Answer to Question #312499 in Statistics and Probability for Ningning

Question

Consider all samples of size 5 from this population:6 8 10 12 13

1. Compute the mean and variance of the population

2.list all samples of size 4 and compute the mean for each sample

3.construct the sampling distribution of the sample means

4.Calculate the mean of the sampling distribution of the sample means and compare this to the mean of the population

Answer 1

Solution

Population size $N=5$

1.(a) mean

$\mu=\dfrac{\sum X_i}N$

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

$\mu=9.8$

(b) Variance

$\def\arraystretch{1.5}\begin{array}{c:c:c}X_i &X_i -\bar X& (X_i -\bar X)^2\\\hline6&-3.8&14.44\\\hdashline8&-1.8&3.24\\\hdashline10&0.2&0.04\\\hdashline12&2.2&4.84\\\hdashline13&3.2&10.24\\\hline\end{array}$

$\sigma^2=\dfrac{\sum( X_i-\bar X)}N=\dfrac{32.8}5$

$\sigma^2=6.56$

2. Samples of size 4 and their means

Sample size $n=4$

Possible samples

$=C_5^4= \binom{5}{4}=5$ samples

$\def\arraystretch{1.5}\begin{array}{c:c:c}Sample~mean& frequency & Probability \\\hline9.00&1&1/5\\\hdashline9.25&1&1/5\\\hdashline9.75&1&1/5\\\hdashline10.25&1&1/5\\\hdashline10.75&1&1/5 \\\hline\end{array}$$\def\arraystretch{1.5}\begin{array}{c:c:c}no.&Samples&means\\\hline1&6,8,10,12&9.00\\\hdashline2&6,8,10,13&9.25\\\hdashline3&6,8,12,13&9.75\\\hdashline4&6,10,12,13&10.25\\\hdashline5&8,10,12,13&10.75\\\hline\end{array}$

3. Sampling distribution of the sample means

$\def\arraystretch{1.5}\begin{array}{c:c:c}mean& freq & Prob\\\hline9.00&1&1/5\\\hdashline9.25&1&1/5\\\hdashline9.75&1&1/5\\\hdashline10.25&1&1/5\\\hdashline10.75&1&1/5 \\\hline\end{array}$

4. The mean of the sampling distribution

$\bar X=X_1P(1)+X_2P(2)+X_3P(3)+X_4P(4)+X_5P(5)$

$\bar X=9.00(1/5)+9.25(1/5)+9.75(1/5)+10.25(1/5)+10.75(1/5)$

$\bar X=9.8$

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