Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 110​, and the sample standard​ deviation, s, is found to be 10.
​(a) Construct an 80​% confidence interval about μ if the sample​ size, n, is 13.
​(b) Construct an 80​% confidence interval about μ if the sample​ size, n, is 27.
​(c) Construct a 90​% confidence interval about μ if the sample​ size, n, is 13.
​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

****** Need lower bound and upper bound *******

Answer 1

(a)

n = 13

= 110

s =10

= 0.20

SE = s/

= 10/

= 2.7735

ndf =n - 1 = 13 - 1 = 12

From Table, critical values of t = 1.3562

Confidence Interval:

Lower bound = 110 - (1.3562 X 2.7735)

                   = 110 - 3.7614

                   = 106.2386

Upper bound = 110 + (1.3562 X 2.7735)

                   = 110 + 3.7614

                   = 113.7614

So,

Lower bound = 106.2386

Upper bound = 113.7614

(b)

n = 27

= 110

s =10

= 0.20

SE = s/

= 10/

= 1.9245

ndf =n - 1 = 27 - 1 = 26

From Table, critical values of t = 1.3150

Confidence Interval:

Lower bound = 110 - (1.3150 X 1.9245)

                   = 110 - 2.5307

                   = 107.4693

Upper bound = 110 + (1.3150 X 1.9245)

                   = 110 + 2.5307

                   = 112.5307

So,

Lower bound = 107.4693

Upper bound = 112.5307

(c)

n = 13

= 110

s =10

= 0.10

SE = s/

= 10/

= 2.7735

ndf =n - 1 = 13 - 1 = 12

From Table, critical values of t = 1.7823

Confidence Interval:

Lower bound = 110 - (1.7823 X 2.7735)

                   = 110 - 4.9432

                   = 105.0568

Upper bound = 110 + (1.7832 X 2.7735)

                   = 110 + 4.9432

                   = 114.9432

So,

Lower bound = 105.0568

Upper bound = 114.9432

(d)

We could not have computed the confidence intervals in parts (a) - (c) if the population had not been normally distributed, since for allying t distribution, the population must be normal distribution.