Paul W. answered • 10/12/25
PhD in Statistics with 20+ Years of Math / Stat Education Experience
Good question! Here's my answer:
A random sample of size 2 is selected, with replacement, from the set of numbers (0,2,4)
A. List all possible samples and evaluate X(with Bar) and S2
Since we're sampling with replacement we can draw a total of nine possible samples:
(0,0) (0,2) (0,4)
(2,0) (2,2) (2,4)
(4,0) (4,2) (4,4)
We can calculate the mean (Xbar) for each sample by adding up the two values selected and divide that sum by two. This gives us 0 x1, 1 x2, 2 x3, 3 x2, 4 x1 (listed as Xbar, frequency)
We can calculate the mean of the means by adding up all nine values and dividing by 9:
0 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 = 18; 18/9 = 2
And the variance (S2) can be calculated by subtracting the mean from each of those values, squaring the differences, adding them up and then dividing that sum by 8. This gives us:
(4 + 1 + 1 + 0 + 0 + 0 +1 + 1 +4)/8 = 12/8 = 1.5
(Note, some books may say to divide by 9 since this is the whole population of means for samples of size n=2 drawn with replacement -- if this is the case then S2 is 12/9 = 1.33 instead)
B. Determine the sampling distribution of X(With bar)
C. Determine the sampling distribution of S2
The sampling distribution of Xbar is approximately normal with mean 2 and variance 1.5 (or 1.33 if you're using N in the denominator instead of N-1)
The sampling distribution of S2 is a little more tricky. The variance of the population is 8/3 = 2.67. I calculated this by getting the mean of the population (2) and calculating S2 the same way as we did for the set of sample means: ( (0-2)2 + (2-2)2 + (4-2)2 ) / 3 . I used 3 here because this is the full population and not a sample of elements from a larger set. We can call this variance σ2 because it is the true population variance.
To get the sampling distribution of S2 we need to calculate S2 for each sample and then find the mean of those values and then calculate the variance of those values. Fortunately, each sample is going to have a variance of 2 (if the values in the sample are different) or 0 (if the values are the same). There are six samples with different values so we get 2 x6, the other three samples give us a 0. So we add up
2 x 6 + 3 x 0 = 12; the mean is this sum divided by 9 which is 1.33
To get the variance of these S2 values we take (6x(2-1.33)2 + 3x(0-1.33)2)/8 = 0.4356
The distribution of S2 isn't going to be normal because the squaring makes it skewed to the right. We do know that (n-1)S2 / σ2 is χ2 with (n-1) degrees of freedom. I'm not sure how helpful that is in this problem but we do have the sampling mean and sampling variance for the distribution as (1.33, 0.4356).
