Answer to Question #312636 in Statistics and Probability for joli1a

Question

irections. From your family members, form a group of five. (Just in case your

family is composed of less than five members, you can add one from your

neighbors). Get the weight in the kilogram of each member of the group. Draw

random samples of size 𝑛 = 2 from these weights.

1. List all possible samples and compute the mean of each sample.

2. Construct the sampling distribution of the sample means.

3. Find the mean of the population 𝜇.

4. Find the standard deviation of the population 𝜎.

5. Find the mean of the sampling distribution of the sample means 𝜇𝑋̅.

6. Find the standard deviation of the sampling distribution of the sample

means 𝜎𝑋̅.

7. Verify the Central Limit Theorem by:

a. Comparing 𝜇 and 𝜇𝑋̅.

b. Comparing 𝜎 and 𝜎𝑋̅.

Answer 1

Solution

Population size $N=5$

Weights obtained $=52,61,68,78,84$

Sample size $n=2$

1. Possible samples

$=C_5^2= \binom{5}{2}=10$ samples

$~~~~\def\arraystretch{1.5}\begin{array}{c:c:c}no &Sample& Sample ~mean\\\hline1&52,61&56.5\\\hdashline2&52,68&60.0\\\hdashline3&52,78&65.0\\\hdashline4&52,84&68.0\\\hdashline5&61,68&64.5\\\hdashline6&61,78&69.5\\\hdashline7&61,84&72.5\\\hdashline8&68,78&73.0\\\hdashline9&68,84&76.0\\\hdashline10&78,84&81.0\\\hline\end{array}$

2. Sampling distribution of the sample means

$~~~~~\def\arraystretch{1.5}\begin{array}{c:c:c}mean&freq& prob\\\hline56.5&1&1/10\\\hdashline60.0&1&1/10\\\hdashline64.5&1&1/10\\\hdashline65.0&1&1/10\\\hdashline68.0&1&1/10\\\hdashline69.5&1&1/10\\\hdashline72.5&1&1/10\\\hdashline73.0&1&1/10\\\hdashline76.0&1&1/10\\\hdashline81.0&1&1/10\\\hline\end{array}$

3. Mean of the population.

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

$\mu=\dfrac{52+61+68+78+84}{5}$

$\mu =68.6$

4. Standard deviation

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

$~~~\def\arraystretch{1.5}\begin{array}{c:c:c}X_i&X_i-\bar X&(X_i-\bar X)^2\\\hline56.5&-12.1&146.41\\\hdashline60.0&-8.6&73.96\\\hdashline64.5&-4.1&16.81\\\hdashline65.0&-3.6&12.96\\\hdashline68.0&-0.6&0.36\\\hdashline69.5&0.9&0.81\\\hdashline72.5&3.9&15.21\\\hdashline73.0&4.4&19.36\\\hdashline76.0&7.4&54.76\\\hdashline81.0&12.4&153.76\\\hline\sum&&494.40\\\hline\end{array}$

$\sigma=\sqrt{\dfrac{494.40}{5}}=9.944$

5. Mean of the sampling distribution

$\mu_x=\sum XP(X)$

$\mu_x=56.5(0.1)+60.0(0.1)+64.5(0.1)+65.0(0.1)+68.0(0.1)+69.5(0.1)+72.5(0.1)+73.0(0.1)+76.0(0.1)+81.0(0.1)$

$\mu_x=68.6$

6. Standard deviation of the sampling distribution of the sample means.

$\sigma_x=\dfrac{\sigma}{\sqrt N}=\dfrac{9.944}{\sqrt 5}$

$\sigma_x=4.45$

7. Verification

The central limit theory states that if you take samples of larger and larger size from any population, then the mean $\overline{x}$ of the sample tends to get closer and closer to $μ$ and the standard deviation gets smaller.

(a) $\mu=\mu_{\overline x} ~~~(68.6=68.6)$

hence verified.

(b) $\sigma >\sigma_{\overline x}~~~(9.944>4.45)$

hence verified.