Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x bar over x, is found to be 109, and the sample standard deviation, s, is found to be 10.

a) Construct a 96% confidence interval about mu if the sample size, n, is 29

lower bound: __ upper bound: __

b) Re-do, but with a different interval. Construct a 95% confidence interval about mu if sample size n, is 29

lower bound: __ upper bound: __

Compare results with part a. How does decreasing level of confidence affect margin of error size?

c) Construct a 96% confidence interval about mu if the sample size n is 12

lower bound: __ upper bound: __

How does decreasing the sample size affect margin of error, E?

d) Could we have computed the confidence intervals if population hadn't been normally distributed?

Answer 1

Solution-a:

df=n-1=29-1=28

alpha=1-0.96=0.04

alpha/2=0.04/2=0.02

xbar=109

s=10

t critical==T.INV(0.02;28)=2.15393

96% confidence interval for mean is  

109-2.15393*10/sqrt(29),109+2.15393*10/sqrt(29)

105.0003,112.9997

Solution-b:

1-0.95=0.05

alpha/2=0.05/2=0.025

tcrit==T.INV(0.025;28)=2.04841

95% confidence interval for mean is  

109-2.04841*10/sqrt(29),109+2.04841*10/sqrt(29)

105.1962,112.8038

Solution-c

alpha=0.04

alpha/2=0.04/2=0.02

t crit=T.INV(0.02;11)

=2.32814

96% confidence interval for mean is  

109-2.328139833*10/sqrt(12),109+2.328139833*10/sqrt(12)

102.2792,115.7208

decreasing sample size increases margin of error

Solutiond-

No