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.

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