In a survey, 31 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $50 and standard deviation of $16. Construct a confidence interval at a 98% confidence level. Give your answers to one decimal place.
Solution :
Given that,
Z/2 = Z0.01 = 2.326
Margin of error = E = Z/2
* (
/n)
= 2.326 * ( 16 / 31
)
= 6.7
At 98% confidence interval estimate of the population mean is,
± E
= 50 ± 6.7
= ( $ 43.3, $ 56.7 )
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