Question

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2991 occupants not wearing seat​ belts, 28 were killed. Among 7768 occupants wearing seat​ belts, 16 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.

b. Identify the test statistic

c. P-value

d. Confidence interval

Answer 1

a)
p1cap = X1/N1 = 28/2991 = 0.0094
p1cap = X2/N2 = 16/7768 = 0.0021
pcap = (X1 + X2)/(N1 + N2) = (28+16)/(2991+7768) = 0.0041

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

b)
Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.0094-0.0021)/sqrt(0.0041*(1-0.0041)*(1/2991 + 1/7768))
z = 5.31

c)
P-value Approach
P-value = 0.0000
As P-value < 0.05, reject the null hypothesis.
There is sufficient evidence to conclude that the seat belts are effective in reducing fatalities.

d)
Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.0094 * (1-0.0094)/2991 + 0.0021*(1-0.0021)/7768)
SE = 0.0018

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.0094 - 0.0021 - 1.96*0.0018, 0.0094 - 0.0021 + 1.96*0.0018)
CI = (0.0038 , 0.0108)