6.Susan Bellows is comparing the return on equity for two industries. She is convinced that the return on equity for the discount retail industry (DR) is greater than that of the luxury retail (LR) industry. What are the hypotheses for a test of her comparison of return on equity?

A)   H₀: µ_DR = µ_LR versus H_a: µ_DR ≠ µ_LR.

B)   H₀: µ_DR <= µ_LR versus H_a: µ_DR > µ_LR.

C)   H₀: µ_DR ≠ µ_LR versus H_a: µ_DR = µ_LR.

D)   H₀: µ_DR = µ_LR versus H_a: µ_DR < µ_LR.

7.An analyst conducts a two-tailed z-test to determine if small cap returns are significantly different from 10%. The sample size was 200. The computed z-statistic is 2.3. Using a 5% level of significance, which statement is TRUE?

A)   You cannot determine what to do with the information given.

B)   A sample size of 200 indicates that we should fail to reject the null.

C)   Fail to reject the null hypothesis and conclude that small cap returns are close enough to 10% that we cannot say they are significantly different from 10%.

D)   Reject the null hypothesis and conclude that small cap returns are significantly different from 10%.

8.Given the following hypothesis:

§       The null hypothesis is H₀:µ = 5

§       The alternative is H₁:µ does not equal 5

§       The mean of a sample of 17 is 7

§       The population standard deviation is 2.0

What is the calculated Z-statistic?

A)   4.00.

B)   8.00.

C)   4.12.

D)   8.25.

9.An analyst is testing to see if the mean of a population is less than 133. A random sample of 50 observations had a mean of 130. Assume a standard deviation of 5. The test is to be made at the 1% level of significance.

The null hypothesis is:

A)   µ > 133.

B)   µ ≤ 133.

C)   µ ≥ 133.

D)   µ = 133.

10.The calculated test statistic is:

A)   +1.33.

B)   -1.33.

C)   -3.00.

D)   -4.24.

答案和详解如下：

6.Susan Bellows is comparing the return on equity for two industries. She is convinced that the return on equity for the discount retail industry (DR) is greater than that of the luxury retail (LR) industry. What are the hypotheses for a test of her comparison of return on equity?

A)   H₀: µ_DR = µ_LR versus H_a: µ_DR ≠ µ_LR.

B)   H₀: µ_DR <= µ_LR versus H_a: µ_DR > µ_LR.

C)   H₀: µ_DR ≠ µ_LR versus H_a: µ_DR = µ_LR.

D)   H₀: µ_DR = µ_LR versus H_a: µ_DR < µ_LR.

The correct answer was B)

The alternative hypothesis is determined by the theory or the belief. The researcher specifies the null as the hypothesis that she wishes to reject (in favor of the alternative). Note that this is a one-sided alternative because of the “greater than” belief.

7.An analyst conducts a two-tailed z-test to determine if small cap returns are significantly different from 10%. The sample size was 200. The computed z-statistic is 2.3. Using a 5% level of significance, which statement is TRUE?

A)   You cannot determine what to do with the information given.

B)   A sample size of 200 indicates that we should fail to reject the null.

C)   Fail to reject the null hypothesis and conclude that small cap returns are close enough to 10% that we cannot say they are significantly different from 10%.

D)   Reject the null hypothesis and conclude that small cap returns are significantly different from 10%.

The correct answer was D)

At the 5% level of significance the critical z-statistic for a two-tailed test is 1.96 (assuming a large sample size). The null hypothesis is H₀: x = 10%. The alternative hypothesis is H_A: x ≠ 10%. Because the computed z-statistic is greater than the critical z-statistic (2.33 > 1.96), we reject the null hypothesis and we conclude that small cap returns are significantly different than 10%.

8.Given the following hypothesis:

§       The null hypothesis is H₀:µ = 5

§       The alternative is H₁:µ does not equal 5

§       The mean of a sample of 17 is 7

§       The population standard deviation is 2.0

What is the calculated Z-statistic?

A)   4.00.

B)   8.00.

C)   4.12.

D)   8.25.

The correct answer was C)

The Z-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. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)^1/2 = (X-μ) / (σ / n^1/2) = (7 - 5) / (2 / 17^1/2) = (2) / (2 / 4.1231) = 4.12.

9.An analyst is testing to see if the mean of a population is less than 133. A random sample of 50 observations had a mean of 130. Assume a standard deviation of 5. The test is to be made at the 1% level of significance.

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06
0.0	0.0000	0.0040	0.0080	0.0120	0.0160	0.0199	0.0239
0.1	0.0398	0.0438	0.0478	0.0517	0.0557	0.0596	0.0636
0.2	0.0793	0.0832	0.0871	0.0910	0.0948	0.0987	0.1026
0.3	0.1179	0.1217	0.1255	0.1293	0.1331	0.1368	0.1406
|	|	|	|	|	|	|	|
1.7	0.4554	0.4564	0.4573	0.4582	0.4591	0.4599	0.4608
1.8	0.4641	0.4649	0.4656	0.4664	0.4671	0.4678	0.4686
1.9	0.4713	0.4719	0.4726	0.4732	0.4738	0.4744	0.4750
2.0	0.4772	0.4778	0.4783	0.4788	0.4793	0.4798	0.4803
2.1	0.4821	0.4826	0.4830	0.4834	0.4838	0.4842	0.4846
2.2	0.4861	0.4864	0.4868	0.4871	0.4875	0.4878	0.4881
2.3	0.4893	0.4896	0.4898	0.4901	0.4904	0.4906	0.4909
2.4	0.4918	0.4920	0.4922	0.4925	0.4927	0.4929	0.4931

The null hypothesis is:

A)   µ > 133.

B)   µ ≤ 133.

C)   µ ≥ 133.

D)   µ = 133.

The correct answer was C)

The null hypothesis is the hypothesis that the researcher wants to reject. Here the hypothesis that is being looked for is that the mean of a population is less than 133. The null hypothesis is that the mean is greater than or equal to 133. The question is whether the null hypothesis will be rejected in favor of the alternative hypothesis that the mean is less than 133.

10.The calculated test statistic is:

A)   +1.33.

B)   -1.33.

C)   -3.00.

D)   -4.24.

The correct answer was D)

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. Here, the test statistic = (sample mean – hypothesized mean) / ((sample standard deviation / (sample size)^1/2)) = (130 – 133) / (5 / 50^1/2) = (-3) / (5 / 7.0711) = -4.24.