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%.

---

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

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.

---

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

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.

---

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

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.

---

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

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.

---

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

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.

---

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

torontoanalyst

The variance of 100 daily stock returns for Stock A is 0.0078.  The variance of 90 daily stock returns for Stock B is 0.0083.  Using a 5% level of significance, the critical value for this test is 1.61. The most appropriate conclusion regarding whether the variance of Stock A is different from the variance of Stock B is that the:

A) variances are equal.

B) variance of Stock B is significantly greater than the variance of Stock A.

C) variances are not equal.

---

A test of the equality of variances requires an F-statistic. The calculated F-statistic is 0.0083/0.0078 = 1.064. Since the calculated F value of 1.064 is less than the critical F value of 1.61, we cannot reject the null hypothesis that the variances of the 2 stocks are equal.

torontoanalyst

Which of the following statements about the variance of a normally distributed population is least accurate?

A) The Chi-squared distribution is a symmetric distribution.

B) The test of whether the population variance equals σ02 requires the use of a Chi-squared distributed test statistic, [(n − 1)s2] / σ02.

C) A test of whether the variance of a normally distributed population is equal to some value σ02, the hypotheses are: H0: σ2 = σ02, versus Ha: σ2 ≠ σ02.

---

The Chi-squared distribution is not symmetrical, which means that the critical values will not be numerically equidistant from the center of the distribution, though the probability on either side of the critical values will be equal (that is, if there is a 5% level of significance and a two-sided test, 2.5% will lie outside each of the two critical values).

torontoanalyst

A test of the population variance is equal to a hypothesized value requires the use of a test statistic that is:

A) F-distributed.

B) t-distributed.

C) Chi-squared distributed.

---

In tests of whether the variance of a population equals a particular value, the chi-squared test statistic is appropriate.

torontoanalyst

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

A) z-distributed.

B) Chi-squared distributed.

C) F-distributed.

---

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