Based on the program \'samples.do\' (described below), randomly (with replacemen

Question

Based on the program 'samples.do' (described below), randomly (with replacement) generate 500 sample means (xbar) and variances (varx) for the variable dage using samples of sizes n=5, 10, 50. 1. 2. Demonstrate that: a. the expectation of the sample means is equal to the population mean (assume the original mean for donor age, N=144, is the population); of dage divided by the sample size; and dage. b. the variance of the sample means is approximately equal to the original variance c. the expectation of the sample variance is equal to the population variance of . sum m5 v5 m10 v10 m50 v50 dage Variable Obs Mean Std. Dev. Min Max m5 500 33. 692 4.79141 500 130.5108 92.88259 33.796 3.598283 21.4 4.3 24.4 47 538.5 m10 v10 m50 500 500 133.9856 62.97953 14.32222 411.3778 500 33.51444 1.296592 30.44 37.7 v50 dage 500 129.9597 21.12727 66. 78204 183.4139 144 33.63194 11.46277 18 69 I DO NOT UNDERSTAND THE FOLLOWING SOLUTION TO PART 2B, PLEASE EXPLAIN THANK YOU: 2b) The variance of the sample means -variance 33.692,33796,33.5 1444,each taken 500 times)-0.0 135 13 And the variance of dage- 11.462772 131.395 and when divided by sample size of 144*65, the equal to 0.014. hence the variance of sample means and the actual variance divided by the sample size are almost equal.

Explanation / Answer

Took

m5

m10

m50

sample sizes are 500,500,500

Took their mean values which are

33.692,33.796,33.51444

Now take the variance now

which comes to 0.013513

rounding to thousandth=0.014

LHS=0.014

varinace of dage=sd^2=11.46277^2/144*65=0.014

why 65 is total of 5+10+50=65=65*144

RHS=0.014

LHS=RHS

VARIANCE OF SAMPLE MEANS=ACTUAL VARIANCE

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