Question

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2997 occupants not wearing seat​ belts, 32 were killed. Among 7649 occupants wearing seat​ belts, 10 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities.

- Identify the test statistic?

- Identify the p value?

- Test the claim by constructing an appropriate confidence level?

- What is the conclusion base on the hypothesis test?

- What is the conclusion base on the confidence level?

Answer 1

Two sample proportion test will be applied here.

Z-test statistic=6.94

p-value=0.000

MINITAB OUTPUT:

Test and CI for Two Proportions

Sample X N Sample p
1 32 2997 0.010677
2 10 7649 0.001307

Difference = p (1) - p (2)
Estimate for difference: 0.00936998
95% lower bound for difference: 0.00620804
Test for difference = 0 (vs > 0): Z = 6.94 P-Value = 0.000

Since p-value is 0, we can strongly reject the null. We can do conclude that the seat belts are effective in reducing fatalities.

Since the confidence interval is (0.00620804,0) which does not inclue 0, we support the alternative and can conclude the same