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

​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 19.

​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 13. ​

(c) Construct a 90​% confidence interval about mu if the sample​ size, n, is 19.

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 106

sample standard deviation = s = 10

(a) sample size = n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 95% confidence level

= 1 - 95%

=1 - 0.95 = 0.05

/2 = 0.025

t_/2,df = t_0.025,18 = 2.101

Margin of error = E = t_/2,df * (s /n)

= 2.101 * ( 10 / 19 )

Margin of error = E = 4.82

The 95% confidence interval estimate of the population mean is,

- E < < + E

106 - 4.82 < < 106 + 4.82

101.18 < < 110.82

( 101.18 , 110.82 )

The 95% confidence interval is ( 101.18 , 110.82 )

(b) sample size = n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 95% confidence level

= 1 - 95%

=1 - 0.95 = 0.05

/2 = 0.025

t_/2,df = t_0.025,12 = 2.179

Margin of error = E = t_/2,df * (s /n)

= 2.179 * ( 10 / 13 )

Margin of error = E = 6.04

The 95% confidence interval estimate of the population mean is,

- E < < + E

106 - 6.04 < < 106 + 6.04

99.96 < < 112.04

( 99.96 , 112.04 )

The 95% confidence interval is ( 99.96 , 112.04 )

(c)  sample size = n = 19

Degrees of freedom = df = n - 1 = 19 - 1 =18

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

t_/2,df = t_0.05,18  = 1.734

Margin of error = E = t_/2,df * (s /n)

= 1.734 * ( 10 / 19)

Margin of error = E = 3.98

The 90% confidence interval estimate of the population mean is,

- E < < + E

106 - 3.98 < < 106 + 3.98

102.02 < < 109.98

( 102.02 , 109.98 )

The 90% confidence interval is ( 102.02 , 109.98 )

(d) It is neccessary to be a normally distributed.