Question

A simple random sample of 60 items resulted in a sample mean of 67. 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).

(  ,  )

c. What is the effect of a larger sample size on the margin of error?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 67

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

67-3.04 <   < 67 + 3.04

64.0 <   < 70.0
( 64.0 ,70.0 )

(B)

Solution :

Given that,

Point estimate = sample mean =     = 67

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

67-2.15<   < 67 + 2.15

64.9 <   < 69.2
( 64.9 , 69.2 )

(C) larger sample size provide smaller margin of error and smaller sample size provide larger margin of error