Question

A simple random sample of 60 items resulted in a sample mean of 90. The population standard deviation is 12. 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 =     =90

Population standard deviation =    = 12

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 * ( 12 / 60 )

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

- E < < + E

90- 3.04 <   <90 + 3.04

86.96 <   < 93.04

( 86.96, 93.04 )

(B)

Solution :

Given that,

Point estimate = sample mean =     =90

Population standard deviation =    = 12

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 * ( 12 / 120 )

= 2.15
At 95% confidence interval estimate of the population mean
is,

- E < < + E

90- 2.15 <   <90 + 2.15

87.85 <   < 92.15

( 87.85 , 92.15)