Question

If n=21, ¯xx¯(x-bar)=44, and s=17, 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.

< μμ <

Answer 1

Solution :

Given that,

= 44

s =17

n = 21

Degrees of freedom = df = n - 1 = 21- 1 = 20

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,20 =2.528 ( using student t table)

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

= 2.528 * ( 17/ 21)

= 9.4

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

- E < < + E

44 -9.4 < < 44+ 9.4

34.6 < < 53.4