Question

What is the margin of error (M.E.) for a 95%, two-sided confidence interval on mu when xbar = 820, s = 72, and n = 16? (Assume the sample was drawn from a normal population & the fpc factor is NOT needed.)

Answer #1

Solution :

Given that,

= 820

s =72

n =16

Degrees of freedom = df = n - 1 =16 - 1 = 15

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2= 0.05 / 2 = 0.025

t
/2,df = t0.025,15 = **2.120** ( using student t
table)

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

= **2.120** * ( 72/
16)

= 38.16

Margin of error = E =38.16

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