Chapter 10.1, Problem 9P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 9P

The level of significance is 0.05.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.05, and to test the occupations and personality preferences are independent or not.

Hence, the level of significance is 0.05.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The variables occupations and personality preferences are independent.

Alternative hypothesis:

$H_1:$ The variables occupations and personality preferences are not independent.

(b)

To determine

Find the value of the chi-square statistic for the sample.

Check whether the expected frequencies for all cells greater than 5 or not.

Identify the sampling distribution to be used.

Find the degrees of freedom.

Answer to Problem 9P

The value of the chi-square statistic for the sample is 8.649.

Yes, all the expected frequencies are greater than 5.

The chi-square distribution is used.

The degrees of freedom are 2.

Explanation of Solution

Calculation:

Test statistic:

The sample chi-square test statistic is,

${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

In the formula O is the observed frequency, E is the expected frequency, with degrees of freedom $d.f.=\left(R-1\right)\left(C-1\right)$, R is the number of rows in contingency table and C is the number of columns in contingency table,

$E=\frac{\left(\text{Row total}\right)\left(\text{Column total}\right)}{\text{Sample size}}$

The expected values are,

Occupation	Personality Preference Type		Row Total
	E	I
Clergy (all denominations)	$\frac{107\times 186}{406}=49.02$	$\frac{107\times 220}{406}=57.98$	107
M.D.	$\frac{162\times 186}{406}=74.22$	$\frac{162\times 220}{406}=87.78$	162
Lawyer	$\frac{137\times 186}{406}=62.76$	$\frac{137\times 220}{406}=74.24$	137
Column Total	186	220	406

It is clear that all the expected values are greater than 5, this shows that the chi-square distribution can be used. There are 3 rows and 2 columns.

The value of the chi-square statistic for the sample is,

$\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\left[\begin{array}{l}\frac{{\left(62-49.02\right)}^2}{49.02}+\frac{{\left(45-57.98\right)}^2}{57.98}+\frac{{\left(68-74.22\right)}^2}{74.22}+\\ \frac{{\left(94-87.78\right)}^2}{87.78}+\frac{{\left(56-62.76\right)}^2}{62.76}+\frac{{\left(81-74.24\right)}^2}{74.24}\end{array}\right]\\ =3.437+2.906+0.521+0.440+0.729+0.616\\ =8.649\end{array}$

Hence, the value of the chi-square statistic for the sample is 8.649.

Substitute 3 for R, and 2 for C in the degrees of freedom formula.

$\begin{array}{c}d.f.=\left(3-1\right)\left(2-1\right)\\ =2\times 1\\ =2\end{array}$

Hence, the degrees of freedom are 2.

(c)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 9P

The P-value is 0.0132.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Chi-Square’ distribution.
• In Degrees of freedom, enter the value as 2.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 8.649.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0132.

Hence, the P-value is 0.0132.

(d)

To determine

Check whether the null hypothesis of independence is rejected or fail to reject.

Answer to Problem 9P

The null hypothesis of independence is rejected.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0132.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0132 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, $0.0132\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the null hypothesis of independence is rejected.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that, there is sufficient evidence that the variables occupations and personality preferences are not independent at level of significance 0.05.