Question

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2771occupants not wearing seat​ belts, 38 were killed. Among 7702

occupants wearing seat​ belts, 15 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.

a) identify test statistic

b) identify the p-value

c) identify confidence interval

Answer 1

= 38/2771 = 0.0137

= 15/7702 = 0.0019

a) The pooled sample proportion(P) = ( * n1 + * n2)/(n1 + n2)

                                                          = (0.0137 * 2771 + 0.0019 * 7702)/(2771 + 7702)

                                                          = 0.005

The test statistic z = ()/sqrt(P(1 - P)(1/n1 + 1/n2))

                             = (0.0137 - 0.0019)/sqrt(0.005 * (1 - 0.005) * (1/2771 + 1/7702))

                             = 7.55

b) P-value = P(Z > 7.55)

                  = 1 - P(Z < 7.55)

                 = 1 - 1 = 0

c) At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval is

() +/- z_0.005 * sqrt(P(1 - P)(1/n1 + 1/n2))

= (0.0137 - 0.0019) +/- 2.58 * sqrt(0.005 * (1 - 0.005) * (1/2771 + 1/7702))

= 0.0118 +/- 0.004

= 0.0078, 0.0158