Question

Construct a 95​% confidence interval to estimate the population mean when x bar =124 and s​ =29 for the sample sizes below.

​a)

n=40     

​b)

n=50       

​c)

n=90

round to 2 decimale places as needed

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 124

sample standard deviation = s = 29

(a)

sample size = n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,39 = 2.023

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

= 2.023 * (29 / 40)

= 9.28

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

- E < < + E

124 - 9.28 < < 124 + 9.28

114.72 < < 133.28

(114.72 , 133.28)

(b)

ample size = n = 50

Degrees of freedom = df = n - 1 = 50 - 1 = 49

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,49 = 2.010

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

= 2.010 * (29 / 50)

= 8.24

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

- E < < + E

124 - 8.24 < < 124 + 8.24

115.76< < 132.24

(115.76 , 132.24)

(c)

ample size = n = 90

Degrees of freedom = df = n - 1 = 90 - 1 = 89

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,89 = 1.987

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

= 1.987 * (29 / 40)

= 6.07

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

- E < < + E

124 - 6.07 < < 124 + 6.07

117.93 < < 130.07

(117.93 , 130.07)