In a survey, 15 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 95% confidence level.
Solution :
Given that,
Point estimate = sample mean = = $40
sample standard deviation = s = $7
sample size = n = 15
Degrees of freedom = df = n - 1 = 15-1= 14
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,14 = 2.145
Margin of error = E = t/2,df * (s /n)
= 2.145 * (7 / 15)
E = 3.9
The margin of error at a 95% confidence level is 3.9
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