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: [tex](0.21,\ 0.27)[/tex]Step-by-step explanation:The confidence interval estimate for the population proportion is given by :-[tex]\hat{p}\pm ME[/tex], where [tex]\hat{p}[/tex] 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 : [tex]\hat{p}=0.24[/tex]Margin of error : [tex]ME=0.03[/tex]Now, the confidence interval estimate for the population mean will be :-[tex]0.24\pm0.03=(0.24-0.03,\ 0.24+0.03)=(0.21,\ 0.27)[/tex]Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution =  [tex](0.21,\ 0.27)[/tex]