Answer to Question #335010 in Statistics and Probability for Kevin

Question

Random samples with size 5 are drawn from the population containing the values 26, 32, 41, 50, 58, and 63.

Determine the standard error of the sample means

Answer 1

We have population values 26, 32, 41, 50, 58, and 63 population size N=6 and sample size n=5.

Mean of population $(\mu)$ = 

$\dfrac{26+32+41+50+58+63}{6}=45$

Variance of population 

$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{1064}{6}$

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{1064}{6}}$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_5=6.$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 26, 32, 41, 50, 58 & 207/5 \\ \hdashline 2 & 26, 32, 41, 50, 63 & 212/5 \\ \hdashline 3 & 26, 32, 41, 63, 58 & 220/5 \\ \hdashline 4 & 26, 32, 63, 50, 58 & 229/5 \\ \hdashline 5 & 26, 63, 41, 50, 58 & 238/5 \\ \hdashline 6 & 63, 32, 41, 50, 58 & 244/5 \\ \hdashline \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 207/5 & 1/6 & 207/30 & 42849/150\\ \hdashline 212/5 & 1/6 & 212/30 & 44944/150\\ \hdashline 220/5 & 1/6 & 220/30 & 48400/150\\ \hdashline 229/5 & 1/6 & 229/30 & 52441/150 \\ \hdashline 238/5 & 1/6 & 238/30 & 56644/150\\ \hdashline 244/5 & 1/6 & 244/30 & 59536/150 \\ \hdashline \end{array}$

Mean of sampling distribution 

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=45=\mu$

The variance of sampling distribution 

$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$ $=\dfrac{304814}{150}-(45)^2=\dfrac{1064}{150}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$

The standard error of the sample means is

$\sigma_{\bar{X}}=\sqrt{\dfrac{1064}{150}}\approx2.6633$