Question

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2964 occupants not wearing seat​ belts, 30 were killed. Among 7706 occupants wearing seat​ belts, 13 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Find the test statistic Z to 3 decimal places, the P-value to 3 decimals, and find the confidence interval.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

p1cap = X1/N1 = 30/2964 = 0.0101
p1cap = X2/N2 = 13/7706 = 0.0017
pcap = (X1 + X2)/(N1 + N2) = (30+13)/(2964+7706) = 0.004

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.0101-0.0017)/sqrt(0.004*(1-0.004)*(1/2964 + 1/7706))
z = 6.16

P-value Approach
P-value = 0

As P-value < 0.05, reject the null hypothesis.

Here, , n1 = 2964 , n2 = 7706
p1cap = 0.0101 , p2cap = 0.0017

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.0101 * (1-0.0101)/2964 + 0.0017*(1-0.0017)/7706)
SE = 0.0019

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.0101 - 0.0017 - 1.96*0.0019, 0.0101 - 0.0017 + 1.96*0.0019)
CI = (0.005 , 0.012)