Question

9. The lifetime of a certain brand of light bulbs are assumed to be normally distributed. A researcher took a random sample of these light bulbs and observed the following failure times: 17 17 24 24 | 25 34 40 Let xay be 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: v=6 v=7 a = 0.975 1.237 1.690 a=0.95a 1.635 1.2.167 = 0.05 a= 0.025 12.592 T 14.449 14.067 16.013 Construct the 95% confidence interval for the variance of the light bulb lifetime.

Answer 1

First we need to find the variance of sample. Following table shows the calculations::

	X	(X-mean)^2
	17	78.44822041
	17	78.44822041
	24	3.44882041
	24	3.44882041
	25	0.73462041
	34	66.30682041
	40	200.0216204
Total	181	430.8571429

Sample size n=7 Sample sum: .-=181 Sample mean: F=2* =181/7 =25.8571 Sample standard deviation: 2(x-7) - Esqrt(430.8571/(7-1)) n-1 =8.474 Variance is 2-{(x-3) -430.8571/(7-1) n-1 =71.8095

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Here we have 52 =71.8095, n = 7 Degree of freedom: df=n-1 =6 Level of significance: Critical values: a =0.05 XR =1.237 X*=14.449 The confidence interval for variance is: la vita so s (((7-1)*71.8095)/14.449) SOS (((7-1)*71.8095)/1.237) 29.82 sos 348.31