Question

In a random sample of 40 refrigerators, the mean repair cost was $150. Assume the population standard deviation is $15.50. Construct a 99% confidence interval for the population mean repair cost. Then change the sample size to n = 60. Which confidence interval has the better estimate?

Answer 1

Answer:  ($143.69, $156.31)

Step-by-step explanation:

Confidence interval to estimate  population mean :

$\overline{x}\ \pm z\dfrac{\sigma}{\sqrt{n}}$

, where $\sigma$ = population standard deviation

n= sample size

$\overline{x}=$ Sample mean

z= critical value.

As per given,

n= 40

$\sigma$ = $15.50

$\overline{x}=$ $150

Critical value for 99% confidence level = 2.576

Then, 99% confidence interval for the population mean:

$150\pm(2.576)\dfrac{15.50}{\sqrt{40}}\\\\\Rightarrow\ 150\pm6.31 \ \ (approx)\\\\\Rightarrow(150-6.31,150+6.31)=(143.69,156.31)$

Hence, the required confidence interval : ($143.69, $156.31)