The number of possible samples to be drawn without replacement is determined by 6C4 or 6!/4!/(6 − 4)! equal to 15.
Samples--------------------Sample Means
14, 19, 26, 31---------------------22.5
14, 19, 26, 48---------------------26.75
14, 19, 26, 53---------------------28
14, 19, 31, 48---------------------28
14, 19, 31, 53---------------------29.25
14, 19, 48, 53---------------------33.5
14, 26, 31, 48---------------------29.75
14, 26, 31, 53---------------------31
14, 26, 48, 53---------------------35.25
14, 31, 48, 53---------------------36.5
19, 26, 31, 48---------------------31
19, 26, 31, 53---------------------32.25
19, 26, 48, 53---------------------36.5
19, 31, 48, 53---------------------37.75
26, 31, 48, 53---------------------39.5
The sampling distribution of the sample means is built as:
X-bar------22.5----26.75----28-----29.25-----29.75-----31-----32.25-----33.5----35.25------36.5------37.75----39.5
P(X-bar)---1/15----1/15----2/15-----1/15-------1/15-----2/15----1/15------1/15-----1/15------2/15------1/15-----1/15
The mean of the sample means is written as
µX-bar = (1/15)(22.5+26.75+29.25+29.75+32.25+33.5+35.25+37.75+39.5) plus (2/15)(28+31+36.5), which goes to 31.8333333333; this is also the quotient of (14+19+26+31+48+53) divided by 6.
Take the variance of the sampling distribution of the sample means as
σX-bar2 = ∑X-bar2P(X-bar) − µX-bar2 or
(1/15)(22.52+26.752+29.252+29.752+32.252+33.52+35.252+37.752+39.52) plus
(2/15)(282+312+36.52) minus 31.83333333332, which reduces to 20.44722243.
The standard deviation (or standard error) of the sampling distribution of
the sample means is given by the square root of the variance of the sampling
distribution of the sample means or √20.44722243 or 4.521860505.
