Question

g A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2950 occupants not wearing seat​ belts, 39 were killed. Among 7822 occupants wearing seat​ belts, 20 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts​ (a) through​ (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of occupants not wearing seat belts and the second sample to be the sample of occupants wearing seat belts. What are the null and alternative hypotheses for the hypothesis​ test? A. Upper H 0​: p 1less than or equalsp 2 Upper H 1​: p 1not equalsp 2 B. Upper H 0​: p 1not equalsp 2 Upper H 1​: p 1equalsp 2 C. Upper H 0​: p 1equalsp 2 Upper H 1​: p 1not equalsp 2 D. Upper H 0​: p 1equalsp 2 Upper H 1​: p 1less thanp 2 E. Upper H 0​: p 1greater than or equalsp 2 Upper H 1​: p 1not equalsp 2 F. Upper H 0​: p 1equalsp 2 Upper H 1​: p 1greater thanp 2 Identify the test statistic. zequals 5.45 ​(Round to two decimal places as​ needed.) Identify the​ P-value. ​P-valueequals 0.000 ​(Round to three decimal places as​ needed.) What is the conclusion based on the hypothesis​ test?

Answer 1

Answer:

The null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

Test statistic z = 6.687

P-value = 0

The null hypothesis is rejected.

There is enough evidence to support the claim that seat belts are effective in reducing fatalities.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that seat belts are effective in reducing fatalities.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

The significance level is assumed to be 0.05.

The sample 1, of size n1=2950 has a proportion of p1=0.0132.

$p_1=X_1/n_1=39/2950=0.0132$

The sample 2, of size n2=7822 has a proportion of p2=0.0026.

$p_2=X_2/n_2=20/7822=0.0026$

The difference between proportions is (p1-p2)=0.0107.

p_d=p_1-p_2=0.0132-0.0026=0.0107

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{39+20}{2950+7822}=\dfrac{59}{10772}=0.0055$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.0055*0.9945}{2950}+\dfrac{0.0055*0.9945}{7822}}\\\\\\s_{p1-p2}=\sqrt{0+0}=\sqrt{0}=0.0016$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.0107-0}{0.0016}=\dfrac{0.0107}{0.0016}=6.687$

This test is a right-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(t>6.687)=0$

As the P-value (0) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that seat belts are effective in reducing fatalities.