Reading 11: Hypothesis Testing - LOS e, (Part 1) ～ Q6-9

答案和详解如下：

6.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 percent, he will receive a 25 percent bonus. If he can reduce turnover to less than 25 percent, he will receive a 50 percent bonus (using a significance level of 10 percent). The population of turnover rates is normally distributed. The population standard deviation of turnover rates is 1.5 percent. A recent sample of 100 branch offices resulted in an average turnover rate of 24.2 percent. Which of the following statements is TRUE?

A)   Brickley should not receive either bonus.

B)   For the 25% bonus level, the test statistic is -10.66.

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

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

The correct answer was D)

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. a = 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 a 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.

7.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 likely 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)   Mak should use a two-tailed hypothesis test at a 5% level of significance.

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

The correct answer was B)

Mak can conclude with 95% confidence that the average share of viewers for the show this season exceeds 8.5 and thus she can 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 three incorrect choices, you can select the correct response to this question without performing the calculations.

8.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)   -1.685 and 1.685.

B)   -1.96 and 1.96.

C)   1.685.

D)   -2.023 and 2.023.

The correct answer was D)

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%.

9.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)   0.96.

B)   -10.75.

C)   10.75.

D)   -0.96.

The correct answer was C)

With a large sample size (125) and an unknown variance, either the t-stat or the Z-statistic could be used. Using the Z-stat, 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) / (population standard deviation / (sample size)1/2 = (X-µ) / (σ / n^1/2) = (65,000 – 62,500) / (2,600 / 125^1/2) = (2,500) / (2,600 / 11.18) = 10.75.