Question

Assuming the population has an approximate normal distribution, if a sample size n=15n=15 has a sample mean ¯x=49x¯=49 with a sample standard deviation s=7s=7, find the margin of error at a 90% confidence level. Round the answer to two decimal places.

Answer 1

Given that,

= 49

s =7

n = 15

Degrees of freedom = df = n - 1 =15 - 1 = 14

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,14 = 1.761    ( using student t table)

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

=1.761 * ( 7/ 15)

E =3.1828