Question

In a random sample of

50

​refrigerators, the mean repair cost was

​$132.00

and the population standard deviation is

​$15.10

A

95%

confidence interval for the population mean repair cost is

(127.81,136.19).

Change the sample size to

n=100

Construct a

95

confidence interval for the population mean repair cost. Which confidence interval is​ wider? Explain.

Answer 1

Solution :

Given that,

= 132.00

= 15.10

a ) n = 50

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 = 1.960

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

= 1.960 * (15.10 / 50 )

= 4.19

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

- E < < + E

132.00 - 4.19 < < 132.00 + 4.19

127.81 < < 136.19

(127.81 , 136.19)

b ) n = 100

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 = 1.960

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

= 1.960 * (15.10 / 100 )

= 2.96

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

- E < < + E

132.00 - 2.96 < < 132.00 + 2.96

129.04 < < 134.96

(129.04 , 134.96 )

The sample size 50 is confidence is wider.

The sample size decreases the confidence interval is increases.