Q.32 - The Practice of Statistics for AP

Question

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.

Step-by-Step Solution

Answer

(a) The chi-square test can be applied to the given dataset.

(b) The p-values is $0.0042$.

(c) 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.

(d) There is enough data to establish that black, Hispanic, and white parents have different parenting styles. 

Part (a)Step 1: Given information

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 .

Part(a) Step 2: Explanation

Given:

For both questions, use the provided observation table.

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	$24.0397$	$24.0397$	$23.9207$
Total	$21.3686$	$21.3686$	$21.2628$

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.

Part(b) Step 1: Given information

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.

Part (b) Step 2: Explanation

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

$d f=(n-1)(m-1)$

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

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

$d f=(5-1)(3-1)$

$=8$

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

The result is,

Part(c) Step 1: Given information

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

Part(c) Step 2: Explanation

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" style="max-width: none;" $H_1:\left(\begin{array}{l}\text{ There is a difference between Black, }\\ \text{ Hispanic, and White parents }\end{array}\right)$

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.

Part(d) Step 1: Given information

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

Part (d) Step 2: Explanation

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.