Question

Assuming the population has an approximate normal distribution, if a sample size n=17n=17 has a sample mean ¯x=36x¯=36 with a sample standard deviation s=4s=4, find the margin of error at a 98% confidence level. Round the answer to two decimal places.

Answer 1

Solution :

Given that,

= 36

s =4

n =17 Degrees of freedom = df = n - 1 = 17- 1 =16

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,16 =2.583 ( using student t table)

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

= 2.583* ( 4/ 17)

E = 2.51

Margin of error = E = 2.51