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