Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x​, is found to be 100,and the sample standard​ deviation, s, is found to be 8.

​(a) Construct a 95​% confidence interval about μ if the sample​ size, n, is 24.

Lower​ bound:________​;

Upper​ bound:________

​(Round to one decimal place as​ needed.)

​(b) Construct a 95​% confidence interval about μ if the sample​ size, n, is 17. ​(Round to one decimal place as​ needed.)

​(c) Construct a 90​% confidence interval about μ if the sample​ size, n, is 24.​ (Round to one decimal place as​ needed.)

​(d) Should the confidence intervals in parts​ (a)-(c) have been computed if the population had not been normally​ distributed?

Answer 1

a)

Sample size = n = 24

Sample mean = = 100

Standard deviation = s = 8

We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 24 - 1 = 23

Level of significance = 0.05

tc = 2.069   ( Using t table)

So confidence interval is ( 100 - 3.3781 , 100 + 3.3781) = > ( 96.6 , 103.4)

Lower​ bound: 96.6

Upper​ bound: 103.4

b)

Sample size = n = 17

Sample mean = = 100

Standard deviation = s = 8

We have to construct 95% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 17 - 1 = 16

Level of significance = 0.05

tc = 2.120 ( Using t table)

So confidence interval is ( 100 - 4.1132 , 100 + 4.1132) = > ( 95.9 , 104.1)

Lower​ bound: 95.9

Upper​ bound: 104.1

c)

Sample size = n = 24

Sample mean = = 100

Standard deviation = s = 8

We have to construct 90% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 24 - 1 = 23

Level of significance = 0.10

tc = 1.714 ( Using t table)

So confidence interval is ( 100 - 2.7987 , 100 + 2.7987) = > ( 97.2 , 102.8)

Lower​ bound: 97.2

Upper​ bound: 102.8

d)

We can not compute the confidence interval if the population had not been normally distributed.