If n=18, ¯ x x ¯ (x-bar)=30, and s=2, find the margin of error at a 98% confidence level.
Assume the t distribution must be used. Give your answer to two decimal places.
Solution :
Given that,
= 30
s =2
n =18
Degrees of freedom = df = n - 1 = 18 - 1 = 17
At 98% confidence level the t is
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02/ 2 = 0.01
t /2,df = t0.01,17 = 2.567 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.567* ( 2/ 18)
= 1.21
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