Q2. Joe Sutton is evaluating the effects of the 1987 market decline on the volume of trading. Specifically, he wants to test whether the decline affected trading volume. He selected a sample of 500 companies and collected data on the total annual volume for one year prior to the decline and for one year following the decline. What is the set of hypotheses that Sutton is testing?
A) H0: μd = μd0 versus Ha: μd ≠ μd0.
B) H0: μd ≠ μd0 versus Ha: μd = μd0.
C) H0: μd = μd0 versus Ha: μd > μd0.
Correct answer is A)
This is a paired comparison because the sample cases are not independent (i.e., there is a before and an after for each stock). Note that the test is two-tailed, t-test.
Q3. An analyst wants to determine whether the monthly returns on two stocks over the last year were the same or not. What test should she use if she is willing to assume that the returns are normally distributed?
A) A difference in means test only if the variances of monthly returns are equal for the two stocks.
B) A paired comparisons test because the samples are not independent.
C) A difference in means test with pooled variances from the two samples.
Correct answer is B)
A paired comparisons test must be used. The difference in means test requires that the samples be independent. Portfolio theory teaches us that returns on two stocks over the same time period are unlikely to be independent since both have some systematic risk.
Q4. An analyst for the entertainment industry theorizes that betas for most firms in the industry are higher after September 11, 2001. She sampled 31 firms comparing their betas for the one-year period before and after this date. Based on this sample, she found that the mean differences in betas were 0.19, with a sample standard deviation of 0.11. Her null hypothesis is that the betas are the same before and after September 11. Based on the results of her sample, can we reject the null hypothesis at a 5% significance level and why? Null is:
A) not rejected. The critical value exceeds the t-value by 7.58.
B) rejected. The t-value exceeds the critical value by 5.67.
C) rejected. The t-value exceeds the critical value by 7.58.
Correct answer is C)
The t-statistic for paired differences:
t = (d – ud 0) / sd and sd = sd / √n
t = 9.62 from a table with 30 df, the critical value = 2.042
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发表于 2009-5-23 00:38
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