Question

The numbers of successes and the sample sizes for independent simple random samples from two populations are x 1equals32​, n 1equals40​, x 2equals10​, n 2equals20. a. Use the​ two-proportions plus-four​ z-interval procedure to find an​ 95% confidence interval for the difference between the two populations proportions. b. Compare your result with the result of a​ two-proportion z-interval​ procedure, if finding such a confidence interval is appropriate.

Answer 1

X1 = 32

n1 = 40

X2 = 10

n2 = 20

a. the two proportion plus four z-interval

VI+22+2

p1 = (X1 +1)/(n1+2)

= (32+1)/(40+2)

= 0.7857

p2= (X2+1)/(n2+2)

= (10+1)/(20+2)

= 0.5

Z = 1.96 at 95 % confidence level for two tail test

so, (0.7857-0.5) $\pm$ 1.96 * 0.124

= 0.2857 $\pm$ 0.24304

= 0.04266 to 0.52874

b. two proportion z-interval

丬: 2

P1 hat = 32/40 = 0.8

p2 hat = 0.5

n1 = 40

n2 = 20

z = 1.96

so, (0.80-0.50) $\pm$ 1.96 * 0.12845

= 0.3 $\pm$ 0.251762

= 0.048238 to 0.551762

so, both confidence interval are same so, confidence interval is appropriate of two-proportions plus-four​ z-interval