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 random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. Test the claim that men are more safety-conscious than women, at \(\alpha=0.01\). Use the \(P\) -value method.

Short answer

Men are statistically more safety-conscious than women.

Step-by-step solution

State the Hypotheses and Identify the Claim

First, define the populations and claim. We are comparing two proportions: the proportion of men using seat belts (\( p_1 \)) and the proportion of women using seat belts (\( p_2 \)). The null hypothesis (\( H_0 \)) states that \( p_1 = p_2 \). The alternative hypothesis (\( H_a \)) states that \( p_1 > p_2 \), which aligns with the claim that men are more safety-conscious than women.

Calculate Sample Proportions

Calculate the sample proportions for each group. For men: \( \hat{p}_1 = \frac{130}{200} = 0.65 \). For women: \( \hat{p}_2 = \frac{63}{300} = 0.21 \).

Find the Critical Values

For \( \alpha = 0.01 \) and a one-tailed test, find the critical value using the standard normal distribution (z-distribution). The critical value \( z_c \) for \( \alpha = 0.01 \) is approximately 2.33.

Compute the Test Statistic

First, find the pooled proportion \( \hat{p} \): \[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{130 + 63}{200 + 300} = 0.386.\]Then calculate the standard error:\[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.386 \times 0.614 \times \left(\frac{1}{200} + \frac{1}{300}\right)} = 0.043.\]Compute the z-test statistic:\[z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.65 - 0.21}{0.043} \approx 10.0.\]

Make the Decision

Compare the computed z-test value to the critical z-value. Since the computed z (\( 10.0 \)) is greater than the critical value (\( 2.33 \)), we reject the null hypothesis.

Summarize the Results

Since the null hypothesis is rejected, there is enough statistical evidence to support the claim that men are more safety-conscious than women, at the \( \alpha = 0.01 \) level.

Key concepts

Null Hypothesis

The null hypothesis is a fundamental part of hypothesis testing, often denoted as \( H_0 \). It represents the default position or statement that there is no effect or no difference. In this scenario, the null hypothesis asserts that the proportion of men using seat belts is equal to the proportion of women, or mathematically, \( p_1 = p_2 \).

This acts as the starting point for statistical testing. To challenge or reject this hypothesis, we gather evidence through data analysis. Only significant statistical results can lead us to reject the null hypothesis. In our exercise, rejecting \( H_0 \) would imply there's enough evidence to say men (\( p_1 \)) are more safety-conscious than women (\( p_2 \)).

Alternative Hypothesis

Contrary to the null hypothesis, the alternative hypothesis represents what we aim to support with our data. This is often denoted as \( H_a \). For this problem, the alternative hypothesis is that the proportion of men using seat belts is greater than that of women, or \( p_1 > p_2 \).

The alternative hypothesis is what researchers want to prove. It suggests a real effect or difference, supporting the initial claim. To accept \( H_a \), statistical methods compare the observed data to what we'd expect if \( H_0 \) were true. Here, it leads to testing whether there is substantial evidence that men are more safety-conscious.

P-value Method

The \( P \)-value method provides a way to make statistical decisions by determining the probability of observing results as extreme as the ones obtained if \( H_0 \) were true. In hypothesis testing, the \( P \)-value helps decide whether to reject \( H_0 \). A small \( P \)-value, typically less than the chosen significance level \( \alpha \), suggests that the observed effect is statistically significant.

In our exercise, if the \( P \)-value is less than \( \alpha = 0.01 \), the null hypothesis will be rejected. This method gives a probabilistic measure that helps statisticians and researchers confidently state the chances of their observations occurring under the null hypothesis.

Critical Value

A critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject \( H_0 \). Derived from the significance level \( \alpha \), it acts as a threshold. For a \( 0.01 \) significance level in a one-tailed test, the critical z-value is approximately 2.33.

This means that if our test statistic is greater than 2.33, we have enough evidence to reject the null hypothesis under the assumption of statistical significance. Critical values form a barrier which, if crossed by the test statistic, leads to rejecting \( H_0 \).

Test Statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. In many tests, this statistic follows a known distribution and is used to compare against the critical value. For our exercise, a z-test statistic is used to analyze the differences in proportions.

Calculating this involves several steps:

• Find the sample proportions for men and women.
• Calculate the pooled proportion.
• Determine the standard error based on the pooled proportion.
• Compute the z-test statistic using these values.

Finally, the calculated z-value (approximately 10.0) significantly exceeded the critical z-value (2.33), reinforcing the decision to reject the null hypothesis.