32. How are schools doing? Refer to Exercises 28 and 30 .

(a) Check that the conditions for performing the chi-square test are met.

(b) Use Table $\mathrm{C}$to find the $P$-value. Then use your calculator's ${\chi }^2\mathrm{cdf}$command.

(c) Interpret the $P$-value from the calculator in context.

(d) What conclusion would you draw? Justify your answer.

Given in the question that, We need to check whether the conditions for performing the chi-square test are met in exercises 28 and 30 .

Given:

For both questions, use the provided observation table.

The Expected table is,

The chi-suare test must meet the following requirements:

• There are two category variables.
• There are two or more categories for each variable.
• Independent observation
• The anticipated frequency is at least 1 or at least 5 for the vast majority of cells (80%).

Because the information provided meets all of the criteria. As a result, the chi-square test can be used on the data set in question.

Given in the question that, We need to use Table C to find the $P$-value. Then use our calculator's ${\chi }^2\mathrm{cdf}$command.

The test statistics for exercise 30 are $22.4260$, respectively.

$df=(n-1)(m-1)$

$p-$value $=P\left({\chi }^2>{\chi }_{\text{test}}^2\right)$

For both test values, the degree of freedom can be determined as follows:

$df=(5-1)(3-1)$

$=8$

To find $p$-value on TI83, use the following command: $"{\chi }^{2}cdf(22.4260,99,8)"$

The result is,

Given in the question that, We need to interpret the $P$-value from the calculator in context.

The null and alternative hypothesis are numbered 30

$H_0:\left(\begin{array}{l}\text{There is no difference between Black,}\\ \text{Hispanic, and White parents}\end{array}\right)$

width="313" height="48" role="math">H1:There is a difference between Black,Hispanic, and White parents

The p-value shows the possibility of generating results as extreme as the test statistic value if the null hypothesis is true.

The probability of obtaining a sample as extreme as or more extreme than the current sample is $0.0042$if there is no link between black, Hispanic, and white parents.

Given in the question that, we need to justify our conclusion

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

The p-value is smaller than the level of significance in this case. As a result, the null hypothesis is ruled out.

As a result, there is enough data to establish that black, Hispanic, and white parents have different parenting styles.