Chapter 14.3, Problem 36E

Consider the accompanying 2 × 3 table displaying the sample proportions that fell in the various combinations of categories (e.g., 13% of those in the sample were in the first category of both factors).

	1	2	3
1	.13	.19	.28
2	.07	.11	.22

a. Suppose the sample consisted of n = 100 people. Use the chi-squared test for independence with significance level .10.

b. Repeat part (a), assuming that the sample size was n = 1000.

c. What is the smallest sample size n for which these observed proportions would result in rejection of the independence hypothesis?

To determine

Use chi-square test of independence for sample size of 100 at 10% level of significance to test the independence of the factors.

Answer to Problem 36E

There is no sufficient evidence to conclude that two factors are dependent at 10% level of significance.

Explanation of Solution

Given info:

A $2\times 3$ contingency table was given with probabilities in each cell.

Calculation:

The number of observed frequency for each cell is calculated as follows:

	1	2	3
1	$\left(0.13\right)\left(100\right)=13$	$\left(0.19\right)\left(100\right)=19$	$\left(0.28\right)\left(100\right)=28$
2	$\left(0.07\right)\left(100\right)=7$	$\left(0.11\right)\left(100\right)=11$	$\left(0.22\right)\left(100\right)=22$

The claim is to test whether the two factors are dependent. If the claim is rejected, then the two factors are independent of each other.

Testing the hypothesis:

Null hypothesis:

$H_0:p_{1j}=p_{2j}$

That is, the two factors are independent.

Alternative hypothesis:

$H_a:$ The two factors are dependent.

Test statistic:

Software procedure:

Step-by-step procedure to find the chi-square test statistic using MINITAB is given below:

• Choose Stat > Tables > Chi-Square Test (Two-Way Table in Worksheet).
• In Columns containing the table, enter the columns of 1, 2 and 3.
• Click OK.

Output obtained from MINITAB is given below:

Decision rule:

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The P-value is 0.712 and the level of significance is 0.10.

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

That is, $0.712\left(=P\text{-value}\right)>0.10\left(=\alpha \right)$

Hence, the null hypothesis is not rejected.

Thus, there is no sufficient evidence to conclude that two factors are dependent at 10% level of significance.

To determine

Use chi-square test of independence for sample size of 1,000 at 10% level of significance.

Answer to Problem 36E

There is sufficient evidence to conclude that two factors are dependent.

Explanation of Solution

Given info:

A $2\times 3$ contingency table was given with probabilities for each cell.

Calculation:

The claim is to test whether the two factors are dependent. If the claim is rejected, then the two factors are independent of each other.

Testing the hypothesis:

Null hypothesis:

$H_0:p_{1j}=p_{2j}$

That is, the two factors are independent.

Alternative hypothesis:

$H_a:$ The two factors are dependent.

Software procedure:

Step-by-step procedure to find the chi-square test statistic using MINITAB is given below:

• Choose Stat > Tables > Chi-Square Test (Two-Way Table in Worksheet).
• In Columns containing the table, enter the columns of 1, 2 and 3
• Click OK.

Output obtained from MINITAB is given below:

Decision rule:

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The P-value is 0.033 and the level of significance is 0.10.

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

That is, $0.033\left(=P\text{-value}\right)<0.10\left(=\alpha \right)$

Hence, the null hypothesis is rejected.

Thus, there is sufficient evidence to conclude that two factors are dependent.

To determine

Identify the smallest sample size n where the rejection of null hypothesis would happen.

Answer to Problem 36E

The smallest sample size n where the rejection of null hypothesis would happen is 677.

Explanation of Solution

Calculation:

Multiply the sample size n across all cells to get the observed frequency:

	1	2	3
1	$0.13n$	$0.19n$	$0.28n$	0.60n
2	$0.07n$	$0.11n$	$0.22n$	0.40n
Total	0.20n	0.30n	0.50n	1.00n

Expected frequency:

The expected frequency for a cell is calculated by,

${\widehat{e}}_{ij}=\frac{\left(i\text{th row total}\right)\left(j\text{th column total}\right)}{n}$

Where,

n represents the sample size.

$\begin{array}{c}{\widehat{e}}_{11}=\frac{\left(0.60n\right)\left(0.20n\right)}{1.00n}\\ =\frac{0.12n^2}{1.00n}\\ =0.12n\end{array}$

$\begin{array}{c}{\widehat{e}}_{12}=\frac{\left(0.60n\right)\left(0.30n\right)}{1.00n}\\ =\frac{0.18n^2}{1.00n}\\ =0.18n\end{array}$

$\begin{array}{c}{\widehat{e}}_{13}=\frac{\left(0.60n\right)\left(0.50n\right)}{1.00n}\\ =\frac{0.3n^2}{1.00n}\\ =0.3n\end{array}$

$\begin{array}{c}{\widehat{e}}_{21}=\frac{\left(0.40n\right)\left(0.20n\right)}{1.00n}\\ =\frac{0.08n^2}{1.00n}\\ =0.08n\end{array}$

$\begin{array}{c}{\widehat{e}}_{22}=\frac{\left(0.40n\right)\left(0.30n\right)}{1.00n}\\ =\frac{0.12n^2}{1.00n}\\ =0.12n\end{array}$

$\begin{array}{c}{\widehat{e}}_{23}=\frac{\left(0.40n\right)\left(0.50n\right)}{1.00n}\\ =\frac{0.20n^2}{1.00n}\\ =0.20n\end{array}$

Test statistic:

${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$

Where,

$n_i$ represents the observed frequency.

$n{p}_i$ represents the expected frequency.

The table shows the calculation for chi-square test statistic:

Factors	Observed $n_i$	Expected $n{p}_i$	${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$
1/1	0.13n	0.12n	0.000833n
1/2	0.19n	0.18n	0.000556n
1/3	0.28n	0.30n	0.001333n
2/1	0.07n	0.08n	0.00125n
2/2	0.11n	0.12n	0.000833n
2/3	0.22n	0.20n	0.002n
Total			0.006805556n

Critical value:

Use the Table A.7 to find the chi-square critical values.

• Locate 2 under the column of v.
• In the row of $\alpha$ locate 0.10.
• The value intersecting these two numbers gives the critical value corresponding to ${\chi }_{0.10,2}^2$

Thus, the critical value is ${\chi }_{0.10,7}^2$ is 4.605.

Decision rule:

The null hypothesis would be rejected if the P-value is lesser than or equal to the level of significance $\alpha$ and this would occur if the test statistic value of chi-square is greater than or equal to the critical value.

Conclusion:

Here, $H_0$ is rejected if and only if $0.006805556n\ge 4.607$

That is,

$\begin{array}{l}n\ge \frac{4.607}{0.006805556}\\ n\ge \text{676}\text{.95}\end{array}$

Hence, the sample size is 677.