Chapter 28.2, Problem 9EA

(a)

To determine

Find the expected balls to be drawn.

Answer to Problem 9EA

The expected ball to be drawn is 12.

Explanation of Solution

Calculation:

A set of 10 Ping-Pong balls are considered and there are 120 draws from the bowl.

The expected ball to be drawn is,

$\begin{array}{c}\text{Expected}=\frac{120}{10}\\ =12\end{array}$

Hence, the expected ball to be drawn is 12.

(b)

To determine

Explain how the decisions follow from the data.

Explanation of Solution

Calculation:

Chi-square test statistic:

The test statistic for chi-square distribution is,

${\chi }^2=\text{Sum}\text{of}\left[\frac{{\left(\text{observed frequency}-\text{expected frequency}\right)}^2}{\text{expected frequency}}\right]$

With $\text{degrees of freedom}=\text{number of terms in}{\chi }^2-\text{one}$

For Ball set A:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	13	$\frac{120}{10}=12$
1	11	$\frac{120}{10}=12$
2	16	$\frac{120}{10}=12$
3	11	$\frac{120}{10}=12$
4	5	$\frac{120}{10}=12$
5	12	$\frac{120}{10}=12$
6	12	$\frac{120}{10}=12$
7	19	$\frac{120}{10}=12$
8	5	$\frac{120}{10}=12$
9	16	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(13-12\right)}^2}{12}+\frac{{\left(11-12\right)}^2}{12}+\frac{{\left(16-12\right)}^2}{12}+\cdots +\frac{{\left(16-12\right)}^2}{12}\\ =0.08+0.08+1.33+\cdots +1.33\\ =15.17\end{array}$

Hence, the value of ${\chi }^2$ for Ball set A is 15.17.

For Ball set B:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	22	$\frac{120}{10}=12$
1	8	$\frac{120}{10}=12$
2	7	$\frac{120}{10}=12$
3	8	$\frac{120}{10}=12$
4	19	$\frac{120}{10}=12$
5	20	$\frac{120}{10}=12$
6	10	$\frac{120}{10}=12$
7	11	$\frac{120}{10}=12$
8	6	$\frac{120}{10}=12$
9	9	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(22-12\right)}^2}{12}+\frac{{\left(8-12\right)}^2}{12}+\frac{{\left(7-12\right)}^2}{12}+\cdots +\frac{{\left(9-12\right)}^2}{12}\\ =8.33+1.33+2.08+\cdots +0.75\\ =26.67\end{array}$

Hence, the value of ${\chi }^2$ for Ball set B is 26.67.

For Ball set C:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	12	$\frac{120}{10}=12$
1	10	$\frac{120}{10}=12$
2	14	$\frac{120}{10}=12$
3	10	$\frac{120}{10}=12$
4	11	$\frac{120}{10}=12$
5	10	$\frac{120}{10}=12$
6	20	$\frac{120}{10}=12$
7	12	$\frac{120}{10}=12$
8	12	$\frac{120}{10}=12$
9	9	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(12-12\right)}^2}{12}+\frac{{\left(10-12\right)}^2}{12}+\frac{{\left(14-12\right)}^2}{12}+\cdots +\frac{{\left(9-12\right)}^2}{12}\\ =0.00+0.33+0.33+\cdots +0.75\\ =7.50\end{array}$

Hence, the value of ${\chi }^2$ for Ball set C is 7.50.

For Ball set D:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	16	$\frac{120}{10}=12$
1	7	$\frac{120}{10}=12$
2	12	$\frac{120}{10}=12$
3	14	$\frac{120}{10}=12$
4	15	$\frac{120}{10}=12$
5	5	$\frac{120}{10}=12$
6	10	$\frac{120}{10}=12$
7	21	$\frac{120}{10}=12$
8	11	$\frac{120}{10}=12$
9	9	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(16-12\right)}^2}{12}+\frac{{\left(7-12\right)}^2}{12}+\frac{{\left(12-12\right)}^2}{12}+\cdots +\frac{{\left(9-12\right)}^2}{12}\\ =1.33+2.08+0.00+\cdots +0.75\\ =16.50\end{array}$

Hence, the value of ${\chi }^2$ for Ball set D is 16.50.

There are 10 value. The degrees of freedom is,

$\begin{array}{c}\text{degrees of freedom}=\text{number of terms in}{\chi }^2-\text{one}\\ =10-1\\ =9\end{array}$

Hence, the degrees of freedom is 9.

From A CHI-SQUARE TABLE, the value corresponding to the 9 degrees of freedom for 10% level is 14.68, 5% level is 16.92, and the 1% level is 21.67.

It is clear that the ball set A and set D are in between the level 1% to 10% indicates that the sets are marginal and it would be retested.

The ball set B test value is greater than the critical values this indicates that the set B is rejected outright, and ball set C test value is less than the critical value which implies that the set C is accepted.

(c)

To determine

Test the Ball set A and D by using retest values.

Comment on the results.

Explanation of Solution

Calculation:

For Ball set C:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	19	$\frac{120}{10}=12$
1	9	$\frac{120}{10}=12$
2	10	$\frac{120}{10}=12$
3	12	$\frac{120}{10}=12$
4	7	$\frac{120}{10}=12$
5	15	$\frac{120}{10}=12$
6	19	$\frac{120}{10}=12$
7	10	$\frac{120}{10}=12$
8	12	$\frac{120}{10}=12$
9	7	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(19-12\right)}^2}{12}+\frac{{\left(9-12\right)}^2}{12}+\frac{{\left(10-12\right)}^2}{12}+\cdots +\frac{{\left(7-12\right)}^2}{12}\\ =4.08+0.75+0.33+\cdots +2.08\\ =14.50\end{array}$

Hence, the value of ${\chi }^2$ for Ball set C is 14.50.

For Ball set D:

The expected values are,

Ball no.	Observed Frequency	Expected frequency
0	8	$\frac{120}{10}=12$
1	15	$\frac{120}{10}=12$
2	22	$\frac{120}{10}=12$
3	11	$\frac{120}{10}=12$
4	15	$\frac{120}{10}=12$
5	8	$\frac{120}{10}=12$
6	17	$\frac{120}{10}=12$
7	9	$\frac{120}{10}=12$
8	8	$\frac{120}{10}=12$
9	7	$\frac{120}{10}=12$
Total	120

The test statistic is,

$\begin{array}{c}{\chi }^2=\frac{{\left(8-12\right)}^2}{12}+\frac{{\left(15-12\right)}^2}{12}+\frac{{\left(22-12\right)}^2}{12}+\cdots +\frac{{\left(7-12\right)}^2}{12}\\ =1.33+0.75+8.33+\cdots +2.08\\ =18.83\end{array}$

Hence, the value of ${\chi }^2$ for Ball set D is 18.83.

The ball set D test value is greater than the critical values of 5% level this indicates that the set D is rejected, and ball set A test value is less than the critical value which implies that the set A is accepted.