Question

Suppose a simple random sample of 26 male students is asked whether they have pulled an "all-nighter " for academic reasons and 10 answered "yes." Suppose a simple random sample of 25 female students is asked whether they have pulled an "all-nighter" for academic reasons and 8 answered "yes."

Using a 0.06 significance level, construct a confidence interval for the claim that male and female students are equally likely to have pulled an "all-nighter" for academic reasons?

Answer 1

p1cap = X1/N1 = 10/26 = 0.3846
p1cap = X2/N2 = 8/25 = 0.32
pcap = (X1 + X2)/(N1 + N2) = (10+8)/(26+25) = 0.3529

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 = (0.3846-0.32)/sqrt(0.3529*(1-0.3529)*(1/26 + 1/25))
z = 0.48

P-value Approach
P-value = 0.6312
As P-value >= 0.06, fail to reject null hypothesis.

Yes, females are equally likely