Assume that a sample is drawn and z(?/2) = 1.96 and ? = 20. Answer the following questions: (A)If the Maximum Error of Estimate is 0.02 for this sample, what would be the sample size? (B)Given that the sample Size is 400 with this same z(?/2) and ?, what would be the Maximum Error of Estimate? (C)What happens to the Maximum Error of Estimate as the sample size gets smaller? (D)What effect does the answer to C above have to the size of the confidence interval? Solution (A)z = 1.96 s = 10 E = 0.01 n = (z*s/E)^2 = (1.96*10/0.01)^2 = 3841600 (B)n = 400 E = z*s/vn = 1.96*10/v400 = 0.98 (C)As the sample size becomes smaller the maximum error E increases. (D)The width of the confidence interval is 2E and it increases as the sample size decreases..