Question

If n=16, ¯x(x-bar)=32, and s=7, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

< μ <

Answer 1

Solution :

Given that,

= 32

s =7

n =16

Degrees of freedom = df = n - 1 =16 - 1 = 15

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,15 = 2.602 ( using student t table)

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

=2.602 * ( 6/ 16)

=3.9

The 98% confidence interval estimate of the population mean is,

- E < < + E

32 - 3.9 < < 32+3.9  

28.1 < < 35.9