Reading 11: Hypothesis Testing-LOS f 习题精选 2011, concerning, hypothesis, interpret

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Session 3: Quantitative Methods: Application
Reading 11: Hypothesis Testing

LOS f: Identify the appropriate test statistic and interpret the results for a hypothesis test concerning the population mean of both large and small samples when the population is normally or approximately distributed and the variance is 1) known or 2) unknown.

 

 

Brandee Shoffield is the public relations manager for Night Train Express, a local sports team. 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. Attendance is approximately normally distributed. A sample of the attendance at 15 home games results in a mean of 3,150 and a standard deviation of 450. Which of the following statements is most accurate?

A) The calculated test statistic is 1.291.

B) With an unknown population variance and a small sample size, no statistic is available to test Shoffield's hypothesis.

C) Shoffield should use a two-tailed Z-test.

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We will use the process of Hypothesis testing to determine whether Shoffield should reject H_o:

            Step 1:  State the Hypothesis

                        H_o: μ ≤ 3,000

                        H_a: μ > 3,000

            Step 2:  Select Appropriate Test Statistic

Here, 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.) Thus, we will use the t-statistic.

Step 3:  Specify the Level of Significance

Here, the confidence level is 90%, or 0.90, which translates to a 0.10 significance level.

Step 4:  State the Decision Rule

This is a one-tailed test. The critical value for this question will be the t-statistic that corresponds to an α of 0.10, and 14 (n-1) degrees of freedom. Using the t-table , we determine that the appropriate critical value = 1.345. Thus, we will reject the null hypothesis if the calculated test statistic is greater than 1.345.

Step 5:  Calculate sample (test) statistic

The test statistic = t = (3,150 – 3,000) / (450 / √ 15) = 1.291

Step 6:  Make a decision

Fail to reject the null hypothesis because the calculated statistic is less than the critical value. Shoffield cannot state with 90% certainty that the home game attendance exceeds 3,000.

The other statements are false. As shown above, the appropriate test is a t-test, not a Z-test. There is a test statistic for an normally distributed population, an unknown variance and a small sample size – the t-statistic. There is no test for a non-normal population with unknown variance and small sample size.

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1215

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $62,500 per year. What is the test statistic given a sample of 125 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $2,600?

A) -10.75.

B) 10.75.

C) 0.96.

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With a large sample size (125) and an unknown population variance, either the t-statistic or the z-statistic could be used. Using the z-statistic, it 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. The test statistic = (sample mean – hypothesized mean) / (sample standard deviation / (sample size^1/2)) = (X ? μ) / (s / n^1/2) = (65,000 – 62,500) / (2,600 / 125^1/2) = (2,500) / (2,600 / 11.18) = 10.75.

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Ken Wallace is interested in testing whether the average price to earnings (P/E) of firms in the retail industry is 25. Using a t-distributed test statistic and a 5% level of significance, the critical values for a sample of 40 firms is (are):

A) -2.023 and 2.023.

B) -1.96 and 1.96.

C) -1.685 and 1.685.

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There are 40 ? 1 = 39 degrees of freedom and the test is two-tailed. Therefore, the critical t-values are ± 2.023. The value 2.023 is the critical value for a one-tailed probability of 2.5%.

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Simone Mak is a television network advertising executive. One of her responsibilities is selling commercial spots for a successful weekly sitcom. If the average share of viewers for this season exceeds 8.5%, she can raise the advertising rates by 50% for the next season. The population of viewer shares is normally distributed. A sample of the past 18 episodes results in a mean share of 9.6% with a standard deviation of 10.0%. If Mak is willing to make a Type 1 error with a 5% probability, which of the following statements is most accurate?

A) With an unknown population variance and a small sample size, Mak cannot test a hypothesis based on her sample data.

B) Mak cannot charge a higher rate next season for advertising spots based on this sample.

C) The null hypothesis Mak needs to test is that the mean share of viewers is greater than 8.5%.

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Mak cannot conclude with 95% confidence that the average share of viewers for the show this season exceeds 8.5 and thus she cannot charge a higher advertising rate next season.

Hypothesis testing process:

Step 1:  State the hypothesis. Null hypothesis: mean ≤ 8.5%; Alternative hypothesis: mean > 8.5%

Step 2:  Select the appropriate test statistic. Use a t statistic because we have a normally distributed population with an unknown variance (we are given only the sample variance) and a small sample size (less than 30). If the population were not normally distributed, no test would be available to use with a small sample size.

Step 3:  Specify the level of significance. The significance level is the probability of a Type I error, or 0.05.

Step 4:  State the decision rule.  This is a one-tailed test. The critical value for this question will be the t-statistic that corresponds to a significance level of 0.05 and n-1 or 17 degrees of freedom. Using the t-table, we determine that we will reject the null hypothesis if the calculated test statistic is greater than the critical value of 1.74.

Step 5:  Calculate the sample (test) statistic.  The test statistic = t = (9.6 – 8.5) / (10.0 / √ 18) = 0.479 (Note: Remember to use standard error in the denominator because we are testing a hypothesis about the population mean based on the mean of 18 observations.)

Step 6: Make a decision. The calculated statistic is less than the critical value. Mak cannot conclude with 95% confidence that the mean share of viewers exceeds 8.5% and thus she cannot charge higher rates.

Note: By eliminating the two incorrect choices, you can select the correct response to this question without performing the calculations.

1215

Which of the following statements about test statistics is least accurate?

A) In the case of a test of the difference in means of two independent samples, we use a t-distributed test statistic.

B) In a test of the population mean, if the population variance is unknown, we should use a t-distributed test statistic.

C) In a test of the population mean, if the population variance is unknown and the sample is small, we should use a z-distributed test statistic.

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If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown. A t-test is also used to test the difference between two population means while an F-test is used to compare differences between the variances of two populations.

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In a test of the mean of a population, if the population variance is:

A) known, a t-distributed test statistic is appropriate.

B) unknown, a z-distributed test statistic is appropriate.

C) known, a z-distributed test statistic is appropriate.

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If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population variance is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown.

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In order to test whether the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, t_n-1 = 3.4. If you choose a 5% significance level you should:

A) fail to reject the null hypothesis and conclude that the population mean is less than or equal to 100.

B) reject the null hypothesis and conclude that the population mean is greater that 100.

C) fail to reject the null hypothesis and conclude that the population mean is greater than 100.

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At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a one-tailed test and 29 degrees of freedom (sample size of 30 less 1) is 1.699 (with a large sample size the critical z-statistic of 1.645 may be used). Because the calculated t-statistic of 3.4 is greater than the critical t-statistic of 1.699, meaning that the calculated t-statistic is in the rejection range, we reject the null hypothesis and we conclude that the population mean is greater than 100.

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In a two-tailed hypothesis test, Jack Olson observes a t-statistic of -1.38 based on a sample of 20 observations where the population mean is zero. If you choose a 5% significance level, you should:

A) reject the null hypothesis and conclude that the population mean is significantly different from zero.

B) reject the null hypothesis and conclude that the population mean is not significantly different from zero.

C) fail to reject the null hypothesis that the population mean is not significantly different from zero.

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At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a two-tailed test and 19 degrees of freedom (sample size of 20 less 1) is ± 2.093 (with a large sample size the critical z-statistic of 1.960 may be used). Because the critical t-statistic of -2.093 is to the left of the calculated t-statistic of -1.38, meaning that the calculated t-statistic is not in the rejection range, we fail to reject the null hypothesis that the population mean is not significantly different from zero.

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In order to test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken. The sample value of the computed z-statistic = 3.4. 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.

B) reject the null hypotheses and conclude that the population mean is greater than 100.

C) reject the null hypothesis and conclude that the population mean is equal to 100.

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H_o:μ ≤ 100; H_a: μ > 100. Reject the null since z = 3.4 > 1.65 (critical value).

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Maria Huffman is the Vice President of Human Resources for a large regional car rental company. Last year, she hired Graham Brickley as Manager of Employee Retention. 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%, he will receive a 25% bonus. If he can reduce turnover to less than 25%, he will receive a 50% bonus (using a significance level of 10%). The population of turnover rates is normally distributed. The population standard deviation of turnover rates is 1.5%. A recent sample of 100 branch offices resulted in an average turnover rate of 24.2%. Which of the following statements is most accurate?

A) For the 50% bonus level, the critical value is -1.65 and Huffman should give Brickley a 50% bonus.

B) Brickley should not receive either bonus.

C) For the 50% bonus level, the test statistic is -5.33 and Huffman should give Brickley a 50% bonus.

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Using the process of Hypothesis testing:

Step 1: State the Hypothesis. For 25% bonus level - H_o: m ≥ 30% H_a: m < 30%; For 50% bonus level - H_o: m ≥ 25% H_a: m < 25%.

Step 2: Select Appropriate Test Statistic. Here, 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.) Thus, we will use the z-statistic.

Step 3: Specify the Level of Significance. α = 0.10.

Step 4: State the Decision Rule. This is a one-tailed test. The critical value for this question will be the z-statistic that corresponds to an α of 0.10, or an area to the left of the mean of 40% (with 50% to the right of the mean). Using the z-table (normal table), we determine that the appropriate critical value = -1.28 (Remember that we highly recommend that you have the “common” z-statistics memorized!) Thus, we will reject the null hypothesis if the calculated test statistic is less than -1.28.

Step 5: Calculate sample (test) statistics. Z (for 50% bonus) = (24.2 – 25) / (1.5 / √ 100) = ?5.333. Z (for 25% bonus) = (24.2 – 30) / (1.5 / √ 100) = ?38.67.

Step 6: Make a decision. Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is less than the critical value. Thus, Huffman should give Soberg a 50% bonus.

The other statements are false. The critical value of –1.28 is based on the significance level, and is thus the same for both the 50% and 25% bonus levels.