Answer to Question #342345 in Statistics and Probability for max

Question #342345

Construct the sampling distribution of the sample mean for the population that consists of the scores 15, 18, 11, 14, 17, with a sample size of 3. Compute as well the mean of the sampling distribution using the expected value of the distribution and verify the result by computing for the population mean.


1
Expert's answer
2022-05-19T09:33:12-0400

We have population values 15, 18, 11, 14, 17, population size N=5 and sample size n=3.

Mean of population (μ)(\mu) = 15+18+11+14+175=15\dfrac{15+18+11+14+17}{5}=15

Variance of population 


σ2=Σ(xixˉ)2n=0+9+16+1+45=6\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{0+9+16+1+4}{5}=6


σ=σ2=62.4495\sigma=\sqrt{\sigma^2}=\sqrt{6}\approx2.4495

Select a random sample of size 3 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is NCn=5C3=10.^{N}C_n=^{5}C_3=10.

noSampleSamplemean (xˉ)111,14,1540/3211,14,1742/3311,14,1843/3411,15,1743/3511,15,1844/3611,17,1846/3714,15,1746/3814,15,1847/3914,17,1849/31015,17,1850/3\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 11,14,15 & 40/3 \\ \hdashline 2 & 11,14,17 & 42/3 \\ \hdashline 3 & 11,14,18 & 43/3 \\ \hdashline 4 & 11,15,17 & 43/3 \\ \hdashline 5 & 11,15,18 & 44/3 \\ \hdashline 6 & 11,17,18 & 46/3 \\ \hdashline 7 & 14,15,17 & 46/3 \\ \hdashline 8 & 14, 15,18 & 47/3 \\ \hdashline 9 & 14,17,18 & 49/3 \\ \hdashline 10 & 15, 17,18 & 50/3 \\ \hdashline \end{array}



B.


Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)40/31/1040/301600/9042/31/1042/301764/9043/32/1086/303698/9044/31/1044/301936/9046/32/1092/304232/9047/31/1047/302209/9049/31/1049/302401/9050/31/1050/302500/90\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 40/3 & 1/10 & 40/30 & 1600/90 \\ \hdashline 42/3 & 1/10 & 42/30 & 1764/90 \\ \hdashline 43/3 & 2/10 & 86/30 & 3698/90 \\ \hdashline 44/3 & 1/10 & 44/30 & 1936/90 \\ \hdashline 46/3 & 2/10 & 92/30 & 4232/90 \\ \hdashline 47/3 & 1/10 & 47/30 & 2209/90 \\ \hdashline 49/3 & 1/10 & 49/30 & 2401/90 \\ \hdashline 50/3 & 1/10 & 50/30 & 2500/90 \\ \hdashline \end{array}



Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=45030=15=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{450}{30}=15=\mu



The variance of sampling distribution 

Var(Xˉ)=σXˉ2=Xˉi2f(Xˉi)[Xˉif(Xˉi)]2Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2=2034090(15)2=1=σ2n(NnN1)=\dfrac{20340}{90}-(15)^2=1= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})

σXˉ=1=1\sigma_{\bar{X}}=\sqrt{1}=1


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