Question

Find the value of E, the margin of error, for a 98% confidence interval with n = 22 and s = 3.5 assuming that the sample has been selected from a normally distributed population?

a-1.739

b-1.871

c-2.518

d-1.879

Answer 1

Solution :

Given that,

s =3.5

n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,21 =2.518 ( using student t table)

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

= 2.518 * (3.5 / 22)

=1.879