If we intent to state the hypothesis in terms of population means or medians, we need to add an assumptions, the difference is in location (mean or median). Method Case 1. When the samples sizes are equal, n1=n2 1. Put all observation in a single array tagging each observation to differentiate the origin of each observation. 2. Rank the observations in the combined array. 3. Assign the average rank in case of ties. 4. Sum the rank of the first sample (T1) and the rank of the second sample (T2) and compute T = min (T1,T2). 5. Compare T with tabular value (Table A10) 6. Decision Criterion: Reject H0 if T ≤ Ttab. Case 2. When samples sizes are unequal, n1 < n2. 1. Do step 1 to 3 in case 1. 2. Find the total ranks for sample that has the smaller size, n1 (T1). 3. Compute for T2 = n2 (n1+n2) - T1. 4. Determine T = min (T1,T2). 146 5. Compare T with the tabular value. 6. Decision Criterion: Reject H0 if T ≤ Ttab. Illustration: Suppose a researcher wants to compare the daily expenditures of families in rural area with that of the City. Suppose further that there are 15 sample respondents from the City, while there are only 10 respondents from the rural area. Below are the data corresponding to the groups under consideration: DILY EXPENDITURES IN RURAL AREA (PhP) 100 200 186 177 67 74 48 300 244 74 DAILY EXPENDITURES IN CITY (PhP) 250 300 600 134 890 52 570 153 462 115 405 117 334 157 224 Do the data indicate that the daily household expenditures in the urban area are greater than those of the rural area? (Use 5% level of significance) 147 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Solution: 1. H0: The daily household expenditures in the city are the same as those of rural areas. Ha: The daily household expenditures in the city are greater than those of rural areas. 2. Test Statistics: Use Wilcoxon-Mann-Whitney test at 5% level of significance. (case 2) 3. Rejection Criterion: Reject H0 if T ≤ Ttab. 4. Computation: n1 = 10 n2 = 15 A Array 48 Rank 1 B 157 11 A 177 12 B 52 2 A 67 3 A 186 13 A 74 4.5 A 200 14 A 74 4.5 A 244 15.5 A B B B B 100 115 117 134 153 6 7 8 9 10 B 244 15.5 B 250 17 B B B B B B Array 334 405 462 570 600 880 Rank 20 21 22 23 24 25 T1 = 1+3+4.5+4.5+6+12+13+14+15.5+18.5 = 92 T2 = n1 (n1+n2+1) - T1 = 10 (10+15+1) – 92 = 168 148 B 300 18.5 A 300 18.5 T = min (T1,T2) = min (92, 168) = 92 5. Decision: Since T = 92 > T tab = 90, we fail to reject H0. 6. Conclusion: We conclude that the daily household expenditures in the urban are the same as those in the rural area. Related Samples (k – samples case) Friedman Two-Way Analysis of Variance by Ranks The Friedman test (Friedman two-way analysis of variances by ranks) is a nonparametric analogue of the parametric two-way analysis of variance. The objective of this test is to determine if we may conclude from a sample of results that there is difference among treatment effects. The first step in calculating the test statistic is to convert the original results to ranks. Thus, it ranks the algorithms for each problem separately, the best performing algorithm should have the rank of 1, the second best rank 2, etc. In case of ties, average ranks are computed. Let rji be the rank of the jth of k algorithms on the ith of n data sets. The Friedman test needs the computation of the average ranks of algorithms, Rj = 1/nگi rji. Under the null hypothesis, which states that all the algorithms behave similarly and thus their ranks Rj should be equal, the Friedman statistic Function When the data from k matched samples are in at least an ordinal scale, this test is useful for testing the null hypothesis that k samples have been drawn from the same population. 149 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Since k samples are matched, the number of cases is the same in each of the samples. The matching may be achieved by studying the same group of subject under each k conditions. Method The formula to get the value of Friedman statistic denotes Fr is given by: 1. 2. X2F = 12n nگi R2j - k(k+1)2 k(k+1) 4 Where: FR or X2F = Friedman value N = number of rows (subjects) k = number of columns (variables or conditions) Rj = sum of ranks in the jth column (i.e, the sum of ranks for the jth variable) Directs one to sum th squares of the sums of ranks over all conditions The steps in the use of the Friedman two-way analysis of variance by ranks are as follows: 1. Cast the scores in a two-way table having k column (conditions) and N rows (subjects or groups). 2. Rank the scores in each row from 1 to k. 3. Determine the sum of the ranks in each column, Rj. 4. Compute the values of Fr using the formula: 150 5. Compare the Fr value with the critical value of x2 distribution (Appendix Table C). If the probability yielded by the test is equal to or less than α (the set level of significance) then reject H0. That is, if , reject H0. Sample Illustration: In a study on the influence of three different teaching strategies on the extent of learning of statistics, three matched samples (k=3) of 10 students were trained under three strategies of teaching. Matching was achieved by the use of 10 sets of students, three in each set. Prior to the training, a pre-test was given to the 30 statistics students. After the training, the extent of learning was measured giving a post test. Scores ranged from 0 to 50. The table below shows the scores of each of the 30 students as a result of posttest conducted. GROUP A B C D E F G H I J LECTURE 27 30 31 26 25 24 28 30 30 29 Teaching Strategies LECTURE W/ LECTURE W/ POWER FIELD POINT DEMONSTRATION 30 35 31 40 28 42 33 28 35 45 39 45 41 42 42 45 40 42 39 45 With the given data, the researcher wished to test if the teaching strategies have significant effect on the extent of learning statistics among students. 151 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Solution: 1. H0: The different methods of teaching have no differential effect. Ha: The different methods of teaching have differential effect. 2. Test Statistics: Use the non-parametic Friedman two-way analysis of variance because the scores exhibited possible lack of homogeneity of variance and thus the data suggested that one of the basic assumptions of the F-test was unattainable. 3. Level of Significance: Use 5% level of significance with N=10 the number of students in the three matched groups. 4. Rejection Criterion: Reject H0 if 5. Computation: GROUP 1 2 3 4 5 6 7 8 9 10 Rj 152 Methods of Teaching LECTURE W/ LECTURE W/ LECTURE POWER FIELD POINT DEMONSTRATION 1 2 3 1 2 3 1.5 1.5 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 10.5 19.5 30 = 19.05 , reject H0. 6. Decision: Since 7. Conclusion: The different types of training have different effect. Multiple Comparisons Between Groups of Conditions From the Result of the Friedman Two-Way ANOVA When the obtained value of Fr is significant, it indicates that at least one of the conditions differ from at least one other condition. It does not tell the researcher which one is different, nor does it tell the researcher how many of the groups varied from each other. That is, when the obtained value of Fr is significant, we would like to test the hypothesis H0: θu=θv against the alternative hypothesis Ha: θuθv for some conditions of u and v. There is a simple procedure for determining which condition/s differs. Begin by determining the differences | Ru – Rv | for all pairs of conditions or groups. When the sample size is large, the differences are not independent so, the comparison procedure must be adjusted appropriately. Suppose the hypothesis of no difference between k conditions or matched groups was tested and rejected at a significance level. Then we can test the significance of individual pairs of differences by using the following equality. That is, if Or if the data are expressed in terms of average ranks within each condition, and if Then we may reject the hypothesis H0: θu=θv and conclude that θuθv. Thus, if the average between the rank sums (or average 153 O.S. Corpuz 2012 UNDERSTANDING STATISTICS ranks) exceeds the corresponding critical value given in eq. R and eq T, then we may conclude that the two conditions are different. Above which lies , percent of distribution. The value of Z can be obtained from Appendix Table A. Because it is often necessary to obtain values upon extremely small probabilities, especially when k is large, Appendix Table A may be used. This is a table of standard normal distribution which has been arranged so that values used in multiple comparisons may be obtained easily. The table is arranged on the basis of the number of comparisons that can be made. The table values are the upper-tail probabilities associated with values of α. When there are k groups, there are comparison possible. Example: In the example above regarding teaching strategies, the Fr is significant at 5% level, The following total ranks were obtained: RL = 10.5 (Lecture Method), RLH = 19.5 (Lecture with power point presentation, RLD=30 (Lecture with Field Demonstration). We have the following differences: | RL – RLH | = | 10.5 – 19.5 | = 9 | RL – RLD | = | 10.5 – 30.0 | = 19.5 | RLH – RLD | = | 19.5 – 30.0 | = 10.5 We then find the critical differences by using eq. R. since α = 0.05 and k = 3, the number of comparisons, #c, is equal to Referring to Appendix Table AII, we see that the value of Z is 2.394. The critical difference is then 154 Since only the second difference (19.5) exceeds the critical difference, we conclude on the difference between conditions RL and RLD is significant/. Note that the first and third difference, although large, are not a magnitude great enough when using the significance level chosen. Hence, the lecture with power point presentation is significantly differed with the other two teaching strategies influencing learning in statistics. Independent Samples (k-sample case) Chi-square Test for r x k Tables Function and Rationale Research is undertaken because one is interested in the number of subject, object, or responses which fall into various categories. For example, root growth potentials of certain tree species may be categorized according to the frequency of first order lateral roots (FOLR) in seedlings, the hypothesis being that these FOLR will differ in frequency in the seedlings. Or persons may be categorized according to whether they are “in favor of”, “indifferent to” or “opposed to” an opinion to enable the researcher to test the hypothesis that these responses will differ in frequency. The Chi-square test is suitable for analyzing data like these. The number of categories may be two or more. The technique is of goodness-of-fit test, in that it may be used to test whether a significant difference exists between an observed number based upon the null hypothesis. That is, the chi-square test assesses the degree of correspondence between the observed and expected observations in each category. 155 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Method To compare an observed with an expected group of frequencies, we must be able to state what frequencies would be expected. The null hypothesis states the proportion of objects falling in each of the categories in the presumed population. That, from the null hypothesis we may deduce what are the expected frequencies. The chi-square technique gives the probability that the observed frequencies could have been sampled from a population with the given expected values. The null hypothesis may be tested by using the following statistic: Where: = the observed number of cases in the ijth category = the expected number of cases in the ijth category when the null hypothesis is true k = the number of categories If the agreement between the observed and expected frequencies is close, the differences ( ) will be small and the chi-square will be small. However if the divergence is large, the value of chi-square as computed in the equation will likewise be large. Roughly, the bigger the value of chi-square, less likely it is that the observed frequencies came from the population on which the null hypothesis and the expected frequencies are based. Summary Procedure: These are the steps in the used of the Chi-square test for k in dependent samples: 156 1. Cast the observed frequencies into r x k contingencies table, using the k columns for the groups and r rows for the conditions. 2. Calculate the row totals Ri and the column total Cj. 3. Determine the expected frequency for each cell by finding the product of the marginal totals common to it and divide this by N (where N represents the total number of independent observations); thus . 4. Determine the significance of the observed x2 by reference to Appendix Table C with degrees of freedom equal to (r-1)(k-1). If the probability given by Table C is equal to or smaller than α (set level of significance), reject H0. That is, if , reject H0 in favor of Ha. Sample Problem: 1. In an experiment to study the dependence of FOLR in seedlings, the following data were taken on 180 seedlings. Group A Root Class 1 (1-10) 21 Root Class 2 (11-20) 36 Root Class 3 (21 & Above) 30 B 48 26 19 Test the hypothesis that the seedlings group is independent of FOLR: Solution: a. H0: The seedling group is independent on FOLR. Ha: The FOLR is dependent on seedling group. b. c. d. e. Test-Statistic: Use Chi-square test. Significance Level: Use 0.05 level of significance. Rejection Criterion: Rejects Computation: 157 O.S. Corpuz 2012 Group UNDERSTANDING STATISTICS A R. Class 1 (1-10) 21 R. Class 2 (11-20) 36 R.Class 3 (21 +) 30 B 48 26 19 93 Total 69 62 49 180 Total 87 Note: The numbers written in italics are expected frequencies determined by: Solution: = 4.57 + 1.21 + 1.68 + 4.28 + 1.14 + 1.58 = 14.46 f. Decision: Since 158 reject the H0. g. Conclusion: The result shows that the FOLR is dependent on seedling group. Kruskal – Wallis Test Kruskal-Wallis test is a nonparametric test designed to detect difference among populations that does not require any assumptions about the shape of the populations. The test is an alternative nonparametric test for completely randomized design or one-way analysis of variance. The Kruskal-Wallis test is an analysis of variance that uses the ranks of the observations rather than the data themselves. This assumes, that the observations are on the interval scale. If data are in the form of ranks, used them as they are. The Kruskal-Wallis test is identical to the Mann-Wallis test for comparing k populations, where k is greater than 2. The null hypothesis is that the k populations under study have the same distribution, and the alternative hypothesis is that at least two of the population distributions are different from each other. Although the hypothesis test is in terms of the distributions of the populations of interest, the test is most sensitive to differences in the locations of the populations. Therefore, the test is actually used to test the ANOVA hypothesis of equality of k population means. The only assumption required for the Kruskal-Wallis test are that the k samples are random variables under study and continuous, and the measurement scale is atleast ordinal. Rank data points in the entire set from the smallest to largest, without regard to which sample they come from. Then sum all the ranks from each separate sample. Let n1 be the sample size of population 1, n2 the sample size from population 2, and so on up to nk, which is the sample size of population k. Define n as the total sample 1, R2 as the sum of the ranks from sample 2, and so on up 159 O.S. Corpuz 2012 UNDERSTANDING STATISTICS to Rk as the sum of the ranks from sample k. The Kruskal – Wallis test-statistic is equated as If the average of the ranks in the jth sample is considered, the Kruskal-Wallis test-statistic is given by: The null hypothesis is rejected if H exceeds the critical value of Chi-square at α level of significance with degrees of freedom equal to k-1. Example: The palatability studies are conducted to determine the acceptability of taste of four menus of cake, and different sample score follow. At 0.05 level of significance, test, the hypothesis that the four menus have the same acceptability level. Menu A 50 50 53 58 59 Menu B 59 60 63 65 67 Menu C 45 48 51 54 55 Menu D 62 64 68 70 72 Solution: 1. H0: The four menus have the same acceptability level. Ha: The four menus differ in acceptability level. 160 2. Test-Statistic: Use the Kruskal-Wallis test since there are four independent populations. 3. Level of significance: Use 0.05 level of significance. 4. Rejection Criterion: Reject H0 5. Computation: a. Rank the combine samples from the lowest to the highest. b. For each individual sample, find the number of observations and the sum of the ranks. Menu A 50 50 53 58 59 32.50 6.50 5 Menu B 59 60 63 65 67 69.50 13.90 5 Menu C 45 48 51 54 55 23.00 4.60 5 Menu D 62 64 68 70 72 85.00 17.00 5 c. Compute the value of the test – statistics H: = (0.02857)(2728.1) – 63 = 14.942 6. Decision: Since reject H0. 161 O.S. Corpuz 2012 UNDERSTANDING STATISTICS 7. Conclusion: The four menus differ significantly in their acceptability level. Exercise No. 16 Test yourself: Find the significant different of judging beauty contestants ȇ = 1 – 6D2 n(n2-1) Entry No. 1 2 3 4 5 6 7 8 9 162 1st Judge 9 2 7 4 5 8 6 3 1 2nd Judge 7 6 8 4 9 1 2 5 3 D 2 -4 -1 0 -4 7 4 -2 -2 D2 4 16 1 0 16 49 16 4 4 ANALYSIS OF VARIANCE ONE – WAY CLASSIFICATION DESIGN: CRD CRD is a design wherein the allocation of treatments is done by randomizing the treatments completely over the entire experimental units (eu’s) without any restriction imposed on the units and there is only one criterion for data classification. CRD is commonly used when: 1. The eu’s are sufficiently homogenous (like dishes of culture medium) and 2. Effective local control is assured (as those in laboratories, greenhouse). Randomization and layout Suppose there are: t = 3 treatments, T1, T2, and T3, which are replicated r1 = 2, r2 = 3 and r3 + 4 times respectively; hence, the nos. of eu’s required is n = r1 + r2 + r3 = 9. The randomization (using random no. generator key on calculator) may be as follows: 1. Label the eu’s consecutively from 1 to n = 9. 2. Obtain a sequence of n = 9 random numbers. Rank the numbers in increasing order. Using the sequence of ranks as a randomization of the eu’s, assign the first r1 = 2 eu’s to T1, the next r2 = 3 eu’s to T2 and the last r3 = 4 eu’s to T3. Random no.: 0.678 0.124 Rank (eu no.) 7 3 _______________ T1 0.543 0.667 0.119 5 6 2 _________________________ T2 0.076 0.923 0.876 0.436 1 9 8 4 _________________________________ T3 163 O.S. Corpuz 2012 UNDERSTANDING STATISTICS The layout for this randomization procedure is: 1 2 3 T3 T2 4 T1 5 6 T3 T2 7 T2 8 9 T1 T3 T3 Data Presentation: (One-way Classification) Treatment Observations (Yij) Reps Total Mean ri Yi. Ýi T1 Y11 Y12 … Y1r1 r1 Y1. Ý1 T2 Y21 Y22 … Y2r2 r2 Y2. Ý2 : : : … : : : : Tt Yt1 Yt2 … Ytrt rt Yt. Ýt n Y.. Linear Model: Model = overall population mean = effect of ith treatment ( (NID = normally, independently distributed) = “residual” or error effect of jth measurement of ith treatment = deviation of the ijth observation from the ith treatment mean ) 164 Advantages • Completely flexible for # of replications or treatments • Missing observations are not a problem • Maximum error degree of freedom of any designs Analysis of data: Illustration 1: Three methods of soil analysis, S1, S2, S3, were tried by a research institute. Twelve uniform soil samples were taken from a certain farm and S1 was randomly assigned to 5 of the soil samples, S2 to 3 samples and S3 to 4 samples. The researcher was interested in the time to complete the soil analysis. The data on time (in hours) of the soil analysis were summarized as follows: Method Time to complete analysis (hrs.) S1 2.4 3.8 2.9 4.6 S2 4.8 1.6 0.2 S3 7.2 5.3 2.9 3.5 Total 14.4 10.7 6.0 8.1 G.Mean 3.1 3.1 Reps Total Mean 5 16.8 3.36 3 6.6 2.20 4 18.9 4.72 12 42.3 3.35 At the level of significance Į = 5%, test the hypothesis that there is no difference in the time to complete the soil analysis for the three methods. Check for the satisfaction of the assumptions of ANOVA (to be discussed next). Testing significance of treatments via ANOVA F-test 1. State the statistical hypotheses Ho: μ1 = μ1 = μ1 …… All the method means are equal 165 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Ha: μ1 μi, i 1…. All least two method means are different. 2. Formulate the test statistic and get its critical value at the level of significance Į. Test statistic : Fc = MSTr/MSE Critical value: Ftab = FĮ[(t-1), (n-t)] (obtained from the F-table) 3. State the decision rule Reject Ho and Accept Ha if Fc Ftab; else, accept Ho. 4. Construct the ANOVA table outlined as follows: Source of Variance Due to treatments Experimental error TOTAL df SS MS t-1 TrSS MSTr n-t ESS MSE n-1 TSS FFtab Computed MSTr/MSE FĮ[(t-1), (n-t)] Computations: Let Yij = jth observation on the ith treatment Yi = total of observations on the ith treatment Y.. = grand total of all observations. Then compute a. the sums of squares CF = (Y..)2 = (42.3)2 n 12 t TSS = Yij2 – CF = (2.42 + 3.82 + … + 3.52) – 149.11 = 36.50 i=1 j=1 t 166 ri = 149.11 TrSS = Yi2/ri – CF = (16.82/5 + 6.62/3 + 18.92/4) – 149.11 = 11.16 ESS = TSS – TrSS = 36.50 – 11.16 = 25.34 i=1 b. the mean square MSTr = TrSS/(t-1) = 11.16/2 = 5.58 MSE = ESS/(n-t) = 25.34/9 = 2.82 c. the test statistics: Fc = MSTr/MSE = 5.58/2.82 = 1.98 The critical value: Ftab = F0.05(2,9) = 4.26 d. Set up the ANOVA table: Source of Variance Treatment df SS MS 2 11.16 5.581 Error 9 25.34 2.82 TOTAL 11 36.50 FComputed 1.98ns Ftab 4.26 e. State the decision and make the conclusion. It is concluded that there is no significant differences in the mean time to complete the soil analysis for the three methods Note: The observed differences among the means in the experiment are not large enough to warrant that in the population (or in general) there are atleast two means which are different. Compute other summary statistics 1. Coefficient of variation, CV(%) is the experimental error expressed as percentage of the mean. It measures the degree of 167 O.S. Corpuz 2012 UNDERSTANDING STATISTICS precision of the experiment or as an index of the reliability of the experiment. The higher the CV, the lower is the reliability of the experimental results. CV(%) = ¥MSE x 100 = ¥2.82 x 100 = 47.60% Y.. 3.52 2. Standard error of a treatment mean, s.e.(Ǔi) is the measure of the average error in estimating the true treatment mean. It measures the degree of precision of Ǔi as the estimate of the true treatment mean. s.e.(Ǔi) = ¥MSE/ri 3. Standard error of the difference between two treatment means, s.e.(Ǔi - Ǔi’) is a measure of the average error in estimating the difference between two treatment means. It measures the degree of precision of .(Ǔi - Ǔi’) as the estimate of the difference between the true means of the treatment i and treatment i’. s.e.(Ǔi - Ǔi’) = ¥MSE (1/ri+1/ri’) ….. for un equal replication s.e.d = ¥ 2MSE/r ….. for equal replication Estimates of treatment means and effects: 1. Estimate of true mean of treatment i; ȝi = Yi For treatment 1: ; ȝ1 = Y1 For treatment 2: ; ȝ2= Y2 For treatment 3: ; ȝ3 = Y3 For general mean: ȝ = Y = = = = 3.36 2.20 4.72 3.35 2. Estimate of the effect of treatment i; IJi= Yi. – Yi.. For treatment 1: IJ1 = Y1 – Y.. = 3.36 – 3.35 = 0.01 For treatment 2: IJ2 = Y2 – Y.. = 2.20 – 3.35 = -1.15 168 For treatment 3: IJ3 = Y3 – Y.. = 4.72 – 3.35 1.37 Summarize the treatment means and effects: Method Means Effects S1 3.36 0.01 S2 2.20 -1.15 S3 4.72 1.37 Mean 3.35 0.00 Note: Since the ANOVA test showed no significant effects, the observed effects cannot be generalized to be significant. Let us illustrate the analysis of a CRD experiment using the Statistical Analysis SystemTM (SAS). Data entry is best done using Microsoft excel. You should have a “mother file” in Excel, then save it as a prn file (formatted text [space delimited]) before going to SAS. Although the cards statement in SAS can be used for data entry, it is not practical to use for large data sets. The format of the data entry in Excel should be: ɨ ɨ ɩɬ ɨɰ ɨ ɩ ɩɪ ɨɥ ɨ ɪ ɫɬ ɩɩ ɨ ɫ ɭɰ ɪɫ ɨ ɬ ɬɬ ɩɰ ɨ ɭ ɬɥ ɪɬ ɨ ɮ ɩɩ ɨɰ ɩ ɨ ɨɯ ɩɥ ɩ ɩ ɨɬ ɩɩ ɩ ɪ ɪɪ ɩɫŜɫ ɩ ɫ ɪɫ ɪɩŜɫ ɩ ɬ ɬɩ ɩɩŜɪ ɩ ɭ ɩɮ ɨɬ ɩ ɮ ɬɫ ɮ 169 UNDERSTANDING STATISTICS O.S. Corpuz 2012 ɩ ɨ ɪɥ ɩ ɩ ɫɬ ɪɫ ɩɪ ɩ ɪ ɫɰ ɩɩŜɯ ɪ ϭ ϮϮ͘ϰ ɨɯŜɫ ɪ ɩ ɪɩŜɬ ɩɨŜɪ ɪ ɪ ɫɥ ɩɩŜɩ ɪ ɫ ɪɪŜɩ ɩɬ ɪ ɭ ɨɭŜɬ ɩɭ ɪ ɮ ɩɪ ɩɭ ɫ ɨ ɩɬ ɩɮ ɫ ɩ ɩɬŜɬ ɩɫŜɬ ɫ ɪ ɩɪ ɩɬŜɬ ɫ ɫ ɨɬ ɪɥ The column of the Replication (Rep) is not crucial in CRD, take note though that the columns preceding the response variable are the group or independent variables. The Excel data should then be saved as a prn file (formatted text, space delimited). Close the prn file and then go to SAS. Saved the file under the file name “D:\CRD.prn” In the program editor of SAS, write the following program statements: ʰɮɭɨʰɯɥŚ
Ś ŧśɎŜŨʰɩŚ Ś
Ś
Ś ʰŚ Ś ʰɩ, imply that the data starts at row 2. Make sure that the order of the variables in the input statement is exactly the same as the order inputted in Excel. Click the run icon and if everything is alright, then you will get the following in the output window of SAS (if there is a mistake in the program statements, the program may not run. Check the log window for mistakes and comments on the program statements). 170 ɨɰśɨɨř
ɨɯřɩɥɨɩɨ
ɮɨɩɪɫɬɭɮ ʰɩɮ ɨɰśɨɨř
ɨɯřɩɥɨɩɩ
ś
ʴ ɭɨɬɮɭŜɩɨɩɥɥɥɥɥɩɭɩŜɮɥɩɥɥɥɥɥɨŜɪɰɥŜɩɭɯɬ ɩɥɪɮɰɩŜɪɮɫɭɭɭɭɮɨɯɰŜɭɨɯɮɪɪɪɪ
ɩɭɬɪɭɯŜɬɯɭɭɭɭɭɮ ŞŜŜ ɥŜɩɰɪɬɰɰɫɨŜɩɨɫɫɮɨɪŜɮɮɥɩɨɨɯɨɪɪŜɫɨɨɨɨɨɨɨ
ʴ ɭɨɬɮɭŜɩɨɩɥɥɥɥɥɩɭɩŜɮɥɩɥɥɥɥɥɨŜɪɰɥŜɩɭɯɬ ɨɰśɨɨř
ɨɯřɩɥɨɩɪ
ś
ʴ ɭɪɮɭŜɰɨɮɮɫɥɮɫɭɩŜɯɨɰɭɩɪɫɭɨŜɭɨɥŜɨɰɬɪ ɩɥɮɮɯŜɯɮɭɪɪɪɪɪɪɯŜɰɫɪɯɨɭɭɮ
ɩɭɨɨɬɬŜɮɰɫɥɮɫɥɮ ŞŜŜ ɥŜɪɩɭɨɨɩɩɭŜɫɬɰɫɥɭŜɩɫɥɫɰɯɨɨɩɪŜɬɯɬɨɯɬɨɰ
ʴ ɭɪɮɭŜɰɨɮɮɫɥɮɫɭɩŜɯɨɰɭɩɪɫɭɨŜɭɨɥŜɨɰɬɪ 171 O.S. Corpuz 2012 UNDERSTANDING STATISTICS TRT and ERROR were particularly interesting; these are the sources of variation in CRD. The mean square (MS) for the treatment and error are 931195.8214 and 94773.2143, respectively. The F Value of the treatment (TRT) is 9.83 which is highly significant. If Pr > F is less than 0.01, then the difference among the treatment is highly significant; if it is greater than 0.01 but less that 0.05 it is significant. If Pr > F is greater than 0.05, then there are no significant differences among treatments. The result of the analysis is consistent with Table 2.2 of Gomez and Gomez (1984). Missing Data in CRD The previous analysis had equal replication of all treatment. This is not necessary, and sometimes you may set up an experiment, planning for equal replication, but you cannot collect data from all experiment units. Some data may be lost due to a lot of reason and factors. Missing data should be entered as “.” And not as “0” in Excel. At the end of the infile statement of SAS append the word “missover” to indicate that there were missing data. It would also be better to use proc glm instead of proc anova in the analysis of variance for experiments with missing data. The modified CRD SAS program would then be: ʰɮɭʰɯɥŚ
Ś ŧŪɎŜŨʰɩŚ Ś
Ś
Ś ʰŚ Ś The ANOVA output is: ɨɰśɪɭř
ɨɯřɩɥɨɩɨ
172 ɮɨɩɪɫɬɭɮ ʰɩɯ ɨɩɬ ɩɩɫ ś
Ŝ ɨɰśɪɭř
ɨɯřɩɥɨɩɩ
ś
ʴ ɭɨɫɫɮŜɪɬɥɰɪɪɪɪɩɫɨŜɩɩɬɨɬɬɬɭɨŜɩɫɥŜɪɪɫɬ ɨɯɪɬɨɫŜɪɮɫɭɭɭɭɮɨɰɬŜɩɫɪɥɪɮɥɫ
ɩɫɫɰɭɨŜɮɩɬɭɥɥɥɥ ŞŜŜ ɥŜɩɰɨɮɥɪɫɪŜɩɩɮɮɬɨɪŜɰɮɩɰɪɰɫɭɪɩŜɪɩɫɥɥɥɥɥ
ʴ ɭɨɫɫɮŜɪɬɥɰɪɪɪɪɩɫɨŜɩɩɬɨɬɬɬɭɨŜɩɫɥŜɪɪɫɬ
ʴ ɭɨɫɫɮŜɪɬɥɰɪɪɪɪɩɫɨŜɩɩɬɨɬɬɬɭɨŜɩɫɥŜɪɪɫɬ ɨɰśɪɭř
ɨɯřɩɥɨɩɪ
ś
ʴ ɭɪɯɰŜɬɨɩɬɥɥɥɥɭɫŜɰɨɯɮɬɥɥɥɨŜɫɮɥŜɩɫɮɪ ɨɮɮɬɨŜɫɫɬɯɪɪɪɪɫɫŜɩɥɩɭɰɭɥɯ
ɩɪɨɨɫɥŜɰɬɯɪɪɪɪɪ ŞŜŜ ɥŜɪɫɨɪɰɨɩɮŜɰɫɫɮɥɭŜɭɫɯɬɨɥɯɩɩɪŜɮɰɨɭɭɭɭɮ
ʴ ɭɪɯɰŜɬɨɩɬɥɥɥɥɭɫŜɰɨɯɮɬɥɥɥɨŜɫɮɥŜɩɫɮɪ
ʴ ɭɪɯɰŜɬɨɩɬɥɥɥɥɭɫŜɰɨɯɮɬɥɥɥɨŜɫɮɥŜɩɫɮɪ Example – DBH measurements taken on 5 trunks of trees per plot in a tree species trial. The variation among trunks within a plot is not an estimate of experimental error; it estimates within-plot variation. The variation among mean DBH means of replicated plots of the tree species is the estimate of experimental error. Thus, you cannot plant one replication of the trial, and measure multiple trunks within plots and call this a replicated experiment. You must always replicate the experimental units. An appropriate ANOVA table for sub sampling in a replicate experiment is like this – Measurements made per plot – 173 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Source df Hybrids Within hybrids 9 20 Within plots 120 MS Variation among hybrids Variation among reps within hybrids = experimental error Variation within plots = sampling variation 149 observations Thus, the appropriate F-test to see if there is variation among hybrids is – MS Hybrids/MS within Hybrids MS hybrids/MS within plots is not a valid test of the hypothesis. If you get confused as to which term is experimental error, go back and think what are the experimental units. It is the variation among experimental units which are treated alike that is experimental error. In the case discussed above, it is not the trees that are experimental units but the plots. The treatments were randomly assigned to plots, but not to the trees within plots. All of the trees within a plot were the same treatment. TWO-WAY CLASSIFICATION DESIGN Randomized Complete Block Design (RCBD) “Complete Block” A unit that contains every treatment To use RCBD … (1) Must be able to group treatments into blocks (2) Blocks must be large enough to hold treatments 174 RCBD is a two-way classification. Each observation is classified according to two criteria – treatment effect and block effect Model – Same as RCB, except is effect of jth block: Now: Ti ∼ NID ) βj ∼ NID ) εij ∼ NID ) Randomization Procedure (1) Randomize treatments to experimental units within blocking groups (2) Randomly assign the groups of experimental units to blocks. Blocking Methods (1) (2) (3) Time; e.g., soaking hours, years of conducting the experiment. Space; e.g., different planting distance or areas within fields, different laboratories. People; e.g., each block cared for or measured by a single person. Advantages (1) (2) (3) More precise than CRD if blocking is effective. It is simple to understand (vs. more complex designs). Any number of treatments and replication is allowed. 175 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Disadvantages (1) (2) If blocking is ineffective, provision is lost compared to CRD. Homogeneity within blocks is often reduced as treatment numbers increases (as block size increases). This is not a disadvantage compared to CRD, but compared to more complex designs. By changing from CRD to RCBD, we gain an increase in precision if heterogeneity of experimental units can be effectively partitioned into blocks. is partitioned into Remember, in CRD we said By blocking in RCBD, you are trying to remove reduce , and increase precision. to But we lose degrees of freedom for estimating error – Note: If you have many treatments relative to blocks, this df errors : CR →→→ RCBD loss of df is unimportant! T(r-1)↓ r-1 df (t-1)(r-1) If you don’t gain precision by blocking in RCBD, stay with a CRD. ANOVA Layout Source Definition df Blocks (r-1) Treatments (t-1) 176 SS Calculation MS B x T (error) (r-1)(t-1) Total rt-1 SSTotal - SSβ - ST H0 : Ti = 0 (equivalent to all treatments are equal, or hopefully, MSe is reduces by blocking = 0) Relative Efficiency of RCBD vs. CRD We use the relative efficiency (RE) to estimate the gain precision in addition to the loss of df. • Estimate what MSerror would have been in a CRD. We use a weighted average of the block variance and the error variance. We give the error variance additional weight by including the treatment df in the weighting factor. • The relative efficiency takes into account the difference in erro df as well as the error term itself. As MSeRCB is reduced compared to MSeCR, then RE is increased. How to set-up blocks if we know a gradient exists Example – Fertility gradient in field →→→→→fertility→→→→ 5 7 RIGHT 1 WAY 2 I II 5 →→→→→→→→fertility→→→→→→→ 7 1 2 8 3 4 6 WRONG WAY 177 O.S. Corpuz 2012 8 3 4 6 I UNDERSTANDING STATISTICS III II III Right way→→arrange blocks perpendicular to gradient. This maximized variation among blocks and minimizes variation within blocks. Missing Values in the RCBD What to do when you can’t collect data from a plot? This happens every year – it is expected! One Missing Plot – (1) We use a “least squares estimate” to estimate what we expect the plot would have yielded, based on the rest of the data from the experiment. Note: This estimate is never as good as actually having the data! Where: βi = total of observation in block containing missing plot: T.j = total of observations of treatment with missing plot. So, this formula assumes that we have good estimates of the block effect and the treatment effect and that the effects are additive. (2) Enter the estimate value ( ) into the data table and then perform the ANOVA. But there’s no free lunch! You must pay a penalty! 178 Reduce the total df by 1 Reduce error df by 1 What about more than one missing plot? The more missing plots you have, the more difficult it becomes to estimate them, and the poorer your estimates become. You can do them by hand, but it’s best to use matrix algebra to get simultaneous estimates of the missing plots. In this topic, we’ll let SAS PROC GLM do the matrix calculations. To illustrate the ANOVA of RCBD using SAS, we will work on the following problem: A forester studied the growth of white lauan under various weed defoliage schemes. The experiment was conducted in an RCB design with four locations taken as blocks. [The data is in the file ‘D:RCB.xls’. Again, take note of the way the data were entered.] The SAS program for this problem is: ʰɮɭʰɯɥŚ
Ś ŧśɎŜŨʰɩŚ Ś
Ś
Ś ʰŚ Ś Compared to the SAS program of CRD, the variable blk (i.e. block) was included in both the class and model statements. The SAS output is: ɩɥśɫɰř
ɨɯřɩɥɨɩɨ
ɪɨɩɪ ɫɨɩɪɫ 179 O.S. Corpuz 2012 UNDERSTANDING STATISTICS ʰɬɥ ɨɫɭ ɩɫɩ ś
Ŝ ɩɥśɫɰř
ɨɯřɩɥɨɩɩ
ś
ʴ ɬɫɭɩɥɰɮŜɭɭɥɭɯɩɥɨɰɩɫɨɰŜɬɪɩɨɪɭɫɥɥŜɰɮɥŜɫɫɬɨ ɫɥɪɮɰɪɬɫɥŜɫɭɪɥɨɪɭɬɰɫɯɪɯŜɬɨɨɬɮɬɪɫ
ɫɬɫɩɬɬɭɪɯŜɨɩɪɭɰɬɭɬ ŞŜŜ ɥŜɨɥɯɬɯɬɩɨŜɰɰɰɫɪɥɮŜɰɬɯɭɨɰɰɨɮɩŜɰɮɭɥɯɭɰɭ
ʴ ɩɨɯɭɨɨɰŜɯɥɨɥɩɯɰɰɰɪɥɬɰŜɰɥɥɬɨɫɫɰɥŜɰɯɥŜɪɯɪɮ ɪɩɮɬɰɮɮŜɯɬɰɭɬɪɥɩɰɨɰɰɩŜɭɨɰɯɯɫɪɫɥŜɰɮɥŜɫɨɭɫ
ʴ ɩɨɭɬɪɯɰŜɬɥɮɥɭɩɨɨɯɩɭɰɫŜɮɬɪɬɪɨɥɭɥŜɯɮɥŜɫɩɬɰ ɪɩɮɬɰɮɮŜɯɬɰɭɬɪɥɩɰɨɰɰɩŜɭɨɰɯɯɫɪɫɥŜɰɮɥŜɫɨɭɫ ɩɥśɫɰř
ɨɯřɩɥɨɩɪ
ś
ʴ ɬɩɬɪŜɬɩɪɯɮɥɨɭɬɥŜɮɥɫɮɮɫɥɪɥŜɮɫɥŜɬɰɮɨ ɪɭɩɫɬɰŜɮɫɥɨɮɮɫɭɭɯŜɪɩɭɨɨɭɥɫ
ɫɨɩɮɨɪŜɩɭɫɥɫɮɭɩ ŞŜŜ ɥŜɥɰɪɫɪɰɪɬŜɥɫɩɰɫɯŜɩɭɬɰɭɨɩɰɩɪŜɬɯɯɥɰɬɩ
ʴ ɩɩɥɬŜɩɬɪɪɭɪɨɥɨɥɩŜɭɩɭɭɯɨɬɬɨŜɬɥɥŜɩɪɭɪ ɪɫɯŜɩɮɥɬɥɮɥɭɨɭŜɥɰɥɨɭɰɥɩɥŜɩɫɥŜɯɮɨɥ
ʴ ɩɨɰɰŜɩɬɪɭɬɥɯɩɰɰŜɭɩɭɯɩɬɫɨɨŜɫɭɥŜɩɫɭɨ ɪɫɯŜɩɮɥɬɥɮɥɭɨɭŜɥɰɥɨɭɰɥɩɥŜɩɫɥŜɯɮɨɥ Based on the analysis, blocking was not significant (P>0.05) implying that it was not effective in reducing experimental error. Treatment is also insignificant implying the treatment were just the same influence on tree growth. 180 181 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Appendix 1 OUTLINE OF RESEARCH METHODS IN EDUCATION Research – is a scientific investigation of phenomena which includes collection, presentation, analysis , and interpretation of facts that links man’s speculation of reality. Characteristics of Researcher 1. Intellectual curiosity 2. Prudence 3. Healthy criticism 4. Intellectual honesty Qualities of a Good Researcher R – research-oriented E – efficient S – scientific E – Effective A – Active R – resourceful C – creative H – honest E – economical R – religious Characteristics of Research 1. 2. 3. 4. 5. 6. Empirical Logical Cyclical Analytical Replicability Critical Types of Research 1. Pure research 2. Applied research 3. Action research Classification of Research 1. Library research 2. Field research 3. Laboratory research 182 The Variable Types: 1. Independent variable – stimulus variable 2. Dependent variable – response variable 3. Moderate variable – special type of independent variable that may alter or modify the relationship of the independent and dependent variable 4. Control variable – a variable controlled by the researcher which the effect can be neutralized by eliminating or removing the variable 5. Intervening variable – a variable that interferes with the independent and dependent variable which may either strengthen or weaken the two variables Schematic Diagram of the Research Process Problem/Objectives Theoretical/Conceptual Framework Assumptions Hypothesis Review of Related Literature Research Design Data Collection Data Processing/Statistical Treatment Analysis and Interpretations Summary, Conclusions and Recommendations 183 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Research Designs 1. Historical Designs – Steps: a. Collection of data b. Criticism of the data collected c. Presentation of the facts 2. Descriptive Designs Types: a. Descriptive-survey b. Descriptive-normative survey c. Descriptive-status d. Descriptive-analysis e. Descriptive-classification f. Descriptive-evaluative g. Descriptive-comparative h. Correlational survey i. Longitudinal survey 3. Case Study Design Steps: a. Recognition and determination of the status of the problem to be investigated b. Collection of data c. Diagnosis or identification of the causal factors d. Application of remedial or adjustment measures e. Subsequent follow-up to determine the effectiveness of the corrective or developmental measures applied Qualities of a Good Research Instrument 1. Validity a. Content validity b. Concurrent validity 2. Reliability 3. Usability Sampling Designs 1. Scientific sampling a. Random sampling b. Systematic sampling - Stratified sampling design - Multi-stage sampling design - Clustering 2. Non-scientific sampling a. Purposive sampling b. Incidental sampling c. Quota sampling 184 Data ProcessingParts of Thesis 1. Chapter 1 – The Problem and Its Background a. Statement of the Problem b. Theoretical/Conceptual Framework c. Assumptions d. Hypothesis e. Significance of the Study f. Scope and Limitations of the Study g. Operational Definition of Terms 2. Chapter 2 – Review of Related Literature a. Related readings b. Related literature c. Related studies d. Justification of the present study 3. Chapter 3 – Methodology a. Research Design b. Determination of Sample Size c. Sampling Design and Technique d. The Subject e. The Research Instrument f. Validation of the Research Instrument g. Data Gathering Procedure h. Data Processing Method i. Statistical Treatment 4. 5. 6. 7. 8. Chapter 4 – Results, Analysis and Interpretations Chapter 5 – Summary, Conclusions and Recommendations Bibliography/Literature Cited/References Appendices Curriculum Vitae 185 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Appendix 2 Multiple Comparisons Between Groups of Conditions From the Results of the Kruskal-Wallis One-Way ANOVA Table A Areas Under the Normal Curve Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -3.4 -3.3 -3.2 -3.1 -3.0 0.0003 0.0005 0.0007 0.0010 0.0013 0.0003 0.0005 0.0007 0.0009 0.0013 0.0003 0.0007 0.000 0.0009 0.0013 0.0003 0.0004 0.0006 0.0008 0.0012 0.0003 0.0004 0.0006 0.0008 0.0012 0.0003 0.0004 0.0006 0.0008 0.0011 0.0003 0.0004 0.0005 0.0008 0.0011 0.0003 0.0004 0.0005 0.0008 0.0011 0.0003 0.0004 0.0005 0.0007 0.0010 0.0002 0.0003 0.0005 0.0007 0.0010 -2.9 -2.8 -2.7 -2.6 -2.5 0.0019 0.0026 0.0035 0.0047 0.0062 0.0018 0.0025 0.0034 0.0045 0.0060 0.0018 0.0025 0.0034 0.0045 0.0060 0.0017 0.0023 0.0032 0.0043 0.0057 0.0016 0.0023 0.0031 0.0041 0.0051 0.0016 0.0022 0.0030 0.0040 0.0054 0.0015 0.0021 0.0029 0.0039 0.0052 0.0015 0.0021 0.0028 0.0038 0.0051 0.0014 0.0020 0.0027 0.0037 0.0049 0.0014 0.0019 0.0026 0.0036 0.0048 -2.4 -2.3 -2.2 -2.1 -2.0 0.0082 0.0107 0.0139 0.0179 0.0228 0.0080 0.0104 0.0136 0.0174 0.0222 0.0080 0.0104 0.0136 0.0174 0.0222 0.0075 0.0099 0.0129 0.0166 0.0212 0.0073 0.0096 0.0125 0.0162 0.0207 0.0071 0.0094 0.012 0.0158 0.0202 0.0069 0.0091 0.0119 0.0154 0.0197 0.0068 0.0089 0.0116 0.0154 0.0192 0.0066 0.0087 0.0113 0.0146 0.0188 0.0064 0.0084 0.0110 0.0143 0.0183 -1.9 -1.8 -1.7 -1.6 -1.5 0.0287 0.0359 0.0446 0.0548 0.0668 0.0281 0.0352 0.0436 0.0537 0.0655 0.0281 0.0352 0.0436 0.0537 0.0655 0.0268 0.0336 0.0418 0.0516 0.0630 0.0262 0.0329 0.0409 0.0505 0.0618 0.0256 0.0322 0.0401 0.0495 0.0606 0.0250 0.0314 0.0392 0.0485 0.0594 0.0244 0.0307 0.0384 0.0475 0.0582 0.0239 0.0301 0.0357 0.0465 0.0571 0.0233 0.0294 0.0367 0.0455 0.0559 -1.4 -1.3 -1.2 -1.1 -1.0 0.0808 0.0968 0.1151 0.1357 0.1587 0.0793 0.0951 0.1131 0.1335 0.1562 0.0793 0.0951 0.1131 0.1335 0.1562 0.0764 0.0918 0.1098 0.1292 0.1515 0.0749 0.0901 0.1075 0.1271 0.1492 0.0735 0.0885 0.1056 0.1251 0.1469 0.0722 0.0869 0.1038 0.1230 0.1446 0.0708 0.0853 0.1020 0.1210 0.1423 0.0694 0.0838 0.1003 0.1190 0.1401 0.0681 0.0823 0.0985 0.1170 0.1379 -0.9 -0.8 -0.7 -0.6 -0.5 0.1841 0.2119 0.2420 0.2743 0.3085 0.1841 0.2090 0.2389 0.2709 0.3050 0.1788 0.2061 0.2358 0.2676 0.3015 0.1762 0.2033 0.2327 0.2643 0.2981 0.1736 0.2005 0.2296 0.2611 0.2946 0.1711 0.1977 0.2266 0.2578 0.2912 0.1685 0.1949 0.2236 0.2546 0.2877 0.1660 0.1922 0.2206 0.2514 0.2843 0.1635 0.1894 0.2177 0.2483 0.2810 0.1611 0.1867 0.2148 0.2451 0.2776 -0.4 -0.3 -0.2 -0.1 -0.0 0.3446 0.3821 0.4287 0.4602 0.5000 0.3409 0.3783 0.4168 0.4562 0.4960 0.3372 0.3745 0.4129 0.4522 0.4920 0.3336 0.3707 0.4090 0.4483 0.4880 0.3300 0.3669 0.4052 0.4443 0.4840 0.3264 0.3632 0.4013 0.4404 0.4801 0.3228 0.3594 0.3974 0.4364 0.4761 0.3192 0.3557 0.3936 0.4325 0.4721 0.3156 0.3520 0.3897 0.4286 0.4681 0.3121 0.3483 0.3859 0.4247 0.4641 0.0 0.1 0.2 0.3 0.4 0.5000 0.5398 0.5793 0.6179 0.6554 0.5040 0.5438 0.5832 0.6217 0.6591 0.5080 0.5478 0.5871 0.6255 0.6628 0.5120 0.5517 0.5910 0.6293 0.6664 0.5160 0.5557 0.5948 0.6331 0.6700 0.5199 0.5596 0.5987 0.6368 0.6736 0.5239 0.5636 0.6026 0.6406 0.6772 0.5279 0.5679 0.6064 0.6433 0.6808 0.5319 0.5714 0.6103 0.6480 0.6844 0.5359 0.5753 0.6141 0.6517 0.6879 0.5 0.6 0.7 0.8 0.9 0.6915 0.7257 0.7580 0.7881 0.8159 0.6950 0.7291 0.7611 0.7910 0.8186 0.6985 0.7324 0.7642 0.7939 0.8212 0.7019 0.7357 0.7673 0.7967 0.8238 0.7054 0.7389 0.7704 0.7995 0.8264 0.7088 0.7422 0.7734 0.8023 0.8289 0.7123 0.7454 0.7764 0.8051 0.8315 0.7157 0.7486 0.7794 0.8078 0.8340 0.7190 0.7517 0.7823 0.8106 0.8365 0.7224 0.7549 0.7852 0.8133 0.8389 1.0 1.1 1.2 1.3 1.4 0.8413 0.8643 0.8849 0.9032 0.9192 0.8438 0.8665 0.8869 0.9049 0.9207 0.8461 0.8686 0.8888 0.9066 0.9222 0.8485 0.8708 0.8907 0.9082 0.9236 0.8508 0.8729 0.8925 0.9099 0.9251 0.8531 0.8749 0.8944 0.9115 0.9265 0.8554 0.8770 0.8962 0.9131 0.9278 0.8577 0.8790 0.8980 0.9147 0.9292 0.8599 0.8810 0.8997 0.9162 0.9306 0.8621 0.8830 0.9015 0.9177 0.9319 186 Table A Continued… Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.5 1.6 1.7 1.8 1.9 0.9332 0.9452 0.9554 0.9641 0.9713 0.9345 0.9463 0.9564 0.9649 0.9719 0.9357 0.9474 0.99573 0.9656 0.9726 0.9370 0.9484 0.9582 0.9664 0.9732 0.9382 0.9495 0.9591 0.9371 0.9738 0.9394 0.9505 0.9599 0.9678 0.9744 0.9406 0.9515 0.9608 0.9686 0.9750 0.9418 0.9525 0.9616 0.9693 0.9756 0.9429 0.9535 0.9625 0.9699 0.9761 0.9441 0.9545 0.9633 0.9706 0.9767 2.0 2.1 2.2 2.3 2.4 0.9772 0.9821 0.9861 0.9893 0.9918 0.9778 0.9826 0.9864 0.9896 0.9920 0.9783 0.9830 0.9868 0.9898 0.9922 0.9788 0.9834 0.9871 0.9901 0.9925 0.9793 0.9838 0.9875 0.9904 0.9927 0.9798 0.9842 0.9878 0.9906 0.9929 0.9803 0.9846 0.9881 0.9909 0.9931 0.9808 0.9850 0.9884 0.9911 0.9932 0.9812 0.9854 0.9887 0.9913 0.9934 0.9817 0.9857 0.9890 0.9916 0.9936 2.5 2.6 2.7 2.8 2.9 0.9938 0.9953 0.9965 0.9974 0.9981 0.9940 0.9955 0.9966 0.9975 0.9982 0.9941 0.9956 0.9967 0.9976 0.9982 0.9943 0.9957 0.9968 0.9977 0.9983 0.9945 0.9959 0.9969 0.9977 0.9984 0.9946 0.9960 0.9970 0.9978 0.9984 0.9948 0.9961 0.9971 0.9979 0.9985 0.9949 0.9962 0.9972 0.9979 0.9985 0.9951 0.9969 0.9973 0.9980 0.9986 0.9952 0.9964 0.9974 0.9981 0.9986 3.0 3.1 3.2 3.3 3.4 0.9987 0.9990 0.9993 0.9995 0.9997 09987 0.9991 0.9993 0.9995 0.9977 0.9987 0.9991 0.9994 0.9995 0.9997 0.9988 0.9991 0.9994 0.9996 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9990 0.9993 0.9995 0.9996 0.9997 0.9990 0.9993 0.9995 0.9997 0.9998 rd Appendix A is taken from Table A.4, Introduction to Statistics R.E. Walpoke 3 Ed. McMillian Publishing company Inc. Select significance level for the normal distribution Two-tailed α One-tailed α Z 0.20 0.10 0.05 0.02 0.01 0.002 0.001 0.0001 0.00001 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 0.00005 0.000005 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.891 4.417 Taken from Nonparametric Statistics for the Behavioral Sciences, S. Siegel and N. Cstellan, McGrew – Hill Book Company. 187 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Appendix Table AII Critical z Values for #c Multiple Comparisons* #c 1 2 3 4 5 6 7 8 9 10 11 12 15 21 28 • Two-tailed One-tailed 0.30 0.15 0.25 0.125 α 0.20 0.10 0.15 0.075 0.10 0.05 0.05 0.025 1.036 1.440 1.645 1.780 1.881 1.960 2.026 2.080 2.128 2.170 2.208 2.241 2.326 2.450 2.552 1.150 1.534 1.732 1.863 1.960 2.037 2.100 2.154 2.200 2.241 2.278 2.301 2.394 2.515 2.615 1.282 1.645 1.834 1.960 2.054 2.128 2.189 2.241 2.287 2.326 2.362 2.394 2.475 2.593 2.690 1.440 1.780 1.960 2.080 2.170 2.241 2.300 2.350 2.394 2.432 2.467 2.498 2.576 2.690 2.785 1.645 1.960 2.128 2.241 2.326 2.394 2.450 2.498 2.539 2.576 2.608 2.638 2.713 2.823 2.913 1.960 2.241 2.394 2.498 2.576 2.638 2.690 2.734 2.773 2.807 2.838 2.886 2.935 3.038 3.125 #c is the number of comparisons Taken from Nonparametric Statistics for the Behavioral Sciences, S. Siegel and N. Cstellan, McGrew – Hill Book Company. 188 Table A–B Wilcozon’s Paired Signed – Ranks Test Critical Values n 0.05 0.10 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 40 50 1 2 4 6 8 11 14 17 21 26 30 36 41 47 54 60 101 152 287 466 One-Tailed 0.025 0.01 Two-Tailed 0.05 0.02 1 2 4 6 8 11 14 17 21 25 30 35 40 46 52 90 137 264 434 0.005 0.01 0 2 3 5 7 10 13 16 20 24 28 33 38 43 77 120 138 398 0 2 3 5 7 10 13 16 19 23 28 32 37 68 109 221 373 nd Taken from CRC Handbook of Tables for Probability and Statistics, 2 Boca Raton, Florida ed., 1968, CRC Press, 189 UNDERSTANDING STATISTICS O.S. Corpuz 2012 Table A – C Wilcoxon’s Two-Sample Rank Test (The Mann-Whitney Test) (These values or smaller cause rejection, Two-tailed Test. Take n1 ≤ n2) 0.05 Level of Significance n2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 190 2 3 3 3 4 4 4 4 4 4 5 5 5 5 6 6 6 6 6 7 7 7 3 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 4 5 6 7 8 10 11 12 13 14 15 15 16 17 18 19 20 21 21 22 23 24 25 26 27 28 28 29 17 18 20 21 22 23 24 26 27 28 29 31 32 33 34 25 27 28 29 40 42 26 27 29 31 32 34 35 37 38 40 42 43 45 46 48 50 51 53 55 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 49 51 53 55 58 60 63 65 67 70 72 74 77 79 82 n1 9 63 65 68 71 73 76 79 82 84 87 90 93 95 10 11 12 13 14 15 78 81 85 88 91 94 97 100 103 107 110 96 99 103 106 110 114 117 121 124 115 119 123 127 131 135 139 137 141 160 145 164 185 150 169 154 Table A – C Continued Wilcoxon’s Two-Sample Rank Test (The Mann-Whitney Test) (These values or smaller cause rejection, Two-tailed Test. Take n1 ≤ n2) n2 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2 3 3 3 3 3 3 3 3 4 4 3 6 6 6 7 7 7 8 8 8 8 9 9 9 10 10 10 11 11 11 4 10 10 11 11 12 12 13 14 14 15 15 16 16 17 18 18 19 19 20 20 21 5 15 16 17 17 18 19 20 21 22 22 23 24 25 26 27 28 29 29 30 31 32 0.01 Level of Significance n1 6 7 8 9 10 11 23 24 25 26 27 28 30 31 32 33 34 36 37 38 39 40 42 43 44 32 34 35 37 38 40 41 43 44 46 47 49 50 52 53 55 57 43 45 47 49 51 53 54 56 58 60 62 64 66 68 70 56 58 61 63 65 67 70 72 74 76 78 81 83 71 74 76 79 81 84 86 89 92 94 97 87 90 93 96 99 102 105 108 111 12 13 14 15 106 109 112 115 119 122 125 125 129 147 133 151 171 137 155 140 191 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Appendix Table B Critical Values of the t Distribution 0.10 0.05 α 0.025 0.01 0.005 1 2 3 4 5 3.078 1.886 1.638 1.533 1.476 6.314 2.290 2.353 2.132 2.015 12.706 4.303 3.182 2.776 2.571 31.821 6.965 4.541 3.747 3.365 63.657 9.925 5.841 4.604 4.032 6 7 8 9 10 1.440 1.415 1.397 1.383 1.372 1.943 1.895 1.860 1.833 1.812 2.447 2.365 3.306 2.262 2.228 3.143 2.998 2.896 2.821 2.764 3.707 3.499 3.355 3.250 3.169 11 12 13 14 15 1.363 1.356 1.350 1.345 1.341 1.796 1.782 1.771 1.761 1.753 2.201 2.179 2.160 2.145 2.131 2.718 2.681 2.650 2.624 2.602 3.106 3.055 3. 012 2.977 2.947 16 17 18 19 20 1.337 1.333 1.330 1.328 1.325 1.746 1.740 1.734 1.729 1.725 2.120 2.110 2.101 2.093 2.086 2.583 2.567 2.552 2.539 2.528 2.921 2.898 2.878 2.861 2.845 21 22 23 24 25 1.3323 1.321 1.319 1.318 1.316 1.721 1.717 1.714 1.711 1.708 2.080 2.074 2.069 2.064 2.060 2.518 2.508 2.500 2.492 2.485 2.431 2.819 2.07 2.797 2.787 26 27 28 29 Inf. 1.315 1.314 1.313 1.312 1.282 1.706 1.703 1.701 1.699 1.645 2.056 2.052 2.048 2.045 1.960 2.479 2.473 2.467 2.462 2.326 2.779 2.771 2.763 2.756 2.576 df Ref: Basilio et.al. 2003. Fundamental Statistics. Trintas Publishing, Inc.Philippines 192 References Aczel, A. 1989. Complete Business Statistics. Richard D. Irwin, Inc. Anderson, R.L and Bancrot, T.A 1952. Statistical Theory in Research. McGraw-Hill Book Co., Inc. Calmorin, L.P and Calmorin, M.A. 1999. Methods of Research and Thesis Writing. 1st Ed. Rex Book Store, Inc. Manila Philippines Daleon, S., Sanches, L. and marquez, T. 1996. Fundamentals of Statistics. National Book Store, Inc. Draper, N. and Smith, H. 1966. Applied Regression Analysis. John Wiley and Sons, Inc. Gomez, K.A and Gomez, A.A 1984. Statistical Procedures for Agricultural Research. 2nd Ed. An International Rice Research Institute Book. John Wiley and Sons Inc. Iman, L.R. and W.J. Canover, 1983. A Modern Approach to Statistics. John Wiley and Sons Inc. Johnston, J. 1972. Econometrics methods, 2nd Ed. McGraw-Hill Book Co. Inc. Mendenhall, W. and Sincich. 1989. A Second Course in Business Statistics Regression Analysis. Dellen Publishing Company Ostle, B. 1966. Statistics in Research. Iowa State University Press, Ames, Iowa. Pacificador, A. Jr. 1997. Outreach Seminar on “Statistics for Researchers”. Urios College, Butuan City. Parel, P. C. 1996. Introduction to Statistical Methods with Application. Macaraig Publishing Co. Inc., Manila Philippines. Searle, S.R. 1971. Linear Models. New York. Wiley and Sons Inc. Siegel, S. and N.J. Castellan. 1988. Non-Parametric Statistics for Biological Science. McGraw-Hill Book Co. Snedecor, G.W and Cohcran, W.G. 1957. Statistical Methods. 5th Ed. Iowa State University Press. Snedecor, G.W and Cohcran, W.G. 1956. Statistical Methods. Iowa State College Press, Ames Iowa. Spiegel, M.1978. Statistics. McGraw-Hill Book Inc. 193 O.S. Corpuz 2012 UNDERSTANDING STATISTICS Steel, R.G.D and J.H Torre 1960. Principles and Procedures of Statistics. McGraw-Hill Book Co., Inc. New York. Tagaro, C.A and Tagaro A.T. Statistics Made Easy. 11th Edition. University of Southern Mindanao, Kabacan, Cotabato Philippines. Walpole, R.E 1982. Introduction to Statistics. 3rd Ed. McMillan Publishing Co. Inc. 194 Buy your books fast and straightforward online - at one of world’s fastest growing online book stores! Environmentally sound due to Print-on-Demand technologies. Buy your books online at www.get-morebooks.com Kaufen Sie Ihre Bücher schnell und unkompliziert online – auf einer der am schnellsten wachsenden Buchhandelsplattformen weltweit! Dank Print-On-Demand umwelt- und ressourcenschonend produziert. Bücher schneller online kaufen www.morebooks.de VDM Verlagsservicegesellschaft mbH Heinrich-Böcking-Str. 6-8 D - 66121 Saarbrücken Telefon: +49 681 3720 174 Telefax: +49 681 3720 1749
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