Question

If n=15, ¯xx¯(x-bar)=33, and s=10, 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.

Answer 1

( )

Solution :

Given that,

= 33

s =10

n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,14 =2.624 ( using student t table)

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

=2.624 * (10 / 15)

= 6.8

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

- E < < + E

33 - 6.8 < < 33 + 6.8

26.2 < < 39.8