Problem 16 - Basic Statistics for Business and Economics with Student

Question

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

Step-by-Step Solution

Answer

We cannot conclude that single-earner couples spend more time watching TV than dual-earner couples.

Step 1: Formulate Hypotheses

To address the problem, we begin by formulating the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the mean time spent watching television together by single-earner couples (\(\mu_1\)) is equal to that of dual-earner couples (\(\mu_2\)), i.e., \(H_0: \mu_1 = \mu_2\). The alternative hypothesis (\(H_a\)) suggests that single-earner couples spend more time, i.e., \(H_a: \mu_1 > \mu_2\).

Step 2: Determine Test Statistic

We use a two-sample t-test to compare the means of two independent groups. The test statistic is calculated using the formula:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}\]where \(\bar{x}_1 = 61\) and \(\bar{x}_2 = 48.4\) are the sample means, \(s_1 = 15.5\) and \(s_2 = 18.1\) are the standard deviations, and \(n_1 = 15\) and \(n_2 = 12\) are the sample sizes.

Step 3: Calculate Test Statistic

Plug the values into the formula to compute the test statistic:\[ t = \frac{61 - 48.4}{\sqrt{\frac{(15.5)^2}{15} + \frac{(18.1)^2}{12}}}\]\[ t = \frac{12.6}{\sqrt{\frac{240.25}{15} + \frac{327.61}{12}}}\]\[ t = \frac{12.6}{\sqrt{16.0167 + 27.3008}}\]\[ t = \frac{12.6}{\sqrt{43.3175}} = \frac{12.6}{6.5801} \approx 1.915\]

Step 4: Determine the Critical Value

For a significance level of \(\alpha = 0.01\) and degrees of freedom approximated using the smaller of \(n_1 - 1 = 14\) and \(n_2 - 1 = 11\), the critical t-value can be found from t-distribution tables or software as around 2.681 for a one-tailed test.

Step 5: Compare Test Statistic to Critical Value

Compare the calculated t-statistic \(t = 1.915\) to the critical value of \(t = 2.681\). Since \(1.915 < 2.681\), we fail to reject the null hypothesis.

Step 6: Conclusion

Due to the test statistic not exceeding the critical value, we do not have enough evidence to reject the null hypothesis at the 0.01 significance level. Thus, we cannot conclude that single-earner couples, on average, spend more time watching television together than dual-earner couples.

Key Concepts

Null Hypothesis

The null hypothesis, often denoted as \( H_0 \), is a fundamental part of statistical testing. It is essentially a statement of no effect or no difference, which the test seeks to support or refute.
For example, in the study comparing television time between single-earner and dual-earner couples, the null hypothesis posits that there is no difference in the average time spent watching TV together between the two groups.
This is mathematically represented as \( H_0: \mu_1 = \mu_2 \), where \( \mu_1 \) is the mean TV time for single-earner couples and \( \mu_2 \) is for dual-earner couples. The goal of the test is to determine whether enough evidence exists to reject this assumption based on the sample data.

Alternative Hypothesis

The alternative hypothesis, symbolized by \( H_a \), is what researchers aim to support with evidence. It is an assertion of a true effect or difference that is opposed to the null hypothesis.
In the context of our exercise, the alternative hypothesis suggests that single-earner couples spend more time watching TV than dual-earner couples.
This can be written as \( H_a: \mu_1 > \mu_2 \), indicating that the average TV watching time for single-earner couples is greater than that for dual-earner couples. Demonstrating that this hypothesis is more plausible than the null hypothesis is a key objective when conducting hypothesis tests.

Test Statistic

The test statistic plays a critical role in hypothesis testing, acting as the bridge between the data and decision-making. For the two-sample t-test, the objective is to see how much the observed sample means differ, taking into account the variation within each group.
The formula to calculate the test statistic \( t \) is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \]where:

• \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means,
• \( s_1 \) and \( s_2 \) are the standard deviations,
• \( n_1 \) and \( n_2 \) are the sample sizes.

In this example, the calculated test statistic gives us a value that we can compare against a critical value to decide whether or not to reject the null hypothesis.

Significance Level

The significance level, denoted by \( \alpha \), is a threshold set by the researcher to decide when to reject the null hypothesis. It represents the probability of making a Type I error, which means rejecting the null hypothesis when it is actually true.
In our example, a significance level of 0.01, or 1%, is used, indicating a very stringent threshold for evidence against the null hypothesis.
This means there is only a 1% risk of falsely claiming that the single-earner couples spend more time watching television when there is no actual difference.
Choosing an appropriate significance level is important as it balances the risk of errors with the confidence in making a well-informed decision.