Q 7.79. - Elementary Statistics

Question

A variable of a population has mean $\mu$ and standard deviation $\sigma$ For a large sample size $n$, fill in the blanks, Justify your answers.

a. Approximately _  $\%$ of all possible samples have means within $\sigma /\sqrt{n}$ of the population mean, $\mu$.

b. Approximately _  $\%$ of all possible samples have means within $2\sigma /\sqrt{n}$ of the population mean, $\mu$

c. Approximately _ $\%$ of all possible samples have means within $3\sigma /\sqrt{n}$ of the population mean, $\mu$

d. Approximately __  $\%$ of all possible samples have means within $z_{v/2}$ of the population mean, $\mu$

Step-by-Step Solution

Answer

a. Approximately $68.26$ $\%$ of all possible samples have means within $\sigma /\sqrt{n}$ of the population mean $\mu$

b. Approximately $95.44$ $\%$ of all possible samples have means within $2\sigma /\sqrt{n}$ of the population mean, $\mu$.

c. Approximately $99.74$ $\%$ of all possible samples have means within $3\sigma /\sqrt{n}$ of the population mean, $\mu$

d. Approximately $100\left(1-\sigma \right)$$\%$ of all possible samples have means within $z_{\theta /2}$ of the population mean, $\mu$

Part (a) Step 1: Given information

Population mean $\mu$ & population S.D $\sigma$

Sample size $n$ is large

$\therefore$ By using CLT, sample mean $\overline{x}$ approximates normal distribution with mean $\mu \&$${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Step 2: Concept

The formula used: Standard deviation= ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

a Step 1: Calculation

Around $68.26\%$ of all feasible samples have means that are within $\frac{\sigma }{\sqrt{n}}$ of the population mean $\mu$

Justification: let $z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}$

Then $z~N(0,1)$

Now for standard normal variable $z$

$$\begin{gathered} P(-1\le z\le 1)=.6826 \\ \Rightarrow P\left(-1\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le 1\right)=.6826 \\ \Rightarrow P\left(-{\sigma }_{\overline{x}}\le \overline{x}-\mu \le {\sigma }_{\overline{x}}\right)=.6826 \\ \Rightarrow P\left(\mu -{\sigma }_{\overline{x}}\le \overline{x}\le \mu +{\sigma }_{\overline{x}}\right)=.6826 \\ \text{ i.e }P\left(\mu -\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +\frac{\sigma }{\sqrt{n}}\right)=.6826 \end{gathered}$$

Hence the proof of the statement in $\left(a\right)$

b Step 1: Calculation

The population mean $\mu$ is within $\frac{2\sigma }{\sqrt{n}}$ of the means of approximately $95.44\%$ percent of all possible samples.

Justification: 

$$\begin{gathered} P(-2\le z\le 2)=.9544\text{ Where }z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}} \\ \Rightarrow P\left(-2\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le 2\right)=.9544z~N(0,1) \\ \Rightarrow P\left(-2{\sigma }_{\overline{X}}\le \overline{x}-\mu \le 2{\sigma }_{\overline{X}}\right)=.9544 \\ \Rightarrow P\left(\mu -2{\sigma }_{\overline{x}}\le \overline{x}\le \mu +2{\sigma }_{\overline{X}}\right)=.9544 \\ \Rightarrow P\left(\mu -2\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +2\frac{\sigma }{\sqrt{n}}\right)=.9544 \end{gathered}$$

c Step 1: Calculation

$99.74\%$ of all feasible samples have means that are within $\frac{3\sigma }{\sqrt{n}}$ of the population mean $\mu$

Justification: $P(-3\le z\le 3)=.9974 z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}$

$$\begin{gathered} z~N(0,1) \\ \Rightarrow P\left(-3\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le 3\right)=.9974 \\ \Rightarrow P\left(-3{\sigma }_{\overline{x}}\le \overline{x}-\mu \le 3{\sigma }_{\overline{x}}\right)=.9974 \\ \Rightarrow P\left(\mu -3{\sigma }_{\overline{x}}\le \overline{x}\le \mu +3{\sigma }_{\overline{x}}\right)=.9974 \\ \Rightarrow P\left(\mu -3\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +3\frac{\sigma }{\sqrt{n}}\right)=.9974 \end{gathered}$$

d Step 1: Calculation

The means of approximately $100(1-\alpha )\%$ of all feasible samples are within $z_{\frac{\alpha }{2}}·\frac{\sigma }{\sqrt{n}}$ mean $\mu$ of the population

Justification: $$\begin{gathered} P\left(-z_{\frac{\alpha }{2}}\le z\le z_{\frac{\alpha }{2}}\right)=1-\alpha \\ \Rightarrow P\left(-z_{\frac{\alpha }{2}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le z_{\frac{\alpha }{2}}\right)=1-\alpha \end{gathered}$$

d Step 2Step 2: Concept : Calculation

$$\begin{gathered} \Rightarrow P\left(\mu -z_{\frac{a}{2}}{\sigma }_{\overline{x}}\le \overline{x}\le \mu +z_{\frac{\alpha }{2}}·{\sigma }_{\overline{x}}\right)=1-\alpha \\ \Rightarrow P\left(\mu -\frac{z\alpha }{2}·\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +z_{\frac{\alpha }{2}}·\frac{\sigma }{\sqrt{n}}\right)=1-\alpha \end{gathered}$$