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?
1 answer:
Answer:
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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