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 109​, and the sample standard​ deviation, s, is found to be 10.

​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 18.

​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 14.

​(c) Construct a 90​% confidence interval about mu if the sample​ size, n, is 18. ​

(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Ans:

sample mean,x-bar=109

s=10

a)df=18-1=17

critical t value=tinv(0.05,17)=2.110

95​% confidence interval about mu

=109+/-2.110*(10/sqrt(18))

=109+/-4.97

=(104.03, 113.97)

b)

df=14-1=13

critical t value=tinv(0.05,13)=2.110

95​% confidence interval about mu

=109+/-2.160*(10/sqrt(14))

=109+/-5.09

=(103.91, 114.09)

c)

df=18-1=17

critical t value=tinv(0.1,17)=2.110

90% confidence interval about mu

=109+/-1.740*(10/sqrt(18))

=109+/-4.10

=(104.90, 113.10)

d)No,as sample size is small.