chunty

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.

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

chunty

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 difference in means test with pooled variances from the two samples.

C) A paired comparisons test because the samples are not independent.

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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.

chunty

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.

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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.

chunty

A test of a hypothesis that the means of two normally distributed populations are equal based on two independent random samples:

A) is a paired-comparisons test.

B) is based on a Chi Square statistic.

C) is done with a t-statistic.

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We have two formulas for test statistics for the hypothesis of equal sample means. Which one we use depends on whether or not we assume the samples have equal variances. Either formula generates a test statistic that follows a T-distribution.

chunty

Roy Fisher, CFA, wants to determine whether there is a significant difference, at the 5% significance level, between the mean monthly return on Stock GHI and the mean monthly return on Stock JKL. Fisher assumes the variances of the two stocks’ returns are equal. Using the last 12 months of returns on each stock, Fisher calculates a t-statistic of 2.0 for a test of equality of means. Based on this result, Fisher’s test:

A) rejects the null hypothesis, and Fisher can conclude that the means are equal.

B) rejects the null hypothesis, and Fisher can conclude that the means are not equal.

C) fails to reject the null hypothesis.

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The null hypothesis for a test of equality of means is H0: μ1 − μ2 = 0. Assuming the variances are equal, degrees of freedom for this test are (n1 + n2 − 2) = 12 + 12 − 2 = 22. From the table of critical values for Student’s t-distribution, the critical value for a two-tailed test at the 5% significance level for df = 22 is 2.074. Because the calculated t-statistic of 2.0 is less than the critical value, this test fails to reject the null hypothesis that the means are equal.

chunty

Simone Mak is a television network advertising executive. One of her responsibilities is selling commercial spots for a successful weekly sitcom. If the average share of viewers for this season exceeds 8.5%, she can raise the advertising rates by 50% for the next season. The population of viewer shares is normally distributed. A sample of the past 18 episodes results in a mean share of 9.6% with a standard deviation of 10.0%. If Mak is willing to make a Type 1 error with a 5% probability, which of the following statements is most accurate?

A) Mak cannot charge a higher rate next season for advertising spots based on this sample.

B) With an unknown population variance and a small sample size, Mak cannot test a hypothesis based on her sample data.

C) The null hypothesis Mak needs to test is that the mean share of viewers is greater than 8.5%.

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Mak cannot conclude with 95% confidence that the average share of viewers for the show this season exceeds 8.5 and thus she cannot charge a higher advertising rate next season.
Hypothesis testing process:
Step 1: State the hypothesis. Null hypothesis: mean ≤ 8.5%; Alternative hypothesis: mean > 8.5%
Step 2: Select the appropriate test statistic. Use a t statistic because we have a normally distributed population with an unknown variance (we are given only the sample variance) and a small sample size (less than 30). If the population were not normally distributed, no test would be available to use with a small sample size.
Step 3: Specify the level of significance. The significance level is the probability of a Type I error, or 0.05.
Step 4: State the decision rule.  This is a one-tailed test. The critical value for this question will be the t-statistic that corresponds to a significance level of 0.05 and n-1 or 17 degrees of freedom. Using the t-table, we determine that we will reject the null hypothesis if the calculated test statistic is greater than the critical value of 1.74.
Step 5: Calculate the sample (test) statistic.  The test statistic = t = (9.6 – 8.5) / (10.0 / √ 18) = 0.479 (Note: Remember to use standard error in the denominator because we are testing a hypothesis about the population mean based on the mean of 18 observations.)
Step 6: Make a decision. The calculated statistic is less than the critical value. Mak cannot conclude with 95% confidence that the mean share of viewers exceeds 8.5% and thus she cannot charge higher rates.
Note: By eliminating the two incorrect choices, you can select the correct response to this question without performing the calculations.

chunty

Ken Wallace is interested in testing whether the average price to earnings (P/E) of firms in the retail industry is 25. Using a t-distributed test statistic and a 5% level of significance, the critical values for a sample of 40 firms is (are):

A) -2.023 and 2.023.

B) -1.96 and 1.96.

C) -1.685 and 1.685.

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There are 40 − 1 = 39 degrees of freedom and the test is two-tailed. Therefore, the critical t-values are ± 2.023. The value 2.023 is the critical value for a one-tailed probability of 2.5%.

chunty

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $62,500 per year. What is the test statistic given a sample of 125 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $2,600?

A) 10.75.

B) -10.75.

C) 0.96.

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With a large sample size (125) and an unknown population variance, either the t-statistic or the z-statistic could be used. Using the z-statistic, it 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. The test statistic = (sample mean – hypothesized mean) / (sample standard deviation / (sample size1/2)) = (X − µ) / (s / n1/2) = (65,000 – 62,500) / (2,600 / 1251/2) = (2,500) / (2,600 / 11.18) = 10.75.

chunty

In a two-tailed hypothesis test, Jack Olson observes a t-statistic of -1.38 based on a sample of 20 observations where the population mean is zero. If you choose a 5% significance level, you should:

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

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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 -1.38, meaning that the calculated t-statistic is not in the rejection range, we fail to reject the null hypothesis that the population mean is not significantly different from zero.

chunty

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, tn-1 = 3.4. If you choose a 5% significance level you should:

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

B) fail to reject the null hypothesis and conclude that the population mean is less than or equal to 100.

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

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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 calculated t-statistic of 3.4 is greater than the critical t-statistic of 1.699, meaning that the calculated t-statistic is in the rejection range, we reject the null hypothesis and we conclude that the population mean is greater than 100.