Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean,

x overbarx​, is found to be 110​, and the sample standard​ deviation, s, is found to be 10.

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

​(c) Construct a 70​% confidence interval about μ if the sample​ size, n, is 25.

​(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,

a ) = 110

s = 10

b ) n = 13

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

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,12 =2.179  

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

= 2.179  * (10 / 13 )

= 6.04

Margin of error = 6.04

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

- E < < + E

110 - 6.04 < < 110 + 6.04

103.96 < < 116.04

(103.96, 116.04 )

c ) n = 25

Degrees of freedom = df = n - 1 = 25- 1 = 24

At 70% confidence level the t is ,

  = 1 - 70% = 1 - 0.70 = 0.30

/ 2 = 0.30/ 2 = 0.15

t /2,df = t0.15,24=1.059

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

= 1.059  * (10 / 25)

= 2.11

Margin of error = 2.11

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

- E < < + E

110 - 2.11 < < 110 + 2.11

107.88< < 112.11

(107.88, 112.11 )

d ) Yes