leiyou

 

6、A return series with 250 observations has a sample mean of 10 percent and a standard deviation of 15 percent. The standard error of the sample mean is closest to:

A) 0.06.

B) 0.95.

C) 15.80.

D) 3.87.

leiyou

 

The correct answer is B

The standard error of the sample mean is the standard deviation of the sample divided by the square root of the number of observations in the sample. In this case, (15 / √250) = 0.95.

leiyou

 

7、The mean daily return for an equity portfolio over 60 months is 1.5 percent. The standard deviation is 3.0 percent. The value of the test statistic to test the hypothesis that mean monthly return is equal to zero is closest to:

A) 0.50.

B) 30.00.

C) 3.87.

D) 2.19.

leiyou

 

The correct answer is C

z = (1.5% - 0.0%) / [(3.0% / √60)] = 3.87

leiyou

 

8、The mean equity risk premium over a 40-year period is equal to 8.0 percent. The standard deviation of the sample is 12 percent. The standard error of the sample mean is closest to:

A) 0.30%.

B) 1.90%.

C) 1.26%.

D) 8.00%.

leiyou

 

The correct answer is B

Note the size of the sample here is the number of years.

σX = 12 / √40 = 1.90%

leiyou

 

9、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) 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.

leiyou

 

The correct answer is D

Using the process of Hypothesis testing:

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

leiyou

 

10、A bottler of iced tea wishes to ensure that an average of 16 ounces of tea is in each bottle. In order to analyze the accuracy of the bottling process, a random sample of 150 bottles is taken.  Using a t-distributed test statistic of -1.09 and a 5% level of significance, the bottler should:

A) not reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea.

B) reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea.

C) reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea.

D) not reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea.

leiyou

 

The correct answer is A

Ho: μ = 16; Ha: μ ≠ 16. Do not reject the null since |t| = 1.09 < 1.96 (critical value).