Chapter 10.2, Problem 14E

a.

To determine

Find the expected frequencies.

Answer to Problem 14E

The expected frequencies under the null hypothesis is obtained as,

	Number of Children
Siblings	0	1	3	3	4	More Than 4
0	39.626	24.501	37.761	23.776	11.396	6.941
1	152.174	94.090	145.013	91.305	43.763	26.655
2	153	94.6	145.8	95.8	44	26.8
3	129.610	80.138	123.510	77.766	37.273	22.703
4	93.836	58.020	89.421	56.302	26.986	16.437
More Than 4	196.754	121.653	187.495	118.052	56.583	34.464

Explanation of Solution

Calculation:

The contingency table with 6 rows consisting of six classification of number of siblings and6 columns consisting of six classification of number of children in families is given.

The sample of 2,780 is randomly selected.

Contingency table:

A contingency table is obtained as using two qualitative variables. One of the qualitative variable is row variable that has one category for each row of the table another is column variable has one category for each column of the table.

The hypotheses are:

Null Hypothesis:

$H_0:\text{The number of Siblings and the number of Children are independent}\text{.}$

Alternate Hypothesis:

$H_1:\text{The number of Siblings and the number of Children are not independent}.$

Now, it is obtained that,

	Number of Children
Siblings	0	1	3	3	4	More Than 4	Row Total
0	60	19	34	15	10	6	144
1	189	97	163	63	29	12	553
2	179	98	137	81	41	20	556
3	124	85	132	83	31	16	471
4	85	53	88	61	32	22	341
More Than 4	128	121	175	156	77	58	715
Column Total	765	473	729	459	220	134	2,780

Expected frequencies:

The expected frequencies in case of contingency table is obtained as,

$E=\frac{\text{Row total}\cdot \text{Coloumn total}}{\text{Grand total}}$

Now, using the formula of expected frequency it is found that the expected frequency for the number of children of 1 and the number of siblings of 1 is obtained as,

$\begin{array}{c}\frac{\left(144\right)\left(765\right)}{2,780}=\frac{110,160}{2,780}\\ =39.625\end{array}$

Hence, in similar way the expected frequencies are obtained as,

	Number of Children
Siblings	0	1	3	3	4	More Than 4
0	$\begin{array}{l}\frac{\left(144\right)\left(765\right)}{2,780}\\ =39.626\end{array}$	$\begin{array}{l}\frac{\left(144\right)\left(473\right)}{2,780}\\ =24.501\end{array}$	$\begin{array}{l}\frac{\left(144\right)\left(729\right)}{2,780}\\ =37.761\end{array}$	$\begin{array}{l}\frac{\left(144\right)\left(459\right)}{2,780}\\ =23.776\end{array}$	$\begin{array}{l}\frac{\left(144\right)\left(220\right)}{2,780}\\ =11.396\end{array}$	$\begin{array}{l}\frac{\left(144\right)\left(134\right)}{2,780}\\ =6.941\end{array}$
1	$\begin{array}{l}\frac{\left(553\right)\left(765\right)}{2,780}\\ =152.174\end{array}$	$\begin{array}{l}\frac{\left(553\right)\left(473\right)}{2,780}\\ =94.090\end{array}$	$\begin{array}{l}\frac{\left(553\right)\left(729\right)}{2,780}\\ =145.013\end{array}$	$\begin{array}{l}\frac{\left(553\right)\left(459\right)}{2,780}\\ =91.305\end{array}$	$\begin{array}{l}\frac{\left(553\right)\left(220\right)}{2,780}\\ =43.763\end{array}$	$\begin{array}{l}\frac{\left(553\right)\left(134\right)}{2,780}\\ =26.655\end{array}$
2	$\begin{array}{l}\frac{\left(556\right)\left(765\right)}{2,780}\\ =153\end{array}$	$\begin{array}{l}\frac{\left(556\right)\left(473\right)}{2,780}\\ =94.6\end{array}$	$\begin{array}{l}\frac{\left(556\right)\left(729\right)}{2,780}\\ =145.8\end{array}$	$\begin{array}{l}\frac{\left(556\right)\left(459\right)}{2,780}\\ =95.8\end{array}$	$\begin{array}{l}\frac{\left(556\right)\left(220\right)}{2,780}\\ =44\end{array}$	$\begin{array}{l}\frac{\left(556\right)\left(134\right)}{2,780}\\ =26.8\end{array}$
3	$\begin{array}{l}\frac{\left(471\right)\left(765\right)}{2,780}\\ =129.610\end{array}$	$\begin{array}{l}\frac{\left(471\right)\left(473\right)}{2,780}\\ =80.138\end{array}$	$\begin{array}{l}\frac{\left(471\right)\left(729\right)}{2,780}\\ =123.510\end{array}$	$\begin{array}{l}\frac{\left(471\right)\left(459\right)}{2,780}\\ =77.766\end{array}$	$\begin{array}{l}\frac{\left(471\right)\left(220\right)}{2,780}\\ =37.273\end{array}$	$\begin{array}{l}\frac{\left(471\right)\left(134\right)}{2,780}\\ =22.703\end{array}$
4	$\begin{array}{l}\frac{\left(341\right)\left(765\right)}{2,780}\\ =93.836\end{array}$	$\begin{array}{l}\frac{\left(341\right)\left(473\right)}{2,780}\\ =58.020\end{array}$	$\begin{array}{l}\frac{\left(341\right)\left(729\right)}{2,780}\\ =89.421\end{array}$	$\begin{array}{l}\frac{\left(341\right)\left(459\right)}{2,780}\\ =56.302\end{array}$	$\begin{array}{l}\frac{\left(341\right)\left(220\right)}{2,780}\\ =26.986\end{array}$	$\begin{array}{l}\frac{\left(341\right)\left(134\right)}{2,780}\\ =16.437\end{array}$
More Than 4	$\begin{array}{l}\frac{\left(715\right)\left(765\right)}{2,780}\\ =196.754\end{array}$	$\begin{array}{l}\frac{\left(715\right)\left(473\right)}{2,780}\\ =121.653\end{array}$	$\begin{array}{l}\frac{\left(715\right)\left(729\right)}{2,780}\\ =187.495\end{array}$	$\begin{array}{l}\frac{\left(715\right)\left(459\right)}{2,780}\\ =118.052\end{array}$	$\begin{array}{l}\frac{\left(715\right)\left(220\right)}{2,780}\\ =56.583\end{array}$	$\begin{array}{l}\frac{\left(715\right)\left(134\right)}{2,780}\\ =34.464\end{array}$

b.

To determine

Find the value of chi-square statistic.

Answer to Problem 14E

The value of chi-square statistic is 126.444.

Explanation of Solution

Chi-Square statistic:

The chi-square statistic is obtained as ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where $O_1,O_2,\mathrm{...},O_k$ be the observed frequencies and $E_1,E_2,\mathrm{...},E_k$ be the expected frequencies for k number of categories.

The classifications of number of children can be written as,

0	1
1	2
2	3
3	4
4	5
More Than 4	6

Test Statistic:

Software procedure:

Step -by-step software procedure to obtain test statistic using MINITAB software is as follows:

• • Select Stat>Table>Cross Tabulation and Chi-Square.
• • Check the box of Raw data (categorical variables).
• • Under For rows enter Siblings.
• • Under For columns enter Number of Children.
• • Check the box of Count under Display.
• • Under Chi-Square, click the box of Chi-Square test.
• • Select OK.
• Output using MINITAB software is given below:

Thus, the value of chi-square statistic is126.444.

c.

To determine

Find the degrees of freedom.

Answer to Problem 14E

The degrees of freedom is25.

Explanation of Solution

It is known that under the null hypothesis $H_0$ the test statistic ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ follows chi-square distribution with $\left(r-1\right)\left(c-1\right)$ degrees of freedom for r number of rows and c number of columns, provided that all the expected frequencies are greater than or equal to 5.

In the given question there are 6rows and 6 columns.

Hence, the degrees of freedom is

$\begin{array}{c}\left(r-1\right)\left(c-1\right)=\left(6-1\right)\left(6-1\right)\\ =\left(5\right)\left(5\right)\\ =25\end{array}$

Thus, the degrees of freedom is25.

d.

To determine

Perform a test of hypothesis of independence at the level of significance of $\alpha =0.05$.

Draw conclusions.

Answer to Problem 14E

There is no enough evidence to conclude that the number of sibling and the number of children are independent.

Explanation of Solution

Calculation:

It is known that when the null hypothesis $H_0$ is true the chi-square statistic for a test of independence, follows approximately chi-square distribution provided that all the expected frequencies are 5 or more.

In part (a) it is found that all the expected frequencies corresponding to all rows and columns of the given contingency table are more than 5.

Hence, the test of independence is appropriate.

The hypotheses are:

Null Hypothesis:

$H_0:\text{The number of siblings and the number of children are independent}\text{.}$

Alternate Hypothesis:

$H_1:\text{The number of siblings and the number of children are not independent}.$

From parts (b) and (c), the value of test statistic is 126.444with the degrees of freedom 25.

Level of significance:

The level of significance is given as 0.05.

Critical value:

In a test of hypotheses the critical value is the point by which one can reject or accept the null hypothesis.

Software procedure:

Step-by-step software procedure to obtain critical value using MINITAB software is as follows:

• • Select Graph>Probability distribution plot > view probability
• • Select Chi -Square under distribution.
• • In Degrees of freedom, enter 25.
• • Choose Probability Value and Right Tail for the region of the curve to shade.
• • Enter the Probability value as 0.05 under shaded area.
• • Select OK.
• Output using MINITAB software is given below:

Hence, the critical value at $\alpha =0.05$ is 37.65.

  Rejection rule:

If the ${\chi }^2$ value is greater than or equal to critical value, that is ${\chi }^2\ge {\chi }_{\alpha ;k-1}^2$, then reject the null hypothesis $H_0$. Otherwise, do not reject $H_0$.

Conclusion:

Here, the ${\chi }^2$ value is greater than the critical value.

That is, ${\chi }^2\left(=126.444\right)>{\chi }_{0.05,25}^2\left(=37.65\right)$

Thus, the decision is “reject the null hypothesis”.

Thus, there is not enough evidence to conclude that the number of sibling and the number of children are independent.