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Chapter 11: Problem 16

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.

Short Answer

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

Step by step solution

01

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., \(H_a: \mu_1 > \mu_2\).
02

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.
03

Calculate Test Statistic

Plug the values into the formula to compute the test statistic:\[ t = \frac{61 - 48.4}{\sqrt{\frac{(15.5)^2}{15} + \frac{(18.1)^2}{12}}}\]\[ t = \frac{12.6}{\sqrt{\frac{240.25}{15} + \frac{327.61}{12}}}\]\[ t = \frac{12.6}{\sqrt{16.0167 + 27.3008}}\]\[ t = \frac{12.6}{\sqrt{43.3175}} = \frac{12.6}{6.5801} \approx 1.915\]
04

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.
05

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.
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. Assume the population standard deviation for the number of units produced on the day shift is 21 and is 28 on the night shift. A sample of 54 day-shift workers showed that the mean number of units produced was 345\. A sample of 60 night-shift workers showed that the mean number of units produced was \(351 .\) At the .05 significance level, is the number of units produced on the night shift larger?

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1} \leq \pi_{2} \\ H_{1}: \pi_{1}>\pi_{2} \end{array} $$ A sample of 100 observations from the first population indicated that \(X_{1}\) is \(70 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(90 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis? e. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1}=\pi_{2} \\ H_{1}: \pi_{1} \neq \pi_{2} \end{array} $$

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) value and interpret it.

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

As part of a recent survey among dual-wage-earner couples, an industrial psychologist found that 990 men out of the 1,500 surveyed believed the division of household duties was fair. A sample of 1,600 women found 970 believed the division of household duties was fair. At the .01 significance level, is it reasonable to conclude that the proportion of men who believe the division of household duties is fair is larger? What is the \(p\) -value?
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