An 80% confidence interval for the mean (n-1 degrees of freedom) is
xbar +/- t critical value * s/sqrt(n)
InvT(0.10, 28) = -1.3125
40 +/- 1.3125*(7/sqrt(29))
Margin of error is 1.3125*(7/sqrt(29)) = 1.706
Eva P.
asked 03/26/20In a survey, 29 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $40 and standard deviation of $7. Find the margin of error at a 80% confidence level.
Give your answer to two decimal places.
An 80% confidence interval for the mean (n-1 degrees of freedom) is
xbar +/- t critical value * s/sqrt(n)
InvT(0.10, 28) = -1.3125
40 +/- 1.3125*(7/sqrt(29))
Margin of error is 1.3125*(7/sqrt(29)) = 1.706
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