Question

A random sample of 20 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 162.5 and 22.60, respectively. Assume that the population is normally distributed.

a. Construct the 99% confidence interval for the population mean. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

b. Construct the 95% confidence interval for the population mean. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Answer 1

Solution :

Given that,

= 162.5

s = 22.60

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

a ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,19 = 2.861

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

= 2.861 * ( 22.60 / 20)

= 14.46

Margin of error = 14.46

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

- E < < + E

162.5 - 14.46< < 162.5 + 14.46

148.04 < < 176.96

(148.04, 176.09 )

B ) At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,19 = 2.093

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

= 2.093 * ( 22.60 / 20)

= 10.58

Margin of error = 10.58

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

- E < < + E

162.5 - 10.58 < < 162.5 + 10.58

151.92 < < 173.08

(151.92 , 1173.08 )