Question

Determine the margin of error for a confidence interval to estimate the population mean with n=22 and s​ =14.4 for the confidence levels below

a. 80%

b.90%

c. 99%

The margin of error for an 80​% confidence interval is_______?

repeat for 90% and 99%

Answer 1

Solution :

Given that,

s = 14.4

n = 22

Degrees of freedom = df = n - 22 = 22 - 1 = 21

a)

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,21 = 1.323

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

= 1.323 * (14.4 / 22)

E = 4.062

The margin of error for an 80​% confidence interval is 4.062

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_0.05,21 = 1.721

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

= 1.721 * (14.4 / 22)

E = 5.284

The margin of error for an 90​% confidence interval is 5.284

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_0.005,21 = 2.8311

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

= 2.831 * (14.4 / 22)

E = 8.691

The margin of error for an 99​% confidence interval is 8.691