leiyou

 

The correct answer is D

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

leiyou

 

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.

leiyou

 

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.

leiyou

 

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.

leiyou

 

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 = [attach]13864[/attach]

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

leiyou

 

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.

leiyou

 

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 =[attach]13865[/attach] ,

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

leiyou

 

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.

aianjie

 

 

The correct answer is B

This is a two-tailed test of the difference of means with equal population variance. The test statistic is:

 [attach]13866[/attach]
[attach]13867[/attach]

 

 

The degrees of freedom is n₁ + n₂ – 2 = 25 + 36 – 2 = 59.

t-critical = 2.000 with degrees of freedom = 25 + 36 – 2 = 59. Therefore, we reject the null hypothesis (|–3.42| > 2.00). Please note that t-tables seldom report the exact values for higher df like 59. Since 60 df is the closest value that is reported, we use the t-value for 60 df = 2.000.

aianjie

 

Using a 5 percent level of significance for a test of the null of H0: σ1 = σ2 versus the alternative of Ha: σ1 ? σ2, 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.