Chapter 11.2, Problem 32E

a.

To determine

To explain: The conditions for performing the chi-square test are met.

Explanation of Solution

Given:

The given observe table for both questions,

	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

The expected table is,

	Black Parents ( B )	Hispanic Parents ( H )	White Parents ( W )
Excellent	22.7041	22.7041	22.5917
Good	68.4463	68.4463	68.1074
Fair	65.4413	65.4413	65.1174
Poor	24.0397	24.0397	23.9207
Don’t know	21.3686	21.3686	21.2628

The requirments to use chi-suare test is,

• Two categorical variables.
• Two or more categories for each variable
• Independence of observations
• Expected frequency for each cell are at least 1 or it should be at least 5 for the majority (80%) of the cells.

Since, each condition is satisfied by the given information. Therefore, chi-square test it can be used for the give dataset.

b.

To determine

To compute: The p-value of chi-square test.

Answer to Problem 32E

The p-values is 0.0042.

Explanation of Solution

Given:

From exercise 30, the test statistics is 22.4260, respectively.

Formula used:

  $\begin{array}{l}df=\left(n-1\right)\left(m-1\right)\\ p-value=P\left({\chi }^2>{\chi }^2{}_{test}\right)\end{array}$

Calculation:

The degree of freedom for both test values, it can be calculated as

For 27 exercise,

  $\begin{array}{c}df=\left(5-1\right)\left(3-1\right)\\ =8\end{array}$

There is the following commend to find p-value on TI83,

“ ${\chi }^{2}cdf\left(22.4260,99,8\right)$”

The output is,

  

Thus, the p-value is 0.0042.

c.

To determine

To explain: Give an interpretation for the p-value.

Explanation of Solution

From 30, the null and alternative hypothesis,

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

If the null hypothesis is true, then p-value is probability of obtaining results as extreme for test statistic value.

If there is no relation between black, Hispanic, and white parents, then the probability of getting a sample as extreme or more extreme as the given sample is 0.0042.

d.

To determine

To conclude: find the conclusion.

Explanation of Solution

The null hypothesis is rejected when the p-value is less than level of significance.

Here, the p-value is less than level of significance. Thus, the null hypothesis is rejected.

Therefore, there is a sufficient evidence to conclude that there is a difference between black, Hispanic, and white parents.