Answer to Question #243575 in Statistics and Probability for Gine

Question

A population consists of the five numbers 2,3,6,8 and 11.Consider samples of size that can be drawn from this population.

Answer 1

QUESTION

Suppose a population consists of the five measurements: 2, 6, 8, 0, and 1:

1. What is the mean and standard deviation of the population?

2. How many different samples of size n=2 can be drawn from the population? List them with their corresponding means.

SAMPLE MEAN

3. Construct the sampling distribution of the sample means.

4. What is the mean of the sampling distribution of te sample means? Compare this to the mean of the population.

5. What is the standard deviation of the sampling distribution of the sample means? Vompare this to the standard deviation of the population.

SOLUTION

1. Mean

$$\mu=\dfrac{2+6+8+0+1}{5}=3.4$$

Variance

$$\sigma^2=\dfrac{1}{5}\big((2-3.4)^2+(6-2.4)^2+(8-3.4)^2$$ $$(0-3.4)^2+(1-3.4)^2\big)=9.44$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{9.44}\approx3.0725$$

2. We have population values $2,6,8,0,1$ population size $N=5$ and sample size $n=2.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{5}{2}=10$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,6 & 4 \\ \hdashline 2 & 2,8 & 5 \\ \hdashline 3 & 2,0 & 1 \\ \hdashline 4 & 2,1 & 1.5 \\ \hdashline 5 & 6,8 & 7 \\ \hdashline 6 & 6,0 & 3 \\ \hdashline 7 & 6,1 & 3.5 \\ \hdashline 8 & 8,0 & 4 \\ \hdashline 9 & 8,1 & 4.5 \\ \hdashline 10 & 0,1 & 0.5 \\ \hline \end{array}$$

3. The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 1/2 & 1& 1/10 & 1/20 & 1/40 \\ \hdashline 1 & 1 & 1/10 & 2/20 & 4/40 \\ \hdashline 3/2 & 1 & 1/10 & 3/20 & 9/40 \\ \hdashline 3 & 1 & 1/10 & 6/20 & 36/40 \\ \hdashline 7/2 & 1 & 1/10 & 7/20 & 49/40 \\ \hdashline 4 & 2 & 2/10 & 16/20 & 128/40 \\ \hdashline 9/2 & 1& 1/10 & 9/20 & 81/40 \\ \hdashline 5 & 1& 1/10 & 10/20 & 100/40 \\ \hdashline 7 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline Total & 10 & 1 & 68/20 & 604/40 \\ \hline \end{array}$$

4.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{68}{20}=3.4$$

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

the mean of the population.

$$E(\bar{X})=3.4=\mu$$

5.

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$ $$=\dfrac{604}{40}-(\dfrac{68}{20})^2=\dfrac{1416}{400}=\dfrac{354}{100}=3.54$$ $$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{354}{100}}\approx1.8815$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{2}(\dfrac{5-2}{5-1})$$ $$=3.54, True$$