Question

A magazine includes a report on the energy costs per year for​ 32-inch liquid crystal display​ (LCD) televisions. The article states that 14 randomly selected​ 32-inch LCD televisions have a sample standard deviation of ​$3.59.Assume the sample is taken from a normally distributed population. Construct 99​% confidence intervals for​ (a) the population variance sigmaσsquared2 and​ (b) the population standard deviation sigmaσ.Interpret the results.

Answer 1

Solution-A;

99​% confidence intervals for​ (a) the population variance sigmaσsquared2

n=14

df=n-1=14-1=13

s=3.59

alpha=0.01

alpha/2=0.01/2=0.005

1-alpha/2=1=0.005=0.995

=CHISQ.INV(0.005,13)

=3.56503458

=CHISQ.INV(0.995,13)

=29.81947122

99% confidence interval for variance is

(n-1)*s^2/chi sq 1-alpha/2<sigma^2<(n-1)*s^2/chi sq alpha/2

(14-1)*3.59^2/29.81947122<sigma^2<(14-1)*3.59^2/3.56503458

   5.618654<sigma^2<   46.99682

99% lower limit variance=   5.618654

99% upper limit variance=    46.99682

we are 99% conident that the true varaince lies in between 5.618654 and  46.99682

99​% confidence intervals for​ (b) tpopulation standard deviation sigmaσ

sqrt( 5.618654)<sigma< sqrt 46.99682)

2.37037<sigma< 6.855423

99% lower limit standard deviation=    2.37037

99% upper limit standard deviation=    6.855423

we are 99% confident that the true standard deviation lies in between 2.37037 and  6.855423