Question

Construct the indicated confidence interval for the difference in proportions. Assume that the samples are independent and that they have been randomly selected. In a random sample of 300 women, 70% favored stricter gun control legislation. In a random sample of 200 men, 56% favored stricter gun control legislation. Construct a 98% confidence interval for the difference in the proportions of women and men who favor stricter gun control legislation.

Answer 1

Solutions We need to constouct 98%, confidence interval for the difference in the proportions of women and men who favor Strickes gun control legislation. Now, Sample proportion (P, 1:0.70 and sample size (ne) - 300 Sample propostion () = 0.56 and Sample size (m)=200 The critical value for dc 0.02 and Z = Zid = 2326 The corresponding confidence interval is computed as shown below: (1=((R)- Ze v85 (0-2); BACA),(A-2) + 76 / 8 (4:+ C 40) ni h2 0.7 (1-0.7) 300 0.56 (1-0.56) 200 (607–0.5) 22:326 N 0-750 0.78 + 0562070.50) 6 V -(0.038,0.242) Therefore, based on the data provided , the 98%. confidence intervcel for the difference between the population proportion ProPro is 0.038 LP7224 0.242, which indicates that we are 984, confident that the true difference between proportion of women send men favour strickes gan control legislation is contain by the interval (0.038, 0.942).