Chapter 11.2, Problem 30E

a.

To determine

To explain: The null and alternative hypotheses for a significance test.

Explanation of Solution

Given:

The observed value ( O_i ) is:

	Black Parents ( B )	Hispanic Parents ( H )	White Parents ( W )
Excellent	12	34	22
Good	69	55	81
Fair	75	61	60
Poor	24	24	24
Don’t know	22	28	14
Total	202	202	201

To test whether there is a difference in the distributions of opinions about how high schools are doing among black, Hispanic, and white parents.

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:\left(\begin{array}{l}\text{There is no difference between Black,}\\ \text{Hispanic, and White parents}\end{array}\right)\\ H_1:\left(\begin{array}{l}\text{There is a difference between Black,}\\ \text{Hispanic, and White parents}\end{array}\right)\end{array}$

b.

To determine

To find: The expected counts for the given data set.

Explanation of Solution

Formula used:

The expected value ( E_i ), can be calculated as:

  $E_i{}_j=\frac{\text{Row sum}\left(i\right)\times \text{Column Sum}\left(j\right)}{\text{Total sum}}$

Calculation:

From the given table,

	Black Parents ( B )	Hispanic Parents ( H )	White Parents ( W )	Total
Excellent	12	34	22	68
Good	69	55	81	205
Fair	75	61	60	196
Poor	24	24	24	72
Don’t know	22	28	14	64
Total	202	202	201	605

The expected value ( E_i ), can be calculated as:

	Black Parents ( B )	Hispanic Parents ( H )	White Parents ( W )
Excellent	$\frac{202\times 68}{605}=22.7041$	$\frac{202\times 68}{605}=22.7041$	$\frac{201\times 68}{605}=22.5917$
Good	$\frac{202\times 205}{605}=68.4463$	$\frac{202\times 205}{605}=68.4463$	$\frac{201\times 205}{605}=68.1074$
Fair	$\frac{202\times 196}{605}=65.4413$	$\frac{202\times 196}{605}=65.4413$	$\frac{201\times 60}{605}=65.1174$
Poor	$\frac{202\times 72}{605}=24.0397$	$\frac{202\times 72}{605}=24.0397$	$\frac{201\times 24}{605}=23.9207$
Don’t know	$\frac{202\times 64}{605}=21.3686$	$\frac{202\times 64}{605}=21.3686$	$\frac{201\times 64}{605}=21.2628$

c.

To determine

To find: The chi-square statistic.

Answer to Problem 30E

The test statistic value, is 22.4260.

Explanation of Solution

Formula used:

The chi-square test statistics value:

  ${\chi }^2{}_{statistic}={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Calculation:

The chi-square statistic, can be calculated as:

  $\begin{array}{c}{\chi }^2{}_{statistic}=\left(\begin{array}{l}\frac{{\left(12-22.7041\right)}^2}{22.7041}+\frac{{\left(34-22.7041\right)}^2}{22.7041}+\frac{{\left(22-22.5917\right)}^2}{22.5917}+\frac{{\left(69-68.4463\right)}^2}{68.4463}\\ +\frac{{\left(55-68.4463\right)}^2}{68.4463}+\frac{{\left(81-68.1074\right)}^2}{68.1074}+\frac{{\left(75-65.4413\right)}^2}{65.4413}+\frac{{\left(61-65.4413\right)}^2}{65.4413}\\ +\frac{{\left(60-65.1174\right)}^2}{65.1174}+\frac{{\left(24-24.0397\right)}^2}{24.0397}+\frac{{\left(24-24.0397\right)}^2}{24.0397}+\frac{{\left(24-23.9207\right)}^2}{23.9207}\\ +\frac{{\left(22-21.3686\right)}^2}{21.3686}+\frac{{\left(28-21.3686\right)}^2}{21.3686}+\frac{{\left(14-21.2628\right)}^2}{21.2628}\end{array}\right)\\ =22.4260\end{array}$