Question

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.

Answer #1

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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