Question

A survey found that 50% of cats and 60% of dogs have pet licenses in the city of Saskatoon. These results were based on random samples of 100 cats and 100 dogs. Test at the 5% significance level that proportion of cats that have licenses is less than the proportion of dogs that have licenses. What is the p-value and what is your conclusion?

Answer 1

p1cap = X1/N1 = 50/100 = 0.5
p1cap = X2/N2 = 60/100 = 0.6
pcap = (X1 + X2)/(N1 + N2) = (50+60)/(100+100) = 0.55

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 = (0.5-0.6)/sqrt(0.55*(1-0.55)*(1/100 + 1/100))
z = -1.42

P-value Approach
P-value = 0.0778

Not sufficient evidence to conclude that the cats that have licenses is less than the proportion of dogs that have licenses.