Question

1. Twenty motors were put on test under a high-temperature setting. The lifetimes in hours of the motors under these conditions are given below. Suppose we assume that the lifetime of a motor under these conditions, X, has a「(1,0) distribution. 8 11 13 13 18 22 22 43 4457 60 76 98 106 110 138 153 154 204 281 (a) Obtain a frequency distribution and a histogram or a stem-leaf plot of the data. Use the intervals [0, 25), [25,50), . the「(1,6) model is credible? . . Based on this plot, do you think that (b) Obtain the maximu ln likelihood estimate of θ and locate it on your plot. (c) Obtain the sample median of the data, which is an estimate of the median lifetime, of a motor. What parameter is it estimating (i.e., determine the median of X)? (d) Based on the mle, what is another estimate of the median of X?

Answer 1

ナhe.akt ne Xi~ ra,e) , is log n 0 0 -0 20 01,

d) Now here the sample standard deviation be calculated as follows,

	X	d=(X-mean(X))	d^2
	1	-100.15	10030.02
	4	-97.15	9438.123
	5	-96.15	9244.823
	21	-80.15	6424.023
	22	-79.15	6264.723
	28	-73.15	5350.923
	40	-61.15	3739.323
	42	-59.15	3498.723
	51	-50.15	2515.023
	53	-48.15	2318.423
	58	-43.15	1861.923
	67	-34.15	1166.223
	95	-6.15	37.8225
	124	22.85	522.1225
	124	22.85	522.1225
	160	58.85	3463.323
	202	100.85	10170.72
	260	158.85	25233.32
	303	201.85	40743.42
	363	261.85	68565.42
Total	2023		211110.6
Mean	101.15

So the sample variance be,

$S^2=211110.6/19=11111.084$

And the sample standard deviation be,

$S=\sqrt{11111.084}=105.41$

So as we know for large sample, $\frac{\bar{X}-\mu}{S/\sqrt{n}}$ , the standardized variable approximately follows standard normal distribution so the 95% confidence interval be,

$\left ( \bar{X}-1.64.\frac{S}{\sqrt{n}} \ , \ \bar{X}+1.64.\frac{S}{\sqrt{n}} \right )$

Putting the values we get,

$\left ( 101.15-1.64\times \frac{105.41}{\sqrt{20}} \ , \ 101.15+1.64\times \frac{105.41}{\sqrt{20}} \right )$

$=(62.495,139.805)$