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The correct answer is D

The F test is used to test the differences of variance between two samples.

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2、Abby Ness is an analyst for a firm that specializes in evaluating firms involved in mineral extraction. Ness believes that the earnings of copper extracting firms are more volatile than those of bauxite extraction firms. In order to test this, Ness examines the volatility of returns for 31 copper firms and 25 bauxite firms. The standard deviation of earnings for copper firms was $2.69, while the standard deviation of earnings for bauxite firms was $2.92. Ness’s Null Hypothesis is σ12 = σ22. Based on the samples, can we reject the null hypothesis at a 95% confidence level using an F-statistic and why? Null is:

A) rejected. The F-value exceeds the critical value by 0.71.

B) rejected. The F-value exceeds the critical value by 0.849.

C) not rejected. The critical value exceeds the F-value by 0.194.

D) not rejected. The critical value exceeds the F-value by 0.71.

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The correct answer is D

F = s12 / s22 = $2.922 / $2.692 = 1.18

From an F table, the critical value with numerator df = 24 and denominator df = 30 is 1.89.

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3、Two samples were drawn from a normally distributed population. For the first sample, the mean was $50, and the standard deviation was $5. For the second sample, the mean was $55, and the standard deviation was $6. The first sample consists of 25 observations, and the second sample consists of 36 observations. (Note: In the following questions, the subscripts “1” and “2” indicate the first and second sample, respectively.)

Consider the hypotheses structured as H0: μ1 = $48 versus Ha: μ1 ≠ $48. At a 1 percent level of significance, the null hypothesis:

A) cannot be rejected.

B) should be rejected.

C) should neither be rejected nor fail to be rejected.

D) cannot be tested using this sample information provided.

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The correct answer is A

A two-tailed t-test is appropriate. The decision rule is to reject H0 if the t-statistic is outside the range defined by ±t at α/2 = 0.025 with df = 24. The t-statistic =


t24 = 1.gif



±t24 at α/2 = p = 0.025 = ±2.797, therefore, H0 cannot be rejected.

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Consider the hypotheses structured as H0: μ1 ≤ $48 versus Ha: μ1 > $48. At a 5 percent level of significance, the null hypothesis:

A) cannot be rejected.

B) should be rejected.

C) should neither be rejected nor fail to be rejected.

D) cannot be tested using the sample information provided.

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The correct answer is B

A one-tailed t-test is appropriate. The decision rule is to reject H0 if the computed t-statistic > t-critical at α = 0.05 with df = 24. The computed value of the t-statistic = 1.gif

,

and t-critical = t24 = 1.711. Since t > t-critical, H0 should be rejected.


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Assuming equal population variances, consider the hypothesis test formulated as H0: μ1 = μ2 versus μ1 ? μ2. At a 5 percent level of significance, the null hypothesis:

A) cannot be rejected.

B) should be rejected.

C) should neither be rejected nor fail to be rejected.

D) cannot be tested using the sample information provided.

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