Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 110​, and the sample standard​ deviation, s, is found to be 10. ​ ​(b) Construct a 98​% confidence interval about mu if the sample​ size, n, is 26. ​

Answer 1

Solution :

Given that,

= 110

s =10

n = Degrees of freedom = df = n - 1 =26 - 1 = 25

At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,25 =2.485 ( using student t table)

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

= 2.485 * (10 / 26)

= 4.87

The 98% confidence interval is,

- E < < + E

110 - 4.87 < < 110 + 4.87

(105.13 , 114.87)