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Advanced Math/Sampling Distribution of sample mean
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Question
Let a very small population consist of five numbers:10,20,30,40,50,each having probability of being selected equal to .2. Consider all possible samples (selected with replacement of size 2 that could be selected. Find sampling distribution of sample mean, Find mean of ample using your answer to previous question, Find standard error of mean using your answer to sampling distribution of sample mean.
Five numbers: 10,20,30,40,50
Chance of each: 0.2 (equally likely)
Take 2, with replacement.
The result of the mean will be 20 through 100 starting and ending at 1 choice. They will increase by 1 until 60 is reached, and then decrease by 1. 20:1, 30:2, 40: 3, etc. to 90:2, 100:1.
The distribution of the mean is a straight line increase and straight line decrease. The number of each is 1,2,3,4,5,4,3,2,1.
The average can be found by 1(20+2(30)+ ... + 2(90)+1(100). Then again, the max is at 60 and is equally weighted, so the mean is seen to be 60.
By standard error, I think what is meant is standard deviation σ.
Standard deviation is the √variance, σ².
The variance, σ², is defined as σ²=Σ(x-µ)²/(n-1).
Note that in this problem, there it would be
[(20-60)²+2(30-60)²+...+5(60-60)²+4(70-60)²+...+2(90-60)²+(100-60)²]/
(25-1).
This will give you the variance, σ². The standard deviation, σ, is thereby the squareroot of the variance.
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Scott A Wilson
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