Q1. In order to test if Stock A is more volatile than Stock B, prices of both stocks are observed to construct the sample variance of the two stocks. The appropriate test statistics to carry out the test is the:
       

A)   Chi-square test.

B)   F test.

C)   t test.

Q2. 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)   not rejected. The critical value exceeds the F-value by 0.71.

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

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

Q3. The use of the F-distributed test statistic, F = s₁² / s₂², to compare the variances of two populations does NOT require which of the following?

A)   populations are normally distributed.

B)   two samples are of the same size.

C)   samples are independent of one another.

Q4. The test of the equality of the variances of two normally distributed populations requires the use of a test statistic that is:

A)     F-distributed.

B)     z-distributed.

C)     Chi-squared distributed.

[此贴子已经被作者于2009-1-13 16:17:06编辑过]

答案和详解如下：

Q1. In order to test if Stock A is more volatile than Stock B, prices of both stocks are observed to construct the sample variance of the two stocks. The appropriate test statistics to carry out the test is the:

A)   Chi-square test.

B)   F test.

C)   t test.

Correct answer is B)

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

Q2. 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)   not rejected. The critical value exceeds the F-value by 0.71.

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

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

Correct answer is A)

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.

Q3. The use of the F-distributed test statistic, F = s₁² / s₂², to compare the variances of two populations does NOT require which of the following?

A)   populations are normally distributed.

B)   two samples are of the same size.

C)   samples are independent of one another.

Correct answer is B)

The F-statistic can be computed using samples of different sizes. That is, n₁ need not be equal to n₂.

Q4. The test of the equality of the variances of two normally distributed populations requires the use of a test statistic that is:

A)     F-distributed.

B)     z-distributed.

C)     Chi-squared distributed.

Correct answer is A)

The F-distributed test statistic, F = s₁² / s₂², is used to compare the variances of two populations.

Thanks

One comment on Q2, if the hypothesis is σ₁² = σ₂²  which implies a two-sided test, shall we find the rejection point in F-tables at the @/2=(1-95%)/2=0.025 significance level?

[此贴子已经被作者于2009-2-25 21:07:15编辑过]

Q6. If the null hypothesis is H₀: ρ ≤ 0, what is the appropriate alternative hypothesis?

A)     H_a: ρ ≠ 0.

B)     H_a: ρ > 0.

C)     H_a: ρ < 0.

Correct answer is B)

The alternative hypothesis must include the possible outcomes the null does not.

Q7. In order to test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, t_n-1 = 1.2. If you choose a 5% significance level you should:

A)   fail to reject the null hypothesis and conclude that the population mean is not greater than 100.

B)   reject the null hypothesis and conclude that the population mean is greater than 100.

C)   fail to reject the null hypothesis and conclude that the population mean is greater than 100.

Correct answer is A)

At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a one-tailed test and 29 degrees of freedom (sample size of 30 less 1) is 1.699 (with a large sample size the critical z-statistic of 1.645 may be used). Because the critical t-statistic is greater than the calculated t-statistic, meaning that the calculated t-statistic is not in the rejection range, we fail to reject the null hypothesis and we conclude that the population mean is not significantly greater than 100.

Q8. In order to test whether the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, t_n-1 = 3.4. The null and alternative hypotheses are:

A)H₀: μ ≤ 100; H_a: μ > 100.

B)H₀: μ = 100; H_a: μ ≠ 100.

C)H₀: X ≤ 100; H_a: X > 100.

Correct answer is A)

The null hypothesis is that the theoretical mean is not significantly different from zero. The alternative hypothesis is that the theoretical mean is greater than zero.

Q9. In a two-tailed test of a hypothesis concerning whether a population mean is zero, Jack Olson computes a t-statistic of 2.7 based on a sample of 20 observations where the distribution is normal. If a 5% significance level is chosen, Olson should:

A)   reject the null hypothesis and conclude that the population mean is not significantly different from zero.

B)   fail to reject the null hypothesis that the population mean is not significantly different from zero.

C)   reject the null hypothesis and conclude that the population mean is significantly different from zero.

Correct answer is C)

At a 5% significance level, the critical t-statistic using the Student’s t-distribution table for a two-tailed test and 19 degrees of freedom (sample size of 20 less 1) is ± 2.093 (with a large sample size the critical z-statistic of 1.960 may be used). Because the critical t-statistic of 2.093 is to the left of the calculated t-statistic of 2.7, meaning that the calculated t-statistic is in the rejection range, we reject the null hypothesis and we conclude that the population mean is significantly different from zero.

Q10. What kind of test is being used for the following hypothesis and what would a z-statistic of 1.68 tell us about a hypothesis with the appropriate test and a level of significance of 5%, respectively?

H₀: B ≤ 0

H_A: B > 0

A)   One-tailed test; fail to reject the null.

B)   Two-tailed test; fail to reject the null.

C)   One-tailed test; reject the null.

Correct answer is C)

The way the alternative hypothesis is written you are only looking at the right side of the distribution. You are only interested in showing that B is greater than 0. You don't care if it is less than zero. For a one-tailed test at the 5% level of significance, the critical z value is 1.645. Since the test statistic of 1.68 is greater than the critical value we would reject the null hypothesis.

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