Question

In a random sample of 40 ​refrigerators, the mean repair cost was ​$133.00 and the population standard deviation is ​$17.20. A 95​% confidence interval for the population mean repair cost is left parenthesis 127.67 comma 138.33 right parenthesis.

Change the sample size to n=80. Construct a 95​% confidence interval for the population mean repair cost. Which confidence interval is​ wider? Explain. Construct a 95​% confidence interval for the population mean repair cost.

The 95​% confidence interval is ​( _____​, _____ ​). ​(Round to two decimal places as​ needed.)

Answer 1

For large random samples, a confidence interval for a population mean is given by (where sample size n > 30)

Confidence interval = Sample mean ± z * s /√n

Where,

Sample mean = $133.00

Standard deviation of population s = $17.20

Sample size n = 40

The value of z = 1.96 at 95% level of confidence

Therefore,

Confidence interval for a population mean repair cost

= {($133.00 - 1.96 * $17.20/√40), ($133.00 + 1.96 * $17.20/√40)}

= {($133.00 - 5.33), ($133.00 + 5.33)}

= {127.67, 138.33}

Now if Sample size n = 80

Then

Confidence interval for a population mean repair cost at 95%

= {($133.00 - 1.96 * $17.20/√80), ($133.00 + 1.96 * $17.20/√80)}

= {($133.00 - 3.77), ($133.00 + 3.77)}

= {129.23, 136.77}

Therefore the confidence interval with sample size of 40 is wider because of higher value of (z * s /√n) than the confidence interval with sample size of 80 for a population mean repair cost at 95% confidence interval.