Answer to Question #208749 in Statistics and Probability for Jess

Question

Situation: Joe, a DOH COVID-19 tracker, found out that the number of infected persons of COVID-19 in seven low-risk provinces of the country are 1, 9, 2, 12, 8, 7, and 10. Suppose that 5 provinces are drawn as sample.

1. What is the mean (u), variance (²) and standard deviation of the population (o)? 2. How many different samples of size n-5 can be drawn from the population?

List them with their corresponding mean. 3. Construct the sampling distribution of the sample mean.

4. What is the mean () of the sampling distribution of the sample mean? Compare this to the mean of the population.

5. What is the variance of the (o,) of the sampling distribution of the sample mean?

Answer 1

1. We have population values $1,2,7,8,9,10,12$ population size $N=7$ and sample size $n=5.$

Mean

$$\mu=\dfrac{1+2+7+8+9+10+12}{7}=7$$

Variance

$$\sigma^2=\dfrac{1}{7}\big((1-7)^2+(2-7)^2+(7-7)^2$$

$$(8-7)^2+(9-7)^2+(10-7)^2+(12-7)^2\big)$$

$$=\dfrac{100}{7}$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{100}{7}}\approx3.779645$$

2. The number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{7}{5}=21$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,2, 7, 8, 9 & 5.4 \\ \hdashline 2 & 1,2, 7, 8, 10 & 5.6 \\ \hdashline 3 & 1,2, 7,8,12 & 6.0 \\ \hdashline 4 & 1,2, 7, 9, 10 & 5.8\\ \hdashline 5 & 1,2,7,9,12 & 6.2 \\ \hdashline 6 & 1,2, 7, 10,12 & 6.4 \\ \hline 7 & 1,2,8,9,10 & 6.0 \\ \hline 8 & 1,2,8,9,12 & 6.4\\ \hline 9 & 1,2,8,10,12 & 6.6 \\ \hline 10 & 1,2,9,10,12 & 6.8 \\ \hline 11 & 1,7,8,9,10 & 7.0 \\ \hline 12 & 1,7,8,9,12 & 7.4 \\ \hline 13 & 1,7, 8,10,12 & 7.6 \\ \hline 14 & 1,7,9,10,12 & 7.8 \\ \hline 15 & 1,8,9,10,12& 8.0 \\ \hline 16 & 2,7,8,9,10& 7.2 \\ \hline 17 & 2,7,8,9,12 & 7.6 \\ \hline 18 & 2,7,8,10,12 & 7.8\\ \hline 19 & 2,7,9,10,12 & 8.0 \\ \hline 20 & 2,8,9,10,12 & 8.2 \\ \hline 21 & 7,8, 9, 10, 12 & 9.2 \\ \hline \end{array}$$

3.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 5.4 & 1 & 1/21 & 5.4/21 & 29.16/21 \\ \hdashline 5.6 & 1 & 1/21 & 5.6/21 & 31.36/21 \\ \hdashline 5.8 & 1 & 1/21& 5.8/21 & 33.64/21\\ \hdashline 6.0 & 2 & 2/21 & 12.0/21 & 72/21 \\ \hdashline 6.2 & 1 & 1/21 &6.2/21 & 38.44/21\\ \hdashline 6.4 & 2 & 2/21 & 12.8/21 & 81.92/21\\ \hdashline 6.6 & 1 & 1/21 & 6.6/21 & 43.56/21\\ \hdashline 6.8 & 1 & 1/21 & 6.8/21 & 46.24/21 \\ \hdashline 7.0 & 1 & 1/21 & 7.0/21 & 49/21\\ \hdashline 7.2 & 1 & 1/21 & 7.2/21 & 51.84/21 \\ \hdashline 7.4 & 1& 1/21 & 7.4/21 & 54.76/21\\ \hdashline 7.6 & 2 & 2/21 & 15.2/21 & 115.52/21\\ \hdashline 7.8 & 2 & 2/21 & 15.6/21 & 121.68/21\\ \hdashline 8.0 & 2 & 2/21 & 16.0/21 & 128/21\\ \hdashline 8.2 & 1& 1/21 & 8.2/21 & 67.24/21\\ \hdashline 9.2 & 1 & 1/21 & 9.2/21 & 84.64/21\\ \hdashline Total & 21 & 1 & 7 & 1049/21 \\ \hline \end{array}$$

4.

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=7$$

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

the mean of the population.

$$E(\bar{X})=7=\mu$$

5.

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

$$=\dfrac{1049}{21}-(7)^2=\dfrac{20}{21}$$

$$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{20}{21}}\approx0.975900$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{\dfrac{100}{7}}{5}(\dfrac{7-5}{7-1})$$

$$=\dfrac{20}{21}, True$$