Question

A simple random sample of 60 items resulted in a sample mean of 87. The population standard deviation is 17.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

( ,  )

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

( , )

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 87

Population standard deviation =    = 17

Sample size n =60

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.96 * (17 /  60 )

= 4.3
At 95% confidence interval estimate of the population mean
is,

- E < < + E

87 - 4.3 <   < 87+ 4.3

82.7 <   < 91.3

( 82.7, 91.3 )

(b)

Solution :

Given that,

Point estimate = sample mean = = 87

Population standard deviation =    = 17

Sample size n =120

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.96 * (17 /  120 )

= 3.04
At 95% confidence interval estimate of the population mean
is,

- E < < + E

87 - 3.04 <   < 87+ 3.04

83.96 <   < 90.04

(83.96 , 90.04)