Question

A random sample of 88 observations produced a mean x=25.8 and a standard deviation s=2.6.

a. Find a​ 95% confidence interval for mu.

b. Find a​ 90% confidence interval for mu.

c. Find a​ 99% confidence interval for mu.

Answer 1

Solution :

Given that,

a.

t /2,df = 1.988

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

= 1.988 * (2.6 / 88)

Margin of error = E = 0.6

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

- E < < + E

25.8 - 0.6 < < 25.8 + 0.6

25.2 < < 26.4

(25.2 , 26.4)

b.

t /2,df = 1.663

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

= 1.663 * (2.6 / 88)

Margin of error = E = 0.5

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

- E < < + E

25.8 - 0.5 < < 25.8 + 0.5

25.3 < < 26.3

(25.3 , 26.3)

c.

t /2,df = 2.634

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

= 2.634 * (2.6 / 88)

Margin of error = E = 0.7

The 99% confidence interval estimate of the population mean is,

- E < < + E

25.8 - 0.7 < < 25.8 + 0.7

25.1 < < 26.5

(25.1 , 26.5)