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 109​, and the sample standard​ deviation, s, is found to be 10.

​(a) Construct a 90​% confidence interval about mu if the sample​ size, n, is 21.

​(b) Construct a 90​% confidence interval about mu if the sample​ size, n, is 29.

(c) Construct a 95​% confidence interval about mu if the sample​ size, n, is 21.

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

Answer 1

​Part a) 90​% confidence interval about mu if the sample​ size, n, is 21

Sample Size: 21
Sample Mean: 109
Standard Deviation: 10
Confidence Level: 90%

Z_c= 1.645
90% Confidence Interval: 109 ± 3.59

(105.41, 112.59)

Part b)  a 90​% confidence interval about mu if the sample​ size, n, is 29.

Sample Size: 29
Sample Mean: 109
Standard Deviation: 10
Confidence Level: 90%

Zc = 1.645

90% Confidence Interval: 109 ± 3.05
(105.95 , 112.05)

Part c) a 95​% confidence interval about mu if the sample​ size, n, is 21.

Sample Size: 21
Sample Mean: 109
Standard Deviation: 10
Confidence Level: 95%

Zc = 1.96

95% Confidence Interval: 109 ± 4.28
(104.72 , 113.28)

Part d)

No because the requirement for calculating a (1-α)*100% interval for a sample size that is less than 30 is sample data must come from a population that is normally distributed. and in these cases sample size is less than 30.