Chapter 7.3, Problem 19E

(a)

To determine

To calculate: The population mean, $\mu$ and the population standard deviation, $\sigma$ .

Answer to Problem 19E

The population mean, $\mu$ is $75.4$ and the population standard deviation, $\sigma$ is $5.12$ .

Explanation of Solution

Given information:The data is shown below in table.

Date	Temperature
July 21	69
July 22	75
July 23	79
July 24	83
July 24	71

Formula used:

  $\begin{array}{l}\text{Population mean: }\mu =\frac{{\displaystyle \sum x}}{N}\\ \text{Population Standard Deviation: }\sigma =\sqrt{{\sigma }^2}=\sqrt{\frac{{\displaystyle \sum (x-}\mu {)}^2}{N}}\end{array}$

Calculation:

The population mean, $\mu$ and the population standard deviation, $\sigma$ can be calculated by using above formulas as follows:

  $\begin{array}{l}\text{Population mean: }\mu =\frac{{\displaystyle \sum x}}{N}=\frac{69+75+79+83+71}{5}\\ \text{Population mean: }\mu =75.4\\ \text{Population Standard Deviation: }\sigma =\sqrt{{\sigma }^2}=\sqrt{\frac{{\displaystyle \sum (x-}\mu {)}^2}{N}}\\ \text{Population Standard Deviation: }\sigma =\sqrt{\frac{{(69-75.4)}^2+{(75-75.4)}^2+\mathrm{.....}+{(71-75.4)}^2}{5}}\\ \text{Population Standard Deviation: }\sigma =5.12\\ \end{array}$

(b)

To determine

To list: All samples of size, $2$ drawn with replacement.

Answer to Problem 19E

All $25$ samples of size $2$ (with replacement) are listed in Table $1$ .

Explanation of Solution

Given information:The data is shown below in table.

Date	Temperature
July 21	69
July 22	75
July 23	79
July 24	83
July 24	71

Calculation:

There are $25$ samples of size $2$ with replacement as shown in Table $1$ .

Sample
$69,69$
$69,75$
$69,79$
$69,83$
$69,71$
$75,69$
$75,75$
$75,79$
$75,83$
$75,71$
$79,69$
$79,75$
$79,79$
$79,83$
$79,71$
$83,69$
$83,75$
$83,79$
$83,83$
$83,71$
$71,69$
$71,75$
$71,79$
$71,83$
$71,71$

(c)

To determine

To compute:Sample mean of eachof the $25$ samples, mean ${\mu }_{\overline{X}}$ of the sample means and the standard deviation, ${\sigma }_{\overline{X}}$ of the sample means.

Answer to Problem 19E

Sample means of eachof the $25$ samples are shown in Table $2$ .

The mean, ${\mu }_{\overline{X}}$ of the sample meansis $75.4$ .

The standard deviation, ${\sigma }_{\overline{X}}$ of the sample means is $3.6968$ .

Explanation of Solution

Given information:The data is shown below in table.

Date	Temperature
July 21	69
July 22	75
July 23	79
July 24	83
July 24	71

Formula used:

  $\begin{array}{l}\text{The mean of sampling distribution: }{\mu }_{\overline{X}}=\frac{{\displaystyle \sum \overline{X}}}{n}\overline{X}\text{ - sample mean }n\text{ - sample size}\\ \text{The standard deviation of sampling distribution: }{\sigma }_{\overline{X}}=\sqrt{{\sigma }_{\overline{X}}^2}=\sqrt{\frac{{\displaystyle \sum (\overline{X}-}{\mu }_{\overline{X}}{)}^2}{n-1}}\end{array}$

Calculation:

The sample mean can be calculated by adding the two samples together and dividing by the sample size which is $2.$

Sample	Sample Mean, $\overline{X}$
$69,69$	$\frac{69+69}{2}=69$
$69,75$	$\frac{69+75}{2}=72$
$69,79$	$\frac{69+79}{2}=74$
$69,83$	$\frac{69+83}{2}=76$
$69,71$	$\frac{69+71}{2}=70$
$75,69$	$\frac{75+69}{2}=72$
$75,75$	$\frac{75+75}{2}=75$
$75,79$	$\frac{75+79}{2}=77$
$75,83$	$\frac{75+83}{2}=79$
$75,71$	$\frac{75+71}{2}=73$
$79,69$	$\frac{79+69}{2}=74$
$79,75$	$\frac{79+75}{2}=77$
$79,79$	$\frac{79+79}{2}=79$
$79,83$	$\frac{79+83}{2}=81$
$79,71$	$\frac{79+71}{2}=75$
$83,69$	$\frac{83+69}{2}=76$
$83,75$	$\frac{83+75}{2}=79$
$83,79$	$\frac{83+79}{2}=81$
$83,83$	$\frac{83+83}{2}=83$
$83,71$	$\frac{83+71}{2}=77$
$71,69$	$\frac{71+69}{2}=70$
$71,75$	$\frac{71+75}{2}=73$
$71,79$	$\frac{71+79}{2}=75$
$71,83$	$\frac{71+83}{2}=77$
$71,71$	$\frac{71+71}{2}=71$

The mean, ${\mu }_{\overline{X}}$ of the sample means is the average of all sample means which is calculated by adding all $25$ sample means and dividing by $25.$

  $\begin{array}{l}\text{The mean of the sample means: }{\mu }_{\overline{X}}=\frac{{\displaystyle \sum \overline{X}}}{n}=\frac{69+72+74+\mathrm{........}+75+77+71}{25}\\ \text{The mean of the sample means: }{\mu }_{\overline{X}}=75.4\\ \text{The standard deviation of the sample means: }{\sigma }_{\overline{X}}=\sqrt{\frac{{(69-75.4)}^2+\mathrm{.............}+{(71-75.4)}^2}{24}}\\ \text{The standard deviation of the sample means: }{\sigma }_{\overline{X}}=3.6968\end{array}$

(d)

To determine

To verify: The relationship between the population mean, $\mu$ and the mean of the sample means, ${\mu }_{\overline{X}}$

  $({\mu }_{\overline{X}}=\mu )$ and the relationship between the population standard deviation, $\sigma$ and the standard deviation of the sample means, ${\sigma }_{\overline{X}}$

  $({\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{2}})$

Answer to Problem 19E

Based on the calculated values, the relationships of ${\mu }_{\overline{X}}=\mu (=75.4)$ and ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{2}}$ can be verified as true.

Explanation of Solution

Given information:The data is shown below in table.

Date	Temperature
July 21	69
July 22	75
July 23	79
July 24	83
July 24	71

Calculation:

As per the calculations in part (a) and (c), the mean of the sample means, ${\mu }_{\overline{X}}$ and the population mean, $\mu$ are both equal to $75.4$ . So, the relationship of ${\mu }_{\overline{X}}=\mu (=75.4)$ holds true.

Based on the calculations in part (a), the population standard deviation, $\sigma$ is $5.12$ and the standard deviation of the sample means, ${\sigma }_{\overline{X}}$ is $3.6968$ .

  $\frac{\sigma }{\sqrt{2}}=\frac{5.12}{\sqrt{2}}=3.62$

  $\frac{\sigma }{\sqrt{2}}$ is approximately equal to ${\sigma }_{\overline{X}}$ though there is a slight difference in second decimal place. But, approximately, the relationship of $({\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{2}})$ holds true.