Question

A random sample of 92 observations produced a mean xequals=25.9 and a standard deviation s=2.6

a. Find a​ 95% confidence interval forμ.

b. Find a​ 90% confidence interval for μ

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

​(Use integers or decimals for any numbers in the expression. Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 25.9

s = 2.6

n = 92

Degrees of freedom = df = n - 1 = 92 - 1 = 91

A ) At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,24 =1.986

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

= 1.986 * ( 2.6/ 92)

= 0.54

Margin of error = 0.54

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

- E < < + E

25.9- 0.54 < < 25.9 + 0.45

25.36 < < 26.40

(25.36, 26.40)

B) At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,91 =1.662

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

= 1.662* ( 2.6/ 92)

= 0.45

Margin of error = 0.45

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

- E < < + E

25.9- 0.45 < < 25.9 + 0.45

25.45 < < 26.35

(25.45, 26.35)

c )  At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,91 =2.797

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

= 2.631* ( 2.6/ 92)

= 0.71

Margin of error = 0.71

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

- E < < + E

25.9- 0.71 < < 25.9 + 0.71

25.19 < < 26.61

(25.19, 26.61)