Question

If all conditions for using a t distribution model are met, and a 95% confidence interval is being created for a sample mean of 85.0, using a sample of size 30, where the standard deviation of the sample is 4.1, what is the margin of error?

Answer 1

Solution :

Given that,

= 85.0

s =4.1

n =30

Degrees of freedom = df = n - 1 = 30- 1 =29

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,29 = 2.045 ( using student t table)

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

= 2.045* ( 4.1/ 30)

E= 1.5308

Margin of error = E =1.5308