Question

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Leisure Time In a sample of \(150 \mathrm{men}, 132\) said that they had less leisure time today than they had 10 years ago. In a random sample of 250 women, 240 women said that they had less leisure time than they had 10 years ago. At \(\alpha=0.10,\) is there a difference in the proportions? Find the \(90 \%\) confidence interval for the difference of the two proportions. Does the confidence interval contain \(0 ?\) Give a reason why this information would be of interest to a researcher.

Short answer

There is a significant difference in the proportions of men and women with less leisure time. The 90% confidence interval is (-0.129, -0.031), not containing 0.

Step-by-step solution

State the Hypotheses

We begin by stating our null and alternative hypotheses: - Null Hypothesis (H_0): There is no difference in the proportions of men and women who have less leisure time today compared to 10 years ago. Mathematically, this is expressed as \( p_1 = p_2 \) where \( p_1 \) is the proportion of men and \( p_2 \) is the proportion of women.- Alternative Hypothesis (H_a): There is a difference in the proportions, expressed as \( p_1 eq p_2 \). The claim in the context of the problem is that there is a difference in proportions.

Find the Critical Value

At \(\alpha = 0.10\), we will use the standard normal distribution to find the critical value for a two-tailed test. With a significance level of 0.10, each tail will have an area of 0.05.The critical z-value for 0.05 in each tail can be found from the z-table: \( z_{\alpha/2} = \pm 1.645 \). These are the critical values for making our decision.

Compute the Test Value

The sample proportions are:\( \hat{p}_1 = \frac{132}{150} = 0.88 \)\( \hat{p}_2 = \frac{240}{250} = 0.96 \)The pooled proportion \( \hat{p} \) is used as follows:\[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{132 + 240}{150 + 250} = \frac{372}{400} = 0.93\]The test statistic formula:\[z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} = \frac{(0.88 - 0.96)}{\sqrt{0.93 \cdot 0.07 \left(\frac{1}{150} + \frac{1}{250}\right)}}\]Calculating the above expression gives:\( z \approx -2.667 \).

Make the Decision

The calculated test value is \(-2.667\). Since \(-2.667\) falls outside the range \(-1.645\) to \(1.645\), it lies in the rejection region. Therefore, we reject the null hypothesis \( H_0 \).

Summarize the Results

By rejecting the null hypothesis, we conclude that there is a statistically significant difference in the proportions of men and women who have less leisure time today compared to 10 years ago.

Confidence Interval for Difference of Proportions

The formula for a confidence interval for the difference \( p_1 - p_2 \) is:\[(\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\]\( z_{\alpha/2} = 1.645 \)\[(0.88 - 0.96) \pm 1.645 \times \sqrt{\frac{0.88 \times 0.12}{150} + \frac{0.96 \times 0.04}{250}}\]Calculating this, the confidence interval is approximately \((-0.129, -0.031)\).

Analyze the Confidence Interval

The confidence interval \((-0.129, -0.031)\) does not contain 0, indicating a significant difference in proportions. This would be of interest to a researcher as it verifies that the proportions are indeed not equal and any observed differences are significant at this confidence level.

Key concepts

Critical Value

When conducting a hypothesis test, the *critical value* is an essential component that helps to determine the decision boundary. In our example, we're using a significance level, equal to \( \alpha = 0.10 \), which is a common choice for such studies where the strictest precision is not crucial. The critical value is found based on the selected significance level in conjunction with the corresponding distribution, which is the standard normal distribution in this case.
For a two-tailed test like ours, the total significance level \( \alpha = 0.10 \) is split equally between the two tails of the distribution. Thus, each tail will have an area of 0.05.
The critical value that corresponds to this distribution cutoff is derived from the z-table and is found to be \( \pm 1.645 \). This means any calculated test statistic beyond this range will lead us to reject the null hypothesis.

• If the test statistic exceeds 1.645 or is below -1.645, the null hypothesis can be rejected.
• This zone is known as the rejection region.

Test Statistic

The *test statistic* is the calculated value from the sample data that will be compared against the critical value to decide whether to reject the null hypothesis. In the context given, we are comparing two proportions to evaluate if there is a significant difference between them.
For our specific problem, we first derive sample proportions: \( \hat{p}_1 = \frac{132}{150} \) for men and \( \hat{p}_2 = \frac{240}{250} \) for women. These sample proportions are used to calculate the test statistic.
The pooled proportion, \( \hat{p} \), is also calculated from both samples, giving a more accurate test statistic under the null hypothesis assumption:

• Formula used: \[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}\]

The test statistic formula incorporates \( \hat{p}_1 \), \( \hat{p}_2 \), and \( \hat{p} \):\[z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}\]Calculating the above gives us a test statistic value of approximately \( z \approx -2.667 \). Since this value falls outside the range of our critical value (1.645), it indicates a significant difference.

Confidence Interval

A *confidence interval* provides a range in which we can say with a specified level of confidence that the true parameter lies. In hypothesis testing, particularly in comparing proportions, it's a helpful tool.
For our exercise, we are asked to compute a 90% confidence interval for the difference between two proportions. This involves using a formula that accounts for each sample proportion, their respective sample sizes, and the selected confidence level:

• Formula:\[(\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\]
• The critical value, \( z_{\alpha/2} = 1.645 \), is consistent across the test.

Calculating using this formula gives us the interval to be approximately \((-0.129, -0.031)\). Since this interval does not include 0, it signifies a significant difference in the proportions.

Proportions

In hypothesis testing, analyzing *proportions* involves understanding the distribution of a population characteristic when divided into two parts. In the exercise, the focus is on comparing the proportions of men and women who feel they have less leisure time than before.
Proportions are denoted by \( p \), with \( p_1 \) representing the men's proportion and \( p_2 \) representing the women's proportion.

• Men's sample proportion: \( \hat{p}_1 = \frac{132}{150} \equiv 0.88 \)
• Women's sample proportion: \( \hat{p}_2 = \frac{240}{250} \equiv 0.96 \)

Comparing these proportions helps us understand differences in leisure time perceptions. Moreover, it guides us in calculating the test statistic and confidence intervals. If the proportions are identical, they suggest no difference between the groups; deviations indicate potential disparities, a central focus in such analyses.

Significance Level

The *significance level*, often denoted by \( \alpha \), represents the probability of rejecting a true null hypothesis, thus quantifying the risk level of drawing a wrong conclusion. It's a balancing act between sensitivity of a test and avoiding false positives.
In hypothesis testing, this level is an essential parameter. In the given problem, \( \alpha = 0.10 \), indicating a 10% risk of a Type I error. This is a moderate level where strict certainty isn't required, still maintaining the notion of caution.

• The significance level affects the critical value. For a given test, it establishes the threshold.
• A lower \( \alpha \) indicates a stronger threshold for rejecting the null hypothesis, necessitating stronger evidence.

Analyzing results considering the significance level helps you identify if observed data differences result from mere chance or indicate a true difference, refining research insights.