Question

9. The lifetime of a certain brand of lightbulbs are assumed to be normally distributed. A researcher took a random sample of these lightbulbs and observed the following failure times 24 25 17 24 34 40 17 Let xbe the critical value of a t random variable with v degrees of freedom. The following table lists values of x for specific combinations of a and v a 0.05 Q=0.975 a=0.025 O= 0.95 1.237 14.449 v6 1.635 12.592 16.013 1.690 1.2.167 14.067 V = 7 Construct the 95% confidence interval for the variance of the light bulb lifetime

Answer 1

Solution:

Confidence interval for population variance is given as below:

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

We are given

Confidence level = 95%

Sample size = n = 7

Degrees of freedom = n – 1 = 6

Sample standard deviation = S = 8.47405

χ²_{α/2, n – 1} = 14.449

χ²_{1 -α/2, n– 1} = 1.237

(By using chi square table)

(7 – 1)* 8.47405^2 / 14.449 < σ² < (7 – 1)* 8.47405^2 / 1.237

[ 6*71.80952 / 14.449 ] < σ² < [ 6*71.80952 /1.237 ]

29.8192 < σ² < 348.3081

Lower limit = 29.8192

Upper limit = 348.3081

Confidence interval = (29.8192, 348.3081)