Questions: A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2937 occupants not wearing seat belts, 27 were killed. Among 7830 occupants wearing seat belts, 14 were Killed. Use a 0.01 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts (a) through (c) below. H0: P1=P2 H1: P1 ≠ P2 Identify the test statistic. z= (Round to two decimal places as needed.) Identify the P-value. P -value = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P -value is less than the significance level of α=0.01, so reject the null hypothesis. There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts. Test the claim by constructing an appropriate confidence interval. The appropriate confidence interval is <(p1-p2)< . (Round to three decimal places as needed.)

Solution Steps

We are testing the claim that seat belts are effective in reducing fatalities in car crashes. The null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \) are defined as follows:

• \( H_0: P_1 = P_2 \) (The fatality rates for occupants not wearing seat belts and those wearing seat belts are equal.)
• \( H_1: P_1 > P_2 \) (The fatality rate for occupants not wearing seat belts is higher.)

The sample proportions for the two groups are calculated as follows:

\[ p_1 = \frac{27}{2937} \approx 0.0092 \] \[ p_2 = \frac{14}{7830} \approx 0.0018 \]

The pooled proportion \( p_{\text{pool}} \) is calculated as:

\[ p_{\text{pool}} = \frac{27 + 14}{2937 + 7830} \approx 0.0038 \]

The standard error (SE) of the difference between the two proportions is given by:

\[ SE = \sqrt{p_{\text{pool}} \left(1 - p_{\text{pool}}\right) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx 0.0013 \]

The Z-statistic is calculated using the formula:

\[ z = \frac{p_1 - p_2}{SE} \approx \frac{0.0092 - 0.0018}{0.0013} \approx 5.5565 \]

The P-value associated with the Z-statistic is calculated. For a Z-statistic of \( 5.5565 \), the P-value is:

\[ \text{P-value} \approx -9.113 \]

Since the P-value \( -9.113 \) is less than the significance level \( \alpha = 0.01 \), we reject the null hypothesis \( H_0 \). There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.

To construct a confidence interval for the difference between the two population proportions at a \( 99\% \) confidence level, we use the formula:

\[ (p_1 - p_2) \pm z \cdot \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} \]

Using \( z \approx 2.576 \) for \( 99\% \) confidence, we calculate:

\[ (0.0092 - 0.0018) \pm 2.576 \cdot \sqrt{\frac{0.0092(1 - 0.0092)}{2937} + \frac{0.0018(1 - 0.0018)}{7830}} \approx (0.003, 0.012) \]

Final Answer