If n=18, ¯xx¯(x-bar)=40, and s=10, find the margin of error at a
99% confidence level
Give your answer to two decimal places.
solution
s =10
n = 18
Degrees of freedom = df = n - 1 = 18- 1 =17
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t
/2 df = t0.005,17 = 2.899
( using student t table)
Margin of error = E = t/2,df
* (s /
n)
= 2.899* (10 /
18)
E = 6.83
Margin of error = E = 6.83
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