Question #267561

random samples of size 4 are drawn with replacement from a finite population 3,6,9


  1. How many possible samples of size 4 are possible?
  2. Find the mean of the sample means.
  3. Find the variance of the sample.
  4. Find the standard deviation of the sample.

Expert's answer

1.

We have population values 3,6,93,6,9 population size N=3N=3 and sample size n=4.n=4. Thus, the number of possible samples which can be drawn without replacement is


Nn=34=81N^n=3^4=81

2.

In sampling with replacement the mean of all sample means equals the mean of the population:


μXˉ=μ=3+6+93=6\mu_{\bar{X}}=\mu=\dfrac{3+6+9}{3}=6


3.

When sampling with replacement the variance of all sample means equals the variance of the population divided by the sample size


σ2=13((3−6)2+(6−6)2+(9−6)2=6\sigma^2=\dfrac{1}{3}((3-6)^2+(6-6)^2+(9-6)^2=6

Var(Xˉ)=σXˉ2=σ2n=64=1.5Var(\bar{X})=\sigma_{\bar{X}}^2=\dfrac{\sigma^2}{n}=\dfrac{6}{4}=1.5



4.


σXˉ=σXˉ2=σ2n=σn\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{\dfrac{\sigma^2}{n}}=\dfrac{\sigma}{\sqrt{n}}




=64=1.5≈1.224745=\dfrac{\sqrt{6}}{\sqrt{4}}=\sqrt{1.5}\approx1.224745


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