In the city of New York, they randomly surveyed 50 people on how much money they have in their wallet. The results of the survey yielded a mean of $85 and a standard deviation of $52. Find a 90% confidence interval estimate of the mean number of dollars in all of the wallets of New Yorkers.
Solution :
t /2,df = 1.677
Margin of error = E = t/2,df * (s /n)
= 1.677 * (52 / 50)
Margin of error = E = 12.3
The 90% confidence interval estimate of the population mean is,
- E < < + E
85 - 12.3 < < 85 + 12.3
72.7 < < 87.3
(72.7 , 87.3)
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