Chapter 10.3, Problem 10.4E

(a)

To determine

whether the test scores have been selected from a normal population using a stem and leaf plot.

Answer to Problem 10.4E

Yes, the test scores have been selected from a normal population

Explanation of Solution

Given:

The below table describes the test scores on a 100-point test for 20 students:

$71$	$93$	$91$	$86$	$75$
$73$	$86$	$82$	$76$	$57$
$84$	$89$	$67$	$62$	$72$
$77$	$68$	$65$	$75$	$84$

Calculation:

Stem and leaf plot:

Draw a vertical line

Place the digits of the tens to the left of the vertical line and the digits of the ones of every data value to the right of the vertical line

stem	leaf
5	7
6	2 5 7 8
7	1 2 3 5 5 6 7
8	2 4 4 6 6 9
9	1 3

It is reasonable to assume that the test scores have been selected from a normal population, because most of the data values lie roughly in the middle of the distribution indicating a symmetric distribution and there are no outliers visible (data values that are separated from the other data values with a gap.)

Conclusion:

Hence, yes, the test scores have been selected from a normal population

(b)

To determine

To Calculate: the mean and standard deviation of the scores

Answer to Problem 10.4E

The mean of the test scores is $\text{76}.65$

The standard deviation of the test scores is $\text{10}.038$

Explanation of Solution

Given:

The below table describes the test scores on a 100-point test for 20 students:

$71$	$93$	$91$	$86$	$75$
$73$	$86$	$82$	$76$	$57$
$84$	$89$	$67$	$62$	$72$
$77$	$68$	$65$	$75$	$84$

Calculation:

The mean of the test scores is

  $\begin{array}{l}\overline{x}\text{ =}\frac{\Sigma x_i}{n}\\ =\frac{1533}{20}=\text{76}.65\end{array}$

The variance of test scores is

  $\begin{array}{l}{\text{s}}^{\text{2}}\text{ =}\frac{\Sigma x_i{}^2-\frac{{\left(\Sigma x_i\right)}^2}{n}}{n-1}\\ \text{=}\frac{119419-\frac{{\left(1533\right)}^2}{20}}{20-1}\\ =\frac{1914.55}{19}=\text{ 100}.766\end{array}$

The standard deviation of the test scores is

  $\begin{array}{l}s=\sqrt{s^2}\\ =\sqrt{100.766}=\text{ 10}.038\end{array}$

Conclusion:

Hence, the mean of the test scores is $\text{76}.65$ and the standard deviation of the test scores is $\text{10}.038$

(c)

To determine

To find: the $95\%$ confidence interval for the average test score in the population.

Answer to Problem 10.4E

The $95\%$ confidence interval for the average test score in the population is $71.95,81.3480$

Explanation of Solution

Given:

The below table describes the test scores on a 100-point test for 20 students:

$71$	$93$	$91$	$86$	$75$
$73$	$86$	$82$	$76$	$57$
$84$	$89$	$67$	$62$	$72$
$77$	$68$	$65$	$75$	$84$

Calculation:

If we choose a $5\%$ level of significance $\left(\alpha =0\cdot 05\right)$ , the critical value can be found from the values of t from Table 4 or Appendix I. With $df=n-1=19$

The critical value is

  $t_{\frac{\alpha }{2}}=t_{\frac{0\cdot 05}{2}}\text{ =2}\cdot \text{093}$

A $95\%$ confidence interval gives the upper and lower limits for $\mu$ as,

  $\begin{array}{l}\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)\\ =\text{ 76}\text{.65}\pm \text{2}\text{.093}\left(\frac{10.038}{\sqrt{20}}\right)\\ =\text{ 76}\text{.65}\pm 4.698\\ =\left(71.95,81.3480\right)\end{array}$

Conclusion:

Hence, the $95\%$ confidence interval for the average test score in the population is $71.95,81.3480$