TOPIC: [2012]Session 3-Reading 11: HypothesisTesting-LOS i 1. Which of the following statements about the variance of a normally distributed population is least accurate? A) The Chi-squared distribution is a symmetric distribution. B) The test of whether the population variance equals σ02 requires the use of a Chi-squared distributed test statistic, [(n ? 1)s2] / σ02. C) A test of whether the variance of a normally distributed population is equal to some value σ02, the hypotheses are: H0: σ2 = σ02, versus Ha: σ2 ≠ σ02. -------------------------------------------------------------------------------A The Chi-squared distribution is not symmetrical, which means that the critical values will not be numerically equidistant from the center of the distribution, though the probability on either side of the critical values will be equal (that is, if there is a 5% level of significance and a two-sided test, 2.5% will lie outside each of the two critical values). 2. A test of the population variance is equal to a hypothesized value requires the use of a test statistic that is: A) Chi-squared distributed. B) F-distributed. C) t-distributed. -------------------------------------------------------------------------------A In tests of whether the variance of a population equals a particular value, the chi-squared test statistic is appropriate. 3. A munitions manufacturer claims that the standard deviation of the powder packed in its shotgun shells is 0.1% of the stated nominal amount of powder. A sport clay association has reviewed a sample of 51 shotgun shells and found a standard deviation of 0.12%. What is the Chi-squared value, and what are the critical values at a 95% confidence level, respectively? 0144) / 0. The most appropriate conclusion regarding whether the variance of Stock A is different from the variance of Stock B is that the: A) variance of Stock B is significantly greater than the variance of Stock A.357 and 71. In order to test if Stock A is more volatile than Stock B.764 and 79. 34. Using a 5% level of significance. X2 = (n – 1)s2 / σ02 = 50(0. the critical values at the 95% confidence level are 32. -------------------------------------------------------------------------------C A test of the equality of variances requires an F-statistic. B) variances are not equal. 5. The calculated F-statistic is 0.A) 72.064. prices of both stocks are observed to construct the sample variance of the two stocks. .61.61. Since the Chi-squared value is outside this range. B) 70. The variance of 90 daily stock returns for Stock B is 0.064 is less than the critical F value of 1.0083/0. C) variances are equal.420.0078. The variance of 100 daily stock returns for Stock A is 0. Since the calculated F value of 1.505. C) 72. B) F test.420.357 and 71.764 and 67.01 = 72. 4. C) t test.0083. we can reject the hypothesis that the standard deviations are the same. The appropriate test statistics to carry out the test is the: A) Chi-square test. we cannot reject the null hypothesis that the variances of the 2 stocks are equal.490. 34.0078 = 1. the critical value for this test is 1. With 50 df. 32. -------------------------------------------------------------------------------C To compare standard deviations we use a Chi-square statistic. 6. 8. can we reject the null hypothesis at a 95% confidence level using an F-statistic and why? Null is: A) rejected. Ness’s Null Hypothesis is σ12 = σ22. C) The test of the mean of the differences is used when performing a paired comparison.18 From an F table. -------------------------------------------------------------------------------C F = s12 / s22 = $2. B) Nonparametric tests rely on population parameters. The standard deviation of earnings for copper firms was $2.69. .71. the critical value with numerator df = 24 and denominator df = 30 is 1. Which of the following statements about parametric and nonparametric tests is least accurate? A) The test of the difference in means is used when you are comparing means from two independent samples. The F-value exceeds the critical value by 0. -------------------------------------------------------------------------------B Nonparametric tests are not concerned with parameters. Ness believes that the earnings of copper extracting firms are more volatile than those of bauxite extraction firms. they make minimal assumptions about the population from which a sample comes. whereas nonparametric tests are not as strict regarding distributional properties.89. Which of the following statements about parametric and nonparametric tests is least accurate? A) Parametric tests are most appropriate when a population is heavily skewed.922 / $2. 7.-------------------------------------------------------------------------------B The F test is used to test the differences of variance between two samples.71. Abby Ness is an analyst for a firm that specializes in evaluating firms involved in mineral extraction. In order to test this.92. The F-value exceeds the critical value by 0.849. It is important to distinguish between the test of the difference in the means and the test of the mean of the differences. C) not rejected. Based on the samples. while the standard deviation of earnings for bauxite firms was $2. Ness examines the volatility of returns for 31 copper firms and 25 bauxite firms. B) rejected.692 = 1. Also. it is important to understand that parametric tests rely on distributional assumptions. The critical value exceeds the F-value by 0. e. What test should she use if she is willing to assume that the returns are normally distributed? A) A difference in means test only if the variances of monthly returns are equal for the two stocks. he wants to test whether the decline affected trading volume. B) H0: μd = μd0 versus Ha: μd ≠ μd0. What is the set of hypotheses that Sutton is testing? A) H0: μd ≠ μd0 versus Ha: μd = μd0. a nonparametric test may be most appropriate. while parametric tests rely on assumptions regarding the distribution of the population.. An analyst wants to determine whether the monthly returns on two stocks over the last year were the same or not. He selected a sample of 500 companies and collected data on the total annual volume for one year prior to the decline and for one year following the decline. Note that the test is two-tailed. -------------------------------------------------------------------------------A For a distribution that is non-normally distributed. -------------------------------------------------------------------------------- . 9. C) H0: μd = μd0 versus Ha: μd > μd0. C) Nonparametric tests are often used in conjunction with parametric tests. -------------------------------------------------------------------------------B This is a paired comparison because the sample cases are not independent (i. there is a before and an after for each stock). A nonparametric test tends to make minimal assumptions about the population. B) A paired comparisons test because the samples are not independent. 10. Both kinds of tests are often used in conjunction with one another. t-test. Joe Sutton is evaluating the effects of the 1987 market decline on the volume of trading. C) A difference in means test with pooled variances from the two samples.B) Nonparametric tests have fewer assumptions than parametric tests. Specifically. the critical value = 2. Her null hypothesis is that the betas are the same before and after September 11. She sampled 31 firms comparing their betas for the one-year period before and after this date. B) With an unknown population variance and a small sample size. C) rejected. with a sample standard deviation of 0. -------------------------------------------------------------------------------A The t-statistic for paired differences: t = (d – ud 0) / sd and sd = sd / √n t = 9. Based on this sample. The t-value exceeds the critical value by 7.291.11. An analyst for the entertainment industry theorizes that betas for most firms in the industry are higher after September 11. The critical value exceeds the t-value by 7. 2001.67. Brandee Shoffield is the public relations manager for Night Train Express.150 and a standard deviation of 450. she found that the mean differences in betas were 0. C) Shoffield should use a two-tailed Z-test. A sample of the attendance at 15 home games results in a mean of 3. The t-value exceeds the critical value by 5. Which of the following statements is most accurate? A) The calculated test statistic is 1. a local sports team. B) not rejected.62 from a table with 30 df. Portfolio theory teaches us that returns on two stocks over the same time period are unlikely to be independent since both have some systematic risk. 11.58.19.042 12.B A paired comparisons test must be used. Based on the results of her sample. -------------------------------------------------------------------------------A We will use the process of Hypothesis testing to determine whether Shoffield should reject Ho: . The difference in means test requires that the samples be independent.58. Attendance is approximately normally distributed. no statistic is available to test Shoffield's hypothesis. can we reject the null hypothesis at a 5% significance level and why? Null is: A) rejected. Shoffield is trying to sell advertising spots and wants to know if she can say with 90% confidence that average home game attendance is greater than 3.000. 345.10 significance level. The other statements are false. the confidence level is 90%.4.345. The sample value of the computed z-statistic = 3. the appropriate test is a t-test. Shoffield cannot state with 90% certainty that the home game attendance exceeds 3. The critical value for this question will be the t-statistic that corresponds to an α of 0.4 > 1.000 Ha: μ > 3. and 14 (n-1) degrees of freedom. . 13. we will reject the null hypothesis if the calculated test statistic is greater than 1. -------------------------------------------------------------------------------B Ho:μ ≤ 100.000) / (450 / √ 15) = 1. an unknown variance and a small sample size – the t-statistic.150 – 3. B) reject the null hypotheses and conclude that the population mean is greater than 100. or 0. a sample of 30 employees is taken. we determine that the appropriate critical value = 1. There is a test statistic for an normally distributed population. Step 5: Calculate sample (test) statistic The test statistic = t = (3. Using the t-table .Step 1: State the Hypothesis Ho: μ ≤ 3. which translates to a 0.10. As shown above. In order to test if the mean IQ of employees in an organization is greater than 100. The appropriate decision at a 5% significance level is to: A) reject the null hypothesis and conclude that the population mean is not equal to 100. Step 3: Specify the Level of Significance Here. Thus.90. Step 4: State the Decision Rule This is a one-tailed test. Ha: μ > 100.000 Step 2: Select Appropriate Test Statistic Here.000. we have a normally distributed population with an unknown variance (we are given only the sample standard deviation) and a small sample size (less than 30. Reject the null since z = 3. not a Z-test. There is no test for a nonnormal population with unknown variance and small sample size. C) reject the null hypothesis and conclude that the population mean is equal to 100.) Thus. we will use the t-statistic.65 (critical value).291 Step 6: Make a decision Fail to reject the null hypothesis because the calculated statistic is less than the critical value. Here. If he can reduce turnover to less than 25%.034. . Step 6: Make a decision.8%. The population standard deviation of turnover rates is 1. This result most likely suggests that the: A) null hypothesis can be rejected at the 5% significance level. we will use the z-statistic. -------------------------------------------------------------------------------C Using the process of Hypothesis testing: Step 1: State the Hypothesis.5%. he will receive a 50% bonus (using a significance level of 10%).5 / √ 100) = ?5.10. For 25% bonus level . C) smallest significance level at which the null hypothesis can be rejected is 6. Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is less than the critical value. An analyst is testing the hypothesis that the mean excess return from a trading strategy is less than or equal to zero. Step 4: State the Decision Rule.) Thus. For 50% bonus level . 15. C) For the 50% bonus level. she hired Graham Brickley as Manager of Employee Retention.65 and Huffman should give Brickley a 50% bonus. the test statistic is -5. The critical value for this question will be the z-statistic that corresponds to an α of 0. A recent sample of 100 branch offices resulted in an average turnover rate of 24. Part of the compensation package was the chance to earn one of the following two bonuses: if Brickley can reduce turnover to less than 30%. The analyst reports that this hypothesis test produces a p-value of 0. Last year. and is thus the same for both the 50% and 25% bonus levels.67. Z (for 50% bonus) = (24. Z (for 25% bonus) = (24. he will receive a 25% bonus.2 – 25) / (1. Step 2: Select Appropriate Test Statistic.333.10.Ho: m ≥ 30% Ha: m < 30%. This is a one-tailed test.14. Which of the following statements is most accurate? A) For the 50% bonus level. Thus. The critical value of –1. α = 0.2%. or an area to the left of the mean of 40% (with 50% to the right of the mean). The population of turnover rates is normally distributed. B) Brickley should not receive either bonus. B) best estimate of the mean excess return produced by the strategy is 3. we determine that the appropriate critical value = -1.28. we will reject the null hypothesis if the calculated test statistic is less than -1. Step 5: Calculate sample (test) statistics.5 / √ 100) = ?38.33 and Huffman should give Brickley a 50% bonus.2 – 30) / (1. Step 3: Specify the Level of Significance.Ho: m ≥ 25% Ha: m < 25%.28 is based on the significance level. Using the z-table (normal table). the critical value is -1. Huffman should give Soberg a 50% bonus.28 (Remember that we highly recommend that you have the “common” z-statistics memorized!) Thus. we have a normally distributed population with a known variance (standard deviation is the square root of the variance) and a large sample size (greater than 30. The other statements are false. Maria Huffman is the Vice President of Human Resources for a large regional car rental company.4%. the relationship will continue into the future.-------------------------------------------------------------------------------A A p-value of 0. 17. B) A result may be statistically significant. but cannot be rejected at the 1% significance level. but may not be economically meaningful. in their presence we would not be able to assert either statistical or economic significance. 16. The others are all mitigating factors that can cause statistically significant results to be less than economically significant. 18.5% or higher. data errors are not a valid explanation. -------------------------------------------------------------------------------A While data errors would certainly come to bear on the analysis. it must also be economically meaningful.035 means the hypothesis can be rejected at a significance level of 3. the analyst needs to examine the reasons why the economic relationship exists to discern whether it is likely to be sustained in the future. In other words. Thus. B) Transactions costs. C) Adjustment for risk. Of the following explanations. the hypothesis can be rejected at the 10% or 5% significance level. statistical significance does not ensure economic significance. which is least likely to be a valid explanation for divergence between statistical significance and economic significance? A) Data errors. what is a 95% confidence interval for the return next year? . Given a mean of 10% and a standard deviation of 14%. However. -------------------------------------------------------------------------------B It is possible for an investigation to determine that something is both statistically and economically significant. Even if a result is both statistically significant and economically meaningful. Which of the following statements about statistical results is most accurate? A) If a result is statistically significant and economically meaningful. C) If a result is statistically significant. The alternative hypothesis is not Ha: M > 7 because in a two-tailed test the alternative is =. a positive sign would indicate that the mean is more than 7.44% to 37. because the calculated Z-statistic is less than the critical Zstatistic. Which of the following statements is least accurate?: A) The mean observation is significantly different from 7.A) -4. C) The alternative hypothesis would be Ha: mean > 7. The way the null hypothesis is written.44%. 19.96. and is in the rejection region. The calculated test statistic of -2 falls to the left of the critical Zstatistic of -1.7) / (1) = -2. . A test statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. C) the null hypothesis.44%. A goal of an “innocent until proven guilty” justice system is to place a higher priority on: A) avoiding type I errors. The calculated Z is -2.96. B) -17.00%. just that it is not 7. it makes no difference whether the mean is more or less than 7. The analyst wants to determine whether the calculated mean. while the critical value is -1. Here.96) = -17. is significantly different from 7 at the 5% level of significance. while < and > indicate one-tailed tests. B) avoiding type II errors.44% to 37. An analyst calculates that the mean of a sample of 200 observations is 5.14(1. -------------------------------------------------------------------------------C The way the question is worded. the null hypothesis is rejected and the conclusion is that the sample mean of 5 is significantly different than 7. What the negative sign shows is that the mean is less than 7. Thus. 20.00% to 38. the test statistic = (sample mean – hypothesized mean) / (standard error of the sample statistic) = (5 . B) The null hypothesis would be: H0: mean = 7. -------------------------------------------------------------------------------C 10% +/.00%.00% to 24. this is a two tailed test. C) -17. which has a standard error of the sample statistic of 1. B) type II error over a type I error. the null hypothesis is that the accused is innocent. C) type I error over a type II error. the bottler should: A) not reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea. The hypothesis can only be rejected by evidence proving guilt beyond a reasonable doubt. 22. Ha: μ ≠ 16. 21. favoring the avoidance of type I errors.-------------------------------------------------------------------------------A In an “innocent until proven guilty” justice system. then the statement “It is better that the guilty go free. Using a t-distributed test statistic of -1. In order to analyze the accuracy of the bottling process. 23.09 < 1. Do not reject the null since |t| = 1. B) not reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. C) reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. If the null hypothesis is innocence. The power of the test is: A) the probability of rejecting a true null hypothesis. . a random sample of 150 bottles is taken. -------------------------------------------------------------------------------B The statement shows a preference for accepting the null hypothesis when it is false (a type II error).96 (critical value). A bottler of iced tea wishes to ensure that an average of 16 ounces of tea is in each bottle.09 and a 5% level of significance. over rejecting it when it is true (a type I error). -------------------------------------------------------------------------------B Ho: μ = 16. than the innocent are punished” is an example of preferring a: A) higher level of significance. C) the probability of rejecting a false null hypothesis. -------------------------------------------------------------------------------C This is the definition of the power of the test: the probability of correctly rejecting the null hypothesis (rejecting the null hypothesis when it is false). . 24.B) equal to the level of confidence.