Question

A technician randomly sampled repair logs for computers at a university. He wanted to find out how many times the computers were repaired. The proportion of computers that were repaired more than three times in the last two years was 0.24, with a margin of error of 0.03. Construct a confidence interval for the proportion of university computers that were repaired more than three times in the last two years.

Answer 1

Answer: $(0.21,\ 0.27)$

Step-by-step explanation:

The confidence interval estimate for the population proportion is given by :-

$\hat{p}\pm ME$, where $\hat{p}$ is the sample proportion of success and ME  is the margin of error.

Given : The proportion of computers that were repaired more than three times in the last two years : $\hat{p}=0.24$

Margin of error : $ME=0.03$

Now, the confidence interval estimate for the population mean will be :-

$0.24\pm0.03=(0.24-0.03,\ 0.24+0.03)=(0.21,\ 0.27)$

Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution =  $(0.21,\ 0.27)$