a. Since it is a few selected 1000 Americans in the survey, it is more of a sample thing and it's too small to determine for whole of america.
b. Standard deviation = square root of p(1-p)/n, p is 39/100, n = 1000 SD =square root of 0.39 (1 - 0.39) / 1000 => SD = square root of 0.2379 /
1000
SD = 0.0154
c. From the z value table critical avlue for a 90% confidence interval is 1.645
d. From the z value table critical avlue for a 95% confidence interval is 1.96
e. For 90% z = 1.645, P = 0.39, SD = 0.0154
CI = P +- z(SD) => CI = 0.39 +- 1.645(0.154) => 0.39 +- 0.0253
So the interval is (0.364, 0.415)
f. For 95% z = 1.96, P = 0.39, SD = 0.0154
CI = P +- z(SD) => CI = 0.39 +- 1.96(0.154) => 0.39 +- 0.0302
So the interval is (0.359, 0.420)