Chapter 11.3, Problem 9E

To determine

(a)

Null hypothesis and alternative hypothesis.

Answer to Problem 9E

Solution:

The required hypothesis is,

$H_0:{\mu }_d\le 10$

$H_1:{\mu }_d>10$

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book	78	82	90	67	79	83	89	93
Old book	67	70	79	54	68	71	78	82

Calculation:

Let the test scores of second graders studying old book be population 1 and the test scores of second graders studying new book be population 2. Let ${\text{μ}}_1$ be the mean test score of population 1 and ${\text{μ}}_{\text{2}}$ mean test score of population 2. And ${\mu }_d={\mu }_2-{\mu }_1$ The publisher claims that the standardized test scores will increase by more than 10 points after adopting the new book. Hence the alternative hypothesis is ${\mu }_d>10$ and since the null hypothesis is opposite of the alternative hypothesis then,

Thus, the null hypothesis is stated as,

$H_0:{\mu }_d\le 10$

Alternative hypothesis

$H_1:{\mu }_d>10$

To determine

(b)

The type of distribution to use for the test statistics and state the level of significance.

Answer to Problem 9E

Solution:

For the mean of the paired differences t- distribution is used and the level of significance is $\alpha =0.05$.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book	78	82	90	67	79	83	89	93
Old book	67	70	79	54	68	71	78	82

Calculation:

When the samples given are dependent over each other, the standard deviation of both the populations are unknown and all the given sample have equal probability of getting selected then t-distribution is used.

Since all the criteria described in the concept above are met, hence t-distribution will be used in this given question.

To determine

(c)

To Calculate:

The sample statistic and test statistic value.

Answer to Problem 9E

Solution:

The sample statistics are $\overline{d}=11.5$ and $s_d=0.7559$ . And the test statistic is 5.612.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book	78	82	90	67	79	83	89	93
Old book	67	70	79	54	68	71	78	82

Formula used:

When the standard deviation of the populations are unknown and the samples taken are dependent over each other then the test statistic for the hypothesis test for the mean of the paired differences is given by,

$t=\frac{\overline{d}-{\mu }_d}{\left(\frac{s_d}{\sqrt{n}}\right)}$

Where $\overline{d}$ is mean of the paired differences calculated as,

$\overline{d}=\frac{{\displaystyle \sum d_i}}{n}$

${\mu }_d$ is the value of the mean of the paired differences for the two given populations obtained from null hypothesis.

$s_d$ is the standard deviation of the mean of paired differences calculated as,

$s_d=\sqrt{\frac{{\displaystyle \sum \left(d_i-\overline{d}\right)}}{n-1}}$

n is the total number of data given,

And t is the test statistic value.

Calculation:

From the given information:

The null hypothesis is,

$H_0:{\mu }_d\le 10$

The alternative hypothesis is,

$H_1:{\mu }_d>10$

The level of significance $\alpha$ given is 0.05.

Since hypothesis test is for the mean of the paired differences so the distribution which will be used is t- distribution.

To test for the null hypothesis there is a need to calculate the test statistic value.

To calculate test statistic value, t, sample statistic are required.

Old Book	New Book	d	$\left(d_i-\overline{d}\right)$	${\left(d_i-\overline{d}\right)}^2$
67	78	11	$-0.5$	0.25
70	82	12	0.5	0.25
79	90	11	$-0.5$	0.25
54	67	13	1.5	2.25
68	79	11	$-0.5$	0.25
71	83	12	0.5	0.25
78	89	11	$-0.5$	0.25
82	93	11	$-0.5$	0.25
	Sum	92		4

The paired difference is calculated as,

$d_i=x_2-x_1$

Substitute 67 for $x_1$ and 78 for $x_2$ in the above equation.

$\begin{array}{rll}{d}_1& =& 78-67\\ & =& 11\end{array}$

Proceed in the same manner to calculate $d_i$ for the rest of the data and refer table for the rest of the $d_i$ values calculated. The size of each sample is 8 that is $n=8$. Then the mean of the paired differences is calculated as,

$\begin{array}{rll}\overline{d}& =& \frac{{\displaystyle \sum d_i}}{n}\\ & =& \frac{\left(-0.5\right)+0.5+\mathrm{....}+\left(-0.5\right)+\left(-0.5\right)}{8}\\ & =& \frac{92}{8}\\ & =& 11.5\end{array}$

Substitute 11 for $d_1$ and 11.5 for $\overline{d}$ in $\left(d_1-\overline{d}\right)$.

$\begin{array}{rll}\left(d_1-\overline{d}\right)& =& \left(11-11.5\right)\\ \left(d_1-\overline{d}\right)& =& -0.5\end{array}$

Square both sides of the equation.

$\begin{array}{rll}{\left(d_1-\overline{d}\right)}^2& =& {\left(-0.5\right)}^2\\ {\left(d_1-\overline{d}\right)}^2& =& 0.25\end{array}$

Proceed in the same manner to calculate ${\left(d_i-\overline{d}\right)}^2$ for all the $1\le i\le n$ for the rest data and refer table for the rest of the ${\left(d_i-\overline{d}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(d_i-\overline{d}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(d_i-\overline{d}\right)}}^2& =& 0.25+0.25+\mathrm{...}+0.25+0.25\\ & =& 4\end{array}$

The standard deviation is calculated as,

$s_d=\sqrt{\frac{{\displaystyle \sum \left(d_i-\overline{d}\right)}}{n-1}}$

Substitute 4 for ${\displaystyle \sum {\left(d_i-\overline{d}\right)}^2}$ and 9 for n in the above equation

$\begin{array}{rll}{s}_d& =& \sqrt{\frac{4}{7}}\\ & =& \sqrt{0.5714}\\ & =& 0.7559\end{array}$

Then the test statistic value is,

$t=\frac{\overline{d}-{\mu }_d}{\left(\frac{s_d}{\sqrt{n}}\right)}$

Substitute for 11.5 in $\overline{d}$, 10 in ${\mu }_d$, 0.7559 in $s_d$ and 8 in n in the formula mentioned above,

$\begin{array}{rll}t& =& \frac{11.5-10}{\left(\frac{0.7559}{\sqrt{8}}\right)}\\ & =& 5.612\end{array}$

Thus, the t-statistic value is 5.612.

To determine

(d)

To Explain:

The conclusion and interpret the decision.

Answer to Problem 9E

Solution:

The null hypothesis is rejected and hence publisher’s claim is right.

Explanation of Solution

Given:

A local school district is looking at adopting a new textbook that, according to the publishers, will increase standardized test scores of second graders by more than 10 points, on average. Never willing to believe a publisher’s claim without evidence to support it, the school board decides to test the claim. The school board chooses two second-grade classes for the study. One class was assigned the new textbook and the other class used the traditional textbook. Eight children from each class were then paired based on demographics and the ability levels. The following table lists the standardized test scores for the pairs.

New book	78	82	90	67	79	83	89	93
Old book	67	70	79	54	68	71	78	82

Calculation:

The null hypothesis, $H_0$ is rejected if:

One-Tail test

$t\le -t_{\alpha }$ for a left tailed test.

$t\ge t_{\alpha }$ for a right tailed test.

Two Tail test

$\left|t\right|\ge t_{\alpha /2}$ for two tailed test.

The degree of freedom is,

$\begin{array}{rll}df& =& n-1\\ & =& 8-1\\ & =& 7\end{array}$

The critical value is,

$\begin{array}{rll}{t}_{\alpha }& =& t_{0.05}\\ & =& 1.895\end{array}$

So by looking at the test statistic value and the t-distribution value, the test statistic value is greater than the t value, which by right tailed test implies that $H_0$ is rejected. Hence, there is sufficient evidence to reject $H_0$. And the hence the publisher claim that the test scores will increase by 10 points is right.