Question

A 2005 study investigated hand washing behaviour in public washrooms in two major US cities: Atlanta and New York. A random observation of 1588 people in Atlanta found that 74% of them washed their hands before leaving a public washroom, whereas a random observation of 1503 people in New York found that 77% of them washed their hands before leaving a public washroom. Test the claim that the proportion of hand washers is greater in New York than in Atlanta. For each significance level below determine if your null hypothesis should be rejected at the stated value. • For a = 0.01 we Choose... • For a = 0.025 we Choose... keep the null hypothesis. • For a = 0.05 we reject the null hy thesis. • For a = 0.10 we Choose...

Answer 1

Sol:

p1cap = X1/N1 = 1175/1588 = 0.7399
p1cap = X2/N2 = 1161/1503 = 0.7725
pcap = (X1 + X2)/(N1 + N2) = (1175+1161)/(1588+1503) = 0.7557

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 = (0.7399-0.7725)/sqrt(0.7557*(1-0.7557)*(1/1588 + 1/1503))
z = -2.11

P-value Approach
P-value = 0.0174
As P-value < 0.05, reject the null hypothesis.

there is sufficient evidence to conclude that the proportion of hand washers is greater in new york