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20 February, 02:35

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?

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  1. 20 February, 04:17
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    0.03919

    Step-by-step explanation:

    You take 1.984+1.96=3.944

    3.944-0.025=3.919

    3.919 into the percentage = 0.03919
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