Question

A random sample of n = 22 school children yielded the following data on their weight (lbs).
sample mean:

and sample standard deviation:
s = 11.
At a confidence level of 94%, calculate the margin of error in the estimate of the mean weight of this population of school children.

Answer 1

Solution :

Given that,

=

s =11

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

a ) At 94% confidence level the t is ,

t /2,df = t /2,21 = 1.988 ( using student t table)

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

= 1.988* ( 11/ 22)

= 4.66