Question

include null and alternate hypothesis, a p-value, and a conclusion PRIVATE about the null hypothesis

In a recent survey of gun control laws, a random sample of 1000 women showed that 650 were in favor of stricter gun laws. In a random sample of 1000 men, 600 favored stricter gun control laws. Determine a 90% confidence interval for the difference between the proportion of women and men who favor stricter gun laws.

Answer 1

Answer)

N1 = 1000, P1 = 650/1000

N2 = 1000, P2 = 600/1000

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not.

N1*p1 = 650

N1*(1-p1) = 350

N2*p2 = 600

N2*(1-p2) = 400

All the conditions are met so we can use standard normal z table to construct the interval.

Margin of error (MOE) = Z*Standard error

Critical value z from z table, for 90% confidence level is 1.645.

Standard error =√ [{(p1*(1-p1)}/n1 + {(p2*(1-p2)}/n2]

After substitution

MOE = 0.03556777597

Interval is given by

(P1-P2) - moe < (P1-P2) < (P1-P2) + MOE

0.01443222402 < (P1-P2) < 0.08556777597