Question

A random sample of 27 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 113.9 and 20.40, respectively. Assume that the population is normally distributed. [You may find it useful to reference the t table.]

a. Construct the 90% 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,

Point estimate = sample mean = = 113.9

sample standard deviation = s = 20.40

sample size = n = 27

Degrees of freedom = df = n - 1 = 27 - 1 = 26

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

t/2,df = 1.706

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

= 1.706 * ( 20.40/ 27)

Margin of error = E = 6.70

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

  ± E  

113.9  ± 6.70

(107.20 , 120.60)

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

t/2,df = 2.056

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

= 2.056 * ( 20.40/ 27)

Margin of error = E = 8.07

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

  ± E  

113.9  ± 8.07

(105.83 , 121.97)