Question

Let X be the length of life of a 60-watt light bulb manufactured by a certain company. If a random sample of 25 units is tested until they burn out, yielding a sample mean of x = 1478 hours and s 2 x = 1296, compute a confidence 95 % -confidence interval for the mean and the variance.

Answer 1

Mean(x) = 1478

Std Dev(s) = Sqrt(1296) = 36

n = 25

Std Error (SE) = s/n^1/2 = 7.2

At alpha = 0.05,

Z_Critical = 1.96

Hence,

95% CI for Mean = x +/- Z_Critical * SE = 1478 +/- 1.96*7.2

= {1463.89,1492.11}

Confidence Interval for Variance is given by:

df = n-1 = 24

ChiSquare_0.025 = CHINV(0.025,24) = 39.3641

And

ChiSquare_0.975 = CHINV(0.975,24) = 12.4011

Hence 95% CI for Variance = {790.16,2508.16}