Questions: Exercise 11-16 (Algo) (LO11-3) A recent study compared the time watching television by single-and dual-earner households. Based on a sample of 15 single-earner households, the mean amount of time watching television was 57 minutes per day, with a standard deviation of 17.0 minutes. Based on a sample of 12 dual-earner households, the mean number of minutes watching television was 44.4 minutes, with a standard deviation of 19.6 minutes. Using the 0.10 level of significance, verify the assumption that the variances are equal. Using the 0.01 significance level, can we conclude that the single-earner households spend more time watching television? Required: a. State the decision rule. Note: Round your answer to 3 decimal places. The decision rule is to reject H0 if t is b. Compute the value of the test statistic. Note: Round your answer to 3 decimal places.

Solution Steps

To solve this problem, we need to perform two statistical tests. First, we will conduct an F-test to verify if the variances of the two samples are equal. Then, we will perform a two-sample t-test to determine if single-earner households spend more time watching television than dual-earner households. For both tests, we will use the given significance levels to determine the decision rules and compute the test statistics.

To verify if the variances of the two samples are equal, we perform an F-test. The test statistic is calculated as:

\[ F = \frac{s_1^2}{s_2^2} = \frac{17.0^2}{19.6^2} = 0.7523 \]

where \(s_1\) and \(s_2\) are the standard deviations of the single-earner and dual-earner households, respectively. The degrees of freedom are \(df_1 = 14\) and \(df_2 = 11\).

The critical value for the F-test at a significance level of \(\alpha = 0.10\) is:

\[ F_{\text{critical}} = 2.7386 \]

The decision rule is to reject \(H_0\) if \(F > 2.739\) or \(F < 0.365\). Since \(0.7523\) is not greater than \(2.739\) and not less than \(0.365\), we do not reject \(H_0\).

Next, we perform a two-sample t-test to determine if single-earner households spend more time watching television. The pooled standard deviation is:

\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} = 18.1898 \]

The t-statistic is calculated as:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{57 - 44.4}{18.1898 \sqrt{\frac{1}{15} + \frac{1}{12}}} = 1.7885 \]

The degrees of freedom for the t-test is \(df = 25\). The critical value for the t-test at a significance level of \(\alpha = 0.01\) is:

\[ t_{\text{critical}} = 2.4851 \]

The decision rule is to reject \(H_0\) if \(t > 2.485\). Since \(1.7885\) is not greater than \(2.485\), we do not reject \(H_0\).

Final Answer