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)`

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)`