Question

Question Help Let a population consist of the values 10 cigarettes, 18 cigarettes and 19 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have man deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population Sample Mean Absolute Deviation (10,10) {10,18) {10,19) 4.5 (18,10) (18,18) 18,19) [19,10) (19,18) (1919) 0.5

Answer 1

samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population

sample	Mean absolute deviation
i	MADi
10,10	0
10,18	4
10,19	4.5
18,10	4
18,18	0
18,19	0.5
19,10	4.5
19,18	0.5
19,19	0
Total	18
mean	2

The mean of sample mean absolute deviation(Mean(MADi))

2 MADi/n i-1

	population observations(xi)	|xi-mean|
	10	5.666667
	18	2.333333
	19	3.333333
Mean	15.66667	3.777778

The mean absolute deviation of the population (MAD_Pop) =   

= 3.8 M1/3 i=1

where   

Β 15.7 Μ Xi/3 i-1

cleary we have MAD_Pop≠Mean(MADi)

Above results show that samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population

All the calculations given in the table above.