Question

A simple random sample of size n is drawn. The sample mean, , is found to be 19.2, and the sample standard deviation, s, is found to be 4.6. Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about u if the sample size, n, is 35. Lower bound: I; Upper bound: (Use ascending order. Round to two decimal places as needed.) (b) Construct a 95% confidence interval about u if the sample size, n, is 61. Lower bound: ; Upper bound: (Use ascending order. Round to two decimal places as needed.) How does increasing the sample size affect the margin of error, E? A. The margin of error increases B. The margin of error decreases C. The margin of error does not change. (c) Construct a 99% confidence interval about u if the sample size, n, is 35. Lower bound: ; Upper bound: (Use ascending order. Round to two decimal places as needed.)

Answer 1

The solution to this problem is given by

x + tots In I txiga toness (34) Solution Given h = E 19.9 Samble SD S-4.6 ca The 95% CI for M is Given by a t where na 35 19.2 (2.032 4.6 J35 2.032 19.2 I 1058 ( 19.2 – 1:58.119.2.+ 1:58) (17062 Jo:78) Lowel Bound = 17-62 Upper Bound To o78 (bn The 95% CI far er is given by X & nool 160) 19.2 + 2x 4.6 6.025 2.000 xdig where is 15 trig t

20.38 Xt2 un 19.2 + 1.18 (19:22 1.18 19.8+1:18) - ( 18.08, 20.38) Lower Bound 18.02 Upper Bound - Comparing these two intervals we get The margin & elror decreases. (c) The 991. CI for he is given by ł S where txig = tooos 2.728 19.2 79.278 x 4.6 n=35 tooog (34) (19.2 + 1.34) (19.2 - 1.34,19-2 +) ~34) = (17.86, do -54) Lavel Bound - 17-86 Upher Bound 11 20.54 (20.54

with part (a), ua get increases le For B, the conditions Comparing (A) The margin of error must be satisfied is 10). The data muest come from a lodculation that is natinally distributed with outliers. no