Answer all parts 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 $3.95. Assume the sample is taken from a normally distributed population. Construct 98% confidence intervals for (a) the population variance σ² and (b) the population standard deviation σ. Interpret the results.(a) The confidence interval for the population variance is ( [ ] , [ ] )(Round to two decimal places as needed.)[Note: Diagram and graphs are not applicable in this text.]

Given sample size,n = 14 df=n-1= 13 sample standard deviation,s= 3.95 confidence level,c=0.98solution : Significance level, α =1- f0.98, = 0.02 critical value : =χ2_ (0.02/2,13) =χ2_0.01,13 =4.1069 excel formula :=CHIINV(0.01,13) =χ2_ (1-0.02/2,13) =χ2_0.99,13 =27.6882 excel formula :=CHIINV(0.99,13) a) For population variance 98% confidence interval is : =((14-1)x3.95^2/27.6882),((14-1)x3.95^2/4.1069 =(7.33,49.39) Interpretation : we are 98 % confident that population variance lie between 7.33 and 49.39 b) For standard deviation 98 % confidence interval is CI(standard deviation)= ( √7.33,√49.39) =(2.71,7.03) Interpretation : we are 98 % confident that population standard deviation lie between 2.71 and 7.03