Question

A simple random sample of size n is drawn. The sample​ mean,x overbarx​, is found to be 17.8 and the sample standard​ deviation, s, is found to be 4.4

​(a) Construct a 95% confidence interval about μ if the sample​ size, n, is 35

Lower​ bound: ____ Upper​ bound: ______ ​(Use ascending order. Round to two decimal places as​ needed.)

(b)

Construct a 95% confidence interval about μ if the sample​ size, n, is 51

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 does not change.

B.The margin of error increases.

C.The margin of error decreases.

(c) Construct a 99​% confidence interval about μ if the sample​ size, n, is 35

Lower​ bound:_____Upper​ bound: ____(Use ascending order. Round to two decimal places as​ needed.)

Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E?

A.The margin of error decreases.

B. The margin of error does not change.

C. The margin of error increases.

​(d) If the sample size is 15, what conditions must be satisfied to compute the confidence​ interval?

A.The sample must come from a population that is normally distributed and the sample size must be large.

B.The sample size must be large and the sample should not have any outliers.

C. The sample data must come from a population that is normally distributed with no outliers.

Answer 1

Mean, = 17.8

Standard deviation, = 4.4

Corresponding to 95% confidence interval, the z score = 1.96

Corresponding to 99% confidence interval, the z score = 2.575

(a) n = 35

Margin of error = = 1.96*4.4/√35 = 1.46

Thus, lower bound = 17.8 - 1.46 = 16.32

Upper bound = 17.8 + 1.46 = 19.26

(b) n = 51

Margin of error = 1.96*4.4/√51 = 1.21

Thus, lower bound = 17.8 - 1.21 = 16.59

Upper bound = 17.8 + 1.21 = 19.01

Margin of error decreases with increasing sample size

(c) n = 35 and 99% CI

Margin of error = 2.575*4.4/√35 = 1.92

Thus, lower bound = 17.8 - 1.92 = 15.88

Upper bound = 17.8 + 1.92 = 19.72

Margin of error increases on increasing the level of confidence

(d) The sample data must come from a population that is normally distributed with no outliers