Question

In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of 220 men, 60% favored stricter gun control laws. Test the claim that the proportion of women favoring stricter gun control is higher than the proportion of men favoring stricter gun control. Use a significance level of 0.05.

Answer 1

In a random sample of 360 women, 65% favored stricter gun control laws

n1 = 360, p1 = 0.65 and X1 = 234

In a random sample of 220 men, 60% favored stricter gun control laws

n2 = 220, p2 = 0.60 and X2 = 132

significance level of 0.05.

Minitab > Stat > Basic stat > 2 proportions

————— 06-05 21:56:23 ————————————————————

Test and CI for Two Proportions

Sample X N Sample p
1 234 360 0.650000
2 132 220 0.600000

Difference = p (1) - p (2)
Estimate for difference: 0.05
95% lower bound for difference: -0.0182734
Test for difference = 0 (vs > 0): Z = 1.20 P-Value = 0.114

Fisher’s exact test: P-Value = 0.131

P value > 0.05, Do not reject H0

Therefore, there is not enough evidence to conclude that the proportion of women favoring stricter gun control is higher than the proportion of men favoring stricter gun control at 5% significance level.

Formulas:

Z = (P1-P2)/SQRT(Pbar*(1-Pbar)*(1/n1+1/n2))

Pbar = (X1+X2)/(n1+n2)

P value from Calculator or Z table