Answer to Question #256898 in Statistics and Probability for Peochi

Question

1. Consider all samples of size 4 from this population without replacement: 6 8 10 12 13 What is the mean and standard deviation of the Sampling Distribution? 2. Given the population 1, 3, 4, 6, and 8. Suppose samples of size 3 with replacement were drawn from this population. What is the mean and standard deviation of the Sampling Distribution? 3. A random sample of n=100 measurements is obtained from a population of N=120, with

Answer 1

1.

 We have population values $6,8,10,12,13,$ population size $N=5$ and sample size $n=4.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{4}=5$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 6,8,10,12 & 9 \\ \hdashline 2 & 6,8,10,13 & 9.25 \\ \hdashline 3 & 6,8,12,13 & 9.75 \\ \hdashline 4 & 6,10,12,13 & 10.25 \\ \hdashline 5 & 8,10,12,13 & 10.75 \\ \hline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 9 & 1 & 0.2 & 1.8 & 16.2 \\ \hdashline & 9.25 & 1& 0.2 & 1.85 & 17.1125 \\ \hdashline & 9.75 & 1& 0.2 & 1.95 & 19.0125 \\ \hdashline & 10.25 & 1& 0.2 & 2.05 & 21.0125 \\ \hdashline & 10.75 & 1 & 0.2 & 2.15 & 23.1125 \\ \hdashline Total & & 5 & 1 & 9.8 & 96.45 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=9.8$$

$$\mu=\dfrac{6+8+10+12+13}{5}=9.8$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=\mu_{\bar{X}}=9.8=\mu$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$

$$=96.45-(9.8)^2=0.41$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{0.41}\approx0.6403$$

Verification:

$$\sigma^2=\dfrac{1}{5}((6-9.8)^2+(8-9.8)^2+(10-9.8)^2$$

$$+(12-9.8)^2+(13-9.8)^2)=6.56$$

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{6.56}{4}(\dfrac{5-4}{5-1})$$ $$=\dfrac{6.56}{16}=0.41, True$$

2.

 We have population values $1,3,4,6,8,$ population size $N=5$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn with replacement is

$$N^n=5^3=125$$

The mean of the sampling distribution of the sample means is equal to the the mean of the population.

$$\mu_{\bar{X}}=\mu=\dfrac{1+3+4+6+8}{5}=4.4$$

$$\sigma^2=\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2$$

$$+(6-4.4)^2+(8-4.4)^2)=5.84$$

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}=\dfrac{5.84}{4}=1.46$$

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{1.46}\approx1.2083$$

3.

We have population size $N=120$ and sample size $n=100.$ 

The mean of the sampling distribution of the sample means is equal to the the mean of the population.

$$\mu_{\bar{X}}=\mu$$

Variance of the sampling distribution of the sample means is

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\dfrac{\sigma^2}{n}$$

Standard deviation of the sampling distribution of the sample means is

$$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\dfrac{\sigma}{\sqrt{n}}$$