Question

I'm having a pretty difficult time with these types of problems and I'd really appreciate it if someone could show me how to go about doing this one, thank you!

1. Consider the population of {32, 34, 37, 39}

a) Find the mean, variance, and standard deviation

b) Suppose the sample size n=2 is randomly chosen with replacement from this population, List the 16 possible samples of size n=2

c) Fill out the table

Sample Size (n=2)	Sample Mean	Sample Variance	Sample Standard Deviation

d) How do the average of all of the 16 sample means, sample variance, and sample standard deviation compare to the population mean, population variance, and population standard deviation?

Answer 1

1. Here we have population { 32, 34, 37, 39 }

a) Mean :

Variance :

x	(x-)2
32	12.25
34	2.25
37	2.25
39	12.25
Total	29

  

= 9.6667

Standard deviation =s= = = 3.1091

b) Samples of size n = 2 are :

{ (32,32 ), ( 32,34), ( 32,37), (32, 39), (34,32),(34, 34) , (34, 37 ) , (34, 39), (37,32 ), (37, 34), (37, 37), (37,39), ( 39, 32 ), (39, 34 ), (39,37), (39,39) }

c )

Standard deviation =

	sample mean	sample variance	sample standard deviation
(32, 32 )	32	0	0
(32, 34 )	33	2	1.4142
(32, 37)	34.5	12.5	3.5355
(32, 39)	35.5	24.5	4.9497
(34, 32)	33	2	1.4142
(34, 34)	34	0	0
(34,37)	35.5	4.5	2.1213
(34, 39)	36.5	12.5	3.5355
(37, 32)	34.5	12.5	3.5355
(37, 34)	35.5	4.5	2.1213
(37, 37)	37	0	0
(37, 39)	38	2	1.4142
(39, 32 )	35.5	24.5	4.9497
(39, 34)	36.5	12.5	3.5355
(39, 37)	38	2	1.4142
(39,39)	39	0	0
Total	568	116

d)

Mean of sample means =

Variance :

Standard deviation :

Here we can see that population mean and sample mean are same. But sample variance and standard deviation are lower than population variance and standard deviation.