Question

1. In a study of police gunfire reports during a recent year, it was found that among 540 shots fired by New York City police, 182 hit their targets; and among 283 shots fired by Los Angeles police, 77 hit their targets.

a. Use a 0.05 significance level to tes t the claim that New York City police and Los Angeles

police have different proportion of hits.

b. Construct a 90 % confidence interval to estimate the difference between the two

proportions of hits.

Answer 1

#1.
p1cap = X1/N1 = 182/540 = 0.337
p1cap = X2/N2 = 77/283 = 0.2721
pcap = (X1 + X2)/(N1 + N2) = (182+77)/(540+283) = 0.3147

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

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.337-0.2721)/sqrt(0.3147*(1-0.3147)*(1/540 + 1/283))
z = 1.9

P-value Approach
P-value = 0.0574
As P-value >= 0.05, fail to reject null hypothesis.

There is significant evidence to conclude that there is no difference in the two proportions

#2.
Here, , n1 = 540 , n2 = 283
p1cap = 0.337 , p2cap = 0.2721

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.337 * (1-0.337)/540 + 0.2721*(1-0.2721)/283)
SE = 0.0334

For 0.9 CI, z-value = 1.64
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.337 - 0.2721 - 1.64*0.0334, 0.337 - 0.2721 + 1.64*0.0334)
CI = (0.0101 , 0.1197)