Question

A simple random sample of 60 items resulted in a sample mean of 89. The population standard deviation is 18.

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 the larger sample size on the margin of error?

Answer 1

Solution

Given that,

=89

= 18

a ) 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 = 1.960

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

= 1.960 * (18 / 60 )

= 4.5

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

- E < < + E

89 - 4.5 < < 89 +4.5

84.4 < < 93.5

(84.4,93.5)

b ) 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 = 1.960

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

= 1.960 * (18 / 120 )

= 3.2

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

- E < < + E

89 - 3.2< < 89 + 3.2

85.8 < < 92.2

(85.8 , 92.2)

c ) The increasing sample size decrasing margin of error.