Question

A sample of 100 receipts from a restaurant showed that 12 had errors. What is the 95% confidence interval for the proportion of all receipts that have error? Note that z 0.025= 1.96 and t 0.025= 1.984 for 99 degrees of freedom?

Answer 1

0.03919

Explanation:

You take 1.984+1.96=3.944

3.944-0.025=3.919

3.919 into the percentage =0.03919