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If n=12, ¯xx¯ (x-bar)=31, and s=8, find the margin of error at a 99% confidence level...

If n=12, ¯xx¯ (x-bar)=31, and s=8, find the margin of error at a 99% confidence level (use at least three decimal places)

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Answer #1

Given that,

= 31

s =8

n = 12

Degrees of freedom = df = n - 1 =12 - 1 = 11

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df= t0.005,11= 3.106 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 3.106* (8 / 12) = 7.173

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