EXAM #4 Review FALL 2010Name: __________________________ 1. A researcher selects a random sample. A 90% confidence interval for a population mean . A) is an interval with margin of error ± 90%. B) has the property that if we repeatedly selected our random sample in exactly the same way, each time constructing a different 90% confidence interval for , then in the long run 90% of those intervals would contain . C) (a) and (b) are both true. D) is an interval that has width .90. 2. The critical value, z*, used for constructing a 96% confidence interval for a population mean is A) 1.645. B) 2.054. C) 2.326 D) 2.576. 3. A 99% confidence interval for the average weight of a population is computed from a random sample and found to be 6 2, or 4 to 8. We may conclude that A) there is a 99% probability that is between 4 and 8. B) there is a 99% probability that the true mean is 6, and there is a 99% chance that the true margin of error is 2. C) if we took many, many additional random samples, and from each computed a 99% confidence interval for , approximately 99% of these intervals would contain . D) all of the above. Use the following to answer questions 4-5: The population of the scores of all high school seniors that took the SAT-M test (mathematics component of the SAT test) last year followed a Normal distribution, with mean and standard deviation = 100. You read a report that says, “On the basis of a simple random sample of 500 high school seniors that took the SAT-M test this year, a confidence interval for is 512.00 ± 11.52.” 4. The confidence level for this confidence interval is A) 90%. B) 95%. C) 99%. D) unable to be determined 5. A 95% confidence interval for would be A) 503.24 to 520.77 B) 500.48 to 523.52. C) 316 to 708. D) none of the above. The scores of a certain population on the Wechsler Intelligence Scale for Children (WISC) are thought to be Normally distributed with mean and standard deviation = 10. A simple random sample of 25 children from this population is taken and each is given the WISC. The mean of the 25 scores is x = 104.32 6. Based on these data, a 95% confidence interval for is A) 104.32 ± 0.78 B) 104.32 ± 3.29 C) 104.32 ± 3.92 D) 104.32 ± 19.60 05. The researcher gives the SSHA to 100 incoming freshmen and observes a mean score of 119. C) the probability the null hypothesis is true. attitudes. C) the data are statistically significant at level = .” C) a statement about the sample mean.05. Scores range from 0 to 200 and follow (approximately) a Normal distribution. 11. B) 0.4012. D) none of the above Based on the P-value computed in problem 21 above. that the test statistic will take a value at least as extreme as that actually observed. In formulating hypotheses for a statistical test of significance. assuming the null hypothesis is true. D) the data are statistically significant at level = . A researcher suspects that incoming freshmen have a mean .01. 10. B) a statement of “no effect” or “no difference.7. but not at level = . the alternative hypothesis is often A) a statement about the population the researcher suspects is true and for which he/she is trying to find evidence.01. D) 0.05 and not statistically significant at level = . 9.05. The P-value for this test is A) 0. B) a statement of “no effect” or “no difference. but not at level = . D) 0. B) the probability. To verify her suspicion. A) the data are not statistically significant at level = .0164 C) 0.0082. 8. which is different from 115. B) the data are statistically significant at level = . Use the following to answer questions 10-12: The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation. Assume that the scores of all incoming freshmen are approximately Normal with the same standard deviation as the scores of all college students.05. that the test statistic will take a value at least as extreme as that actually observed. with mean of 115 and standard deviation 25.” C) a statement about the sample mean. In formulating hypotheses for a statistical test of significance. the P-value is A) the probability.01. D) the probability the null hypothesis is false. In a test of hypotheses.05 and at level = . she tests the hypotheses H0: = 115 Ha: 115. because they are often excited yet anxious about entering college. . the null hypothesis is often A) a statement about the population the researcher suspects is true and for which he/she is trying to find evidence. and study habits of college students. assuming the null hypothesis is false.01. C) The researcher has proven that the mean SSHA score for freshmen is. After all. 69 62 66 Based on these data. C) 21 ± 0. He then decides that these data are statistically significant. D) the researcher should conclude that terrible errors were made in collecting data. D) 21 ± 0. the margin of error in the 95% confidence interval computed in question 3 above would A) become larger. but the effect must be very large. D) The researcher has proven that the mean SSHA score for freshmen is not 115. The number of days until maturity for each plant is given below 63 16. C) 65. in fact.0001.09.07. is A) 65. C) become smaller. A 95% confidence interval for the mean waiting time for customers is about A) 21 ± 0. which of the following is true with respect to a conclusion reached? A) The researcher has strong enough evidence to conclude that the average freshmen SSHA score differs from that of the average SSHA score for all college students. A simple random sample of four plants of this variety is selected. D) become larger or smaller. B) 65.84 minutes. Since the P-value is so small. B) not change.35. in days. 13. If the researcher deems the data to be statistically significant. a Pvalue such as P = 0.0001 means that if the null hypothesis is true.00 ± 2. the researcher observes data and computes a P-value of P = 0. a sample like this one was virtually impossible to occur by chance. D) 65.00 ± 4. a 99% confidence interval for . Suppose in a test of hypotheses. C) the researcher may now conclude that not only is there an effect. B) the researcher has strong evidence that the claim as stated in the alternative hypothesis is true. It is company policy that the customer wait time should be 20 minutes (or less).05 minutes.54 minutes. The manager of a particular store selects a random sample of 150 customer wait times and observes a mean wait time of 21 minutes. Use the following to answer questions 14-15: The amount of time customers at a “Quick-Change” motor oil store spend waiting for their cars to be serviced has the Normal distribution with mean and standard deviation = 4 minutes. B) 21 ± 7. A) the researcher has proven that the null hypothesis is false. B) The researcher does not have strong enough evidence to conclude that the average freshmen SSHA score differs from that of the average SSHA score for all college students.12. but there is no indication of how big the effect is.00 ± 1. .55.00 ± 3.4. depending on the sample taken. 115. If the manager had selected more customers for her sample. Use the following to answer question 16: The time (in number of days) until maturity of a certain variety of tomato plant is Normally distributed with mean and standard deviation = 2. 14.64 minutes 15. 4 ± 5. 22.1 C) 1. The average for their age group is 200. D) both (b) and (c) are correct.4 ± 2. We are interested in a 99% confidence interval for the population mean birth weight. . The appropriate hypotheses are A) H0: = 200. Scores on the SAT Mathematics test (SAT-M) are believed to be Normally distributed. The one sample t statistic from a sample of n = 25 observations for the two-sided test of H0: = 9. Assume that in the population of all babies born in this hospital.01.84 ounces. 620.10. In the sample. D) H0: = 200. Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The number of observations required is closest to A) 25 B) 30 C) 609 D) 865 Use the following to answer questions 18-20: An SRS of 20 recent birth records at the local hospital were selected. B) 0. with mean .4 ± 3.00 C) 550. 20.00 ± 105. 18.88 B)550. The appropriate degrees of freedom for this test are A) 21 B) 49 C) 51 D) 200 B) H0: = 200. The interval is to have a margin of error of 2 minutes. the birth weights follow a Normal distribution.7 D) 0.63 ounces B) 121.46 ounces B) 4. D) 0. Based on this information A) P-value > 0.00 ± 128. C) 121.025 < P-value < 0.00 ± 173.89 ounces.025.80 ounces.07 ounce. C) we would reject the null hypothesis at = 0.1 B) 6.4 19. Ha: 192 24. with some mean .00 ± 142. with a std deviation of 21.17.4 ± 4.29 ounces D) 121.80 ounces C) 1. The scores of a random sample of three students who recently took the exam are 550.84. Ha: < 200 C) H0: = 192. Ha: 200. The average after the test was 192. Ha: > 200. A 95% confidence interval for based on these data is A) 550. and the amount spent has a Normal distribution with a standard deviation = 30 minutes. The standard error of the mean is A) 27. and 480. the average birth weight was 121. 50 Patients were treated with a new drug thought to reduce cholesterol levels.58 D) 550. Ha: > 9 has the value t = 1. The margin of error associated with the confidence interval is A) 21. A 90% confidence interval for the population mean birth weight based on these data is A) 121.4 ounces and the standard deviation was 7. Is there Evidence that the drug reduced cholesterol levels? 23. 21.5 ounces.05. at least 50. 27.10.01. The P-value would be A) larger B) smaller.69 D) –2. 28. the value of the one-sample t statistic is A) 3.0 ± 33.54 C) 25.08 D) 25. say.25. with the scores of other students using a traditional method of instruction. You are thinking of employing a t-procedure to test hypotheses about the mean of a population using a significance level of 0. D) below 0.0 ± 47. 26. D) unchanged.05.24 29. C) You may use the t-procedure. but you should probably claim the significance level is only 0. C) unchanged. Which of the following statements is correct? A) You should not use the t-procedure.42 B) 2.29 B) 25. with mean . . The P-value for the one-sample t test is A) larger than 0. B) A teacher compares the scores of students using a computer-based method of instruction. because t-procedures are robust to non-Normality for confidence intervals but not for tests of hypotheses.01. because the population does not have a normal distribution. because the difference between x and the hypothesized value = 200 is unchanged.0 ± 43. because the variability measured by the standard deviation stays the same. provided your sample size is large. You suspect the distribution of the population is not Normal and may be moderately skewed. A 90% confidence interval for is A) 25. C) A teacher compares the scores of students in her class on a standardized test with the national average score. D) You may not use the t-procedure.05 and 0. D) A teacher calculates the average of scores of students on a pair of tests and wishes to see if this average is larger than 80%.10 and 0. 30. Which of the following is an example of a matched pairs design? A) A teacher compares the pretest and posttest scores of students. Based on the data. Suppose the mean and standard deviation obtained were based on a sample of 100 patients rather than 50.10 B) between 0. B) You may use the t-procedure. C) between 0.89 C) –2.89.05. Do students tend to improve their math SAT scores the second time they take the test? A random sample of four students who took the test twice received the following scores: Student First score Second score 1 450 440 2 520 600 3 720 720 4 600 630 Assume that the change in math SAT score (second score – first score) for the population of all students taking the test twice is Normally distributed.0 ± 64. 70.95 points. the computed two-sample t statistic is A) 1. Student 1 2 3 4 First SAT score 920 830 960 910 Second SAT score 1010 800 1000 980 To analyze this data.53 to 17. B) between 0. C) the two-sample t test. A 95% confidence interval for the difference in mean reading skill score between children that learned with activities and children that learned without activities is (use Option 2.02. 33. D) Any of the above are valid.45 to 13. C) between 0. It just needs to be some type of t test since is unknown.05.01 and 0.97 to 18. D) 4. we recorded their first SAT score (before the class) and their second SAT score (after the coaching class). one class of second grade children learns with the activities. After some period of time. C) 2. Another class of second grade children serves as the control and learned without the activities.83 to 11.02. For each student. the conservative method for degrees of freedom) A) 0. C) 8. D) 6.025.39 points. In a study of the effectiveness of these activities. The P-value for the test described in question 5 (again.05. . Use the following to answer questions 32-34: A psychologist has developed a set of activities which she hopes will help children develop better reading skills.47 points. Do SAT coaching classes work? Do they help students to improve their test scores? Four students were selected randomly from all of the students that completed an SAT coaching class. B) the matched pairs t test. B) 2.09 points. using Option 2 for degrees of freedom) is A) larger than 0.48 11.025 and 0. D) between 0. B) 2.52 17.84. A summary of these data is: x n s Activities class: 21 51.01 No Activities class: 23 41.15 32. we should use A) the one-sample t test.02 and 0. Based on these data.31. the reading skills of all of these children were assessed. 34. The psychologist suspects that children who learn with activities have higher mean reading skill test scores than children that don't learn with activities.31. 410 supported the property tax levy. 35. Of those surveyed.615 to 0.5105. If we want to estimate p.4463 to 0. B) 0. Of the 850 residents surveyed.4543 to 0.624 to 0. C) 0.4489 to 0. A) 2401 B) 3484 C) 4148 D) 8256 Use the following to answer questions 39-40: A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot.680. D) 0. A 90% confidence interval for p is A) 0. The standard error SE of pˆ . 39. B) 0.357. C) 0.5184.4790.667. .Use the following to answer questions 35-38: Each person in a random sample of 1100 “likely voters” (as defined by a professional polling organization) was questioned about his or her political views.5161.672. 36. D) 0. They randomly selected 850 residents in the community and contacted them by telephone.606 to 0. C) 0.0241. B) 0. how many likely voters need to be surveyed? Assume that you have no idea of the value of p. is A) 0. 38. the sample proportion viewing the economy's state as most urgent. The sample proportion pˆ that felt the economy's state was the most urgent national concern is A) 0.5185. D) 0.663. B) 0. 37. C) 0.620 to 0.500.473. the population proportion of likely voters that believe the economy's state is the most urgent national concern with 99% confidence and a margin of error no greater than 2%.0144. A 99% confidence interval for the proportion p of all likely voters that feel the economy's state is the most urgent national concern is given by (use the plus four confidence interval procedure) A) 0.5159.643. 708 felt that “the economy's state” was the most urgent national concern. Let p represent the proportion of residents in the community that support the property tax levy.4487 to 0. D) 0. How large a sample n would you need to estimate p with margin of error 0. 43. to If X and Y are categorical variables.30. A study in France included a sample of 459 men who were selected at random from men who had visited a health center for a routine checkup. the best way to determine if there is a relation between them is A) B) C) D) calculate the correlation between X and Y. 42. make a two-way table of the X and Y values. C) 4.06. C) 0. B) 0. . D) 6. there is a strong relationship between smoking and education. D) 0.40. The proportion of men with a primary school education that are nonsmokers is A) 0. C) 0. all of the above. This is an r c table. with well-educated people less likely to smoke. B) 0.22. The proportion of former smokers with a university education is A) 0. Use the following to answer questions 42-46: In the United States.38. D) 0.17.12.58.12. Education is classified into three categories corresponding to the highest level of education and smoking status is classified into four categories. A) n = 49 B) n = 750 C) n = 2400 D) n = 4802 41. draw a scatterplot of the X and Y values.04 with 95% confidence? Assume that you don't know anything about the value of p. The number c has value A) 2. 44. B) 3. 32.34. and one toddler. Under the null hypothesis. D) 0.37. 47. B) 8. 46.48. The degrees of freedom for the chi-square test for this two-way table are A) 2. the expected number of times a male adult would be carrying a male toddler according to the table is A) 9. C) 7. 48. B) 0. C) 0. C) 0. .32. C) 6.57.64. D) 5. in which the toddler was being carried. D) 62. the expected number of nonsmokers with a primary school education is A) 42.36.93. D) 0. D) 12.45.64. The data are from 39 groups of three —one adult female. Suppose we wish to test the null hypothesis that the proportion of male adults and the proportion of female adults carrying a toddler is the same.57. B) 6. The proportion of times a male adult was carrying the toddler is A) 0. C) 59. 49. The proportion of all male toddlers that were carried by a male adult is A) 0. Use the following to answer questions 47-50: Do women and men treat female and male children differently? An observational study was conducted near the primate exhibit at the Columbus Zoo on weekends in 1997. Under the null hypothesis. B) 50. regardless of the sex of the toddler. B) 0. Suppose we wish to test the null hypothesis that there is no association between education level and smoking status. one adult male. Recorded below is which adult (male or female) was carrying the toddler by the sex of the toddler.36. in milligrams of phosphate per deciliter of blood. 5.27. C) 3. B) 0.41 milligrams of phosphate per deciliter C) 0.6 milligrams of phosphate per deciliter C) 5.7 4. What is the first quartile for these data? A) 4.94.50. D) 39.32 milligrams of phosphate per deciliter .65 milligrams of phosphate per deciliter C) 5.47 milligrams of phosphate per deciliter D) 4.4 5.4 milligrams of phosphate per deciliter 53.460.75 milligrams of phosphate per deciliter D) 5. What is the mean level of phosphate for the nine clinic visits? A) 6. What is the median level of phosphate for the nine clinic visits? A) 5.6 5. The numerical value of the chi-square statistic for testing independence of sex of the adult and sex of the toddler is A) 0.6 milligrams of phosphate per deciliter 54.4 milligrams of phosphate per deciliter B) 4.15 milligrams of phosphate per deciliter B) 5. made on nine consecutive visits to a clinic.6 milligrams of phosphate per deciliter C) 4.7 milligrams of phosphate per deciliter D) 6.498.9 6.68 milligrams of phosphate per deciliter D) 1.82 milligrams of phosphate per deciliter B) 1.6 milligrams of phosphate per deciliter B) 5.2 51. Below are measurements of the level of phosphate in the blood of a patient.2 4.8 milligrams of phosphate per deciliter 52.9 5. Use the following to answer questions 51-54: The level of various substances in the blood is known to influence our health.7 6.6 4. What is the standard deviation for these data? A) 0. 31. 14. 50. 38. 33. 37. 44. 52. 8. 30. 41. 43. 26. 42. 48. 32. 54. 22. 27. 28. 39.Answer Key 1. 11. 40. 25. 10. B B C C A C B A A B C A B D C C C C B D B A A B C D B B B A B A C D D A D C B D C B A C C B D C A A C A C A . 49. 24. 53. 7. 23. 5. 51. 13. 3. 18. 21. 35. 47. 36. 17. 19. 29. 9. 4. 46. 15. 34. 12. 45. 20. 16. 6. 2.
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