Answer to Question #348860 in Statistics and Probability for CC.....

Question

Answer what is asked in the given problem below. Show your solutions completely. Use an extra sheet of paper for your solutions. NO SOLUTIONS, NO POINTS.

The population is the weight of six pumpkins (in pounds) displayed in carnival "guess the weight" game booth. You are asked to guess the average weight of the six pumpkins by taking a random sample without replacement from the population. Suppose that the sample size of 5 pumpkins were drawn from this population (without replacement), describe the sampling distribution of the sample means.

Pumpkin

A

B

C

D

E

F

Weight in Pounds

19

14

15

9

10

17

What is the mean and variance of the sampling distribution of the sample means?

Compare these values to the mean and variance of the population.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} Pumpkin & Weight\ in\ pounds \\ \hline A & 19 \\ \hdashline B & 14 \\ \hdashline C & 15 \\ \hdashline D & 9 \\ \hdashline E & 10 \\ \hdashline F & 17 \\ \hdashline \end{array}$$

We have population values 9,10,14,15,17,19, population size N=6 and sample size n=5.

Mean of population $(\mu)$ = $\dfrac{9+10+14+15+17+19}{6}=14$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{25+16+0+1+9+25}{6}$$ $$=\dfrac{38}{3}$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{38}{3}}\approx3.56$$

Select a random sample of size 5 without replacement. We have a sample distribution of sample mean.

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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 9,10,14,15,17 & 65/5 \\ \hdashline 2 & 9,10,14,15,19 & 67/5 \\ \hdashline 3 & 9,10,14,17,19 & 69/5 \\ \hdashline 4 & 9,10,15,17,19 & 70/5\\ \hdashline 5 & 9,14,15,17,19 & 74/5 \\ \hdashline 6 & 10,14,15,17,19 & 75/4 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 65/5 & 1/6 & 65/30 & 4225/150 \\ \hdashline 67/5 & 1/6 & 67/30 & 4489/150 \\ \hdashline 69/5 & 1/6 & 69/30 & 4761/150 \\ \hdashline 70/5 & 1/6 & 70/30 & 4900/150 \\ \hdashline 74/5 & 1/6 & 74/30 & 5476/150 \\ \hdashline 75/5 & 1/6 & 75/30 & 5625/150 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{420}{30}=14=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{29476}{150}-(14)^2=\dfrac{38}{75}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$