Question

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. In a sample of 150 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

Reject the null hypothesis; there is a significant difference. The confidence interval does not contain 0, indicating a difference.

Step-by-step solution

State the Hypotheses and Identify the Claim

Start by defining the null and alternative hypotheses. Let \( p_1 \) be the proportion of men who have less leisure time, and \( p_2 \) be the proportion of women who have less leisure time. The null hypothesis \( H_0 \) is that there is no difference in the proportions: \( H_0: p_1 = p_2 \). The alternative hypothesis \( H_1 \) claims there is a difference: \( H_1: p_1 eq p_2 \).

Find the Critical Value(s)

For a two-tailed test with \(\alpha = 0.10\), find the critical value using the standard normal distribution (Z). The critical values for a \(90\%\) confidence level are \(Z = \pm 1.645\).

Compute the Test Value

Calculate the test statistic using the formula:\[Z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2}) }}\]Where \(\hat{p}_1 = \frac{132}{150}\), \(\hat{p}_2 = \frac{240}{250}\), and \(\hat{p}\) is the pooled sample proportion:\[ \hat{p} = \frac{132 + 240}{150 + 250} = \frac{372}{400} = 0.93 \]Plug these into the formula to compute \(Z\):\[Z = \frac{(0.88 - 0.96)}{\sqrt{0.93 \times 0.07 \times \left(\frac{1}{150} + \frac{1}{250}\right)}} = \frac{-0.08}{\sqrt{0.0005824}} \]Complete the calculation to find \(Z\approx -2.60\).

Make the Decision

Compare the calculated \(Z\)-value to the critical values. Since \(-2.60\) is less than \(-1.645\), we reject the null hypothesis \(H_0\).

Summarize the Results

There is sufficient evidence at the \(\alpha = 0.10\) level to conclude that there is a significant difference in the proportion of men and women who report having less leisure time than they did 10 years ago.

Confidence Interval for Difference of Proportions

Calculate the \(90\%\) confidence interval using:\[(p_1 - p_2) \pm Z \times \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\]With \(p_1 = 0.88\) and \(p_2 = 0.96\), plug values to find the interval:\[-0.08 \pm 1.645 \times 0.034\].The resulting confidence interval is \([-0.136, -0.024]\), which does not include 0.

Interpretation for Researchers

The fact that the confidence interval does not contain 0 indicates a difference in the populations. A researcher may be interested in this result to explore underlying social or economic factors contributing to the reported differences in leisure time between genders.

Key concepts

Null and Alternative Hypotheses

The null and alternative hypotheses are foundational concepts in hypothesis testing. They serve as the primary blueprints for testing statistical claims. When you state the null hypothesis (\( H_0 \)), you are assuming that there is no effect or difference. It serves as the default or starting assumption that nothing significant is happening in your data. In mathematical terms, for this exercise, the null hypothesis is:\[H_0: p_1 = p_2\]This suggests that the proportion of men (\( p_1 \)) and women (\( p_2 \)) reporting less leisure time is equal.

On the other hand, the alternative hypothesis (\( H_1 \)) is what you seek evidence for. It represents the claim or the difference we are testing for. In this case, the alternative hypothesis is:\[H_1: p_1 eq p_2\]This indicates that there is, indeed, a difference in proportions between the two groups. Identifying these hypotheses correctly is crucial, as they form the basis of your entire test.

Critical Value

In hypothesis testing, the critical value is like a benchmark. It determines the line between the "safe" region where we fail to reject the null hypothesis, and the "critical" region where we reject it. Critical values are influenced by the level of significance (\( \alpha \)), which is the probability of rejecting the null hypothesis when it is actually true. It often represents the risk you are willing to take to make an incorrect decision.

For a two-tailed test with a significance level of \( \alpha = 0.10 \), the critical values are derived from a standard normal distribution (Z-distribution). A \( 90\% \) confidence level leads to critical values of:

• \( Z = +1.645 \)
• \( Z = -1.645 \)

The test statistic needs to fall beyond these values in either tail of the distribution for the null hypothesis to be rejected. Understanding how to find and use critical values enables you to make informed decisions based on your test results.

Test Statistic

The test statistic is a calculated value that allows you to decide whether to reject the null hypothesis. It essentially measures how far your sample data is from the null hypothesis. In our specific exercise, we deal with proportions, so the test statistic is calculated using the Z-formula for the difference between two proportions:
\[Z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2}) }}\]
Where:

• \( \hat{p}_1 \) and \( \hat{p}_2 \): Sample proportions for men and women respectively
• \( \hat{p} \): Pooled proportion of both samples combined
• \( n_1 \) and \( n_2 \): Sample sizes for men and women respectively

For our scenario, the calculated Z-value is approximately \( -2.60 \). This value will be compared to our critical values to decide whether there is a significant difference in the population proportions.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter to lie, based on our sample data. It offers a probability range, commonly \( 90\% \), \( 95\% \), or \( 99\% \), indicating how confident we are that this interval contains the true population parameter.

For the difference in proportions between men and women regarding leisure time, the confidence interval is calculated as:
\[(p_1 - p_2) \pm Z \times \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\]
In our example, the resulting interval is \([-0.136, -0.024]\) for a \( 90\% \) confidence level. This range does not include 0, suggesting that there is a significant difference in the proportions. This information can be crucial for researchers, providing evidence that the observed difference is not due to random error, but may reflect actual differences in the population.