Question

A simple random sample of 60 items resulted in a sample mean of 84. The population standard deviation is 15.

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).

( , )

How do you find the answers? Please write legibly! Thank you!

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 84

Population standard deviation =    = 15

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 = Z_0.025 = 1.96

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

= 1.96 * (15 /  60 )

= 3.8
At 95% confidence interval estimate of the population mean
is,

- E < < + E

84 - 3.8 <   < 84 + 3.8

80.2<   < 87.8

(  80.2< ,87.8 )

(B)

Solution :

Given that,

Point estimate = sample mean = = 120

Population standard deviation =    = 15

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 = Z_0.025 = 1.96

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

= 1.96 * (15 /  60 )

= 3.8
At 95% confidence interval estimate of the population mean
is,

- E < < + E

120 - 3.8 <   < 120+ 3.8

116.20<   < 123.80

(  116.20 , 123.80 )