Answer to Question #334996 in Statistics and Probability for Kevin

Question

The following are the ages of 8 jeepney passengers in a waiting shed: 12, 15, 18, 19, 21, 23, 35, and 40. The sample with size 7 is chosen to ride in the current jeepney.

Determine the standard error of the sample standard deviations.

Answer 1

We have population values 12, 15, 18, 19, 21, 23, 35, and 40 population size N=8 and sample size n=7.

Mean of population $(\mu)$ = 

$$\dfrac{12+15+18+19+21+23+35+40}{8}=22.875$$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=82.859375$$

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

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{8}C_7=8.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 12, 15, 18, 19, 21, 23, 35 & 143/7 \\ \hdashline 2 & 12, 15, 18, 19, 21, 23, 40 & 148/7 \\ \hdashline 3 & 12, 18, 19, 21, 23, 35,40 & 168/7 \\ \hdashline 4 & 12, 15, 19, 21, 23, 35, 40 & 165/7 \\ \hdashline 5 & 12, 15, 18, 21, 23, 35, 40 & 164/7 \\ \hdashline 6 & 12, 15,18, 19, 23, 35,40 & 162/7 \\ \hdashline 7 & 12,15, 18, 19, 21, 35,40 & 160/7 \\ \hdashline 8 & 15, 18, 19, 21, 23, 35, 40 & 171/7 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 143/7 & 1/8 & 143/56 & 20449/392\\ \hdashline 148/7 & 1/8 & 148/56 & 21904/392\\ \hdashline 168/7 & 1/8 & 168/56 & 28224/392\\ \hdashline 165/7 & 1/8 & 165/56 & 27225/392 \\ \hdashline 164/7 & 1/8 & 164/56 & 26896/392\\ \hdashline 162/7 & 1/8 & 162/56 & 26244/392 \\ \hdashline 160/7 & 1/8 & 160/56 & 25600/392 \\ \hdashline 171/7 & 1/8 & 171/56 & 29241/392 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=22.875=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{205783}{392}-(\dfrac{183}{8})^2=\dfrac{5303}{3136}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{5303}{3136}}\approx1.3004$$

$$SE=\dfrac{\sigma_{\bar{X}}}{\sqrt{n}}=\dfrac{\sqrt{\dfrac{5303}{3136}}}{\sqrt{7}}\approx0.4915$$