Chapter 12, Problem 9CQ

Exercises refer to the following data:

Electric motors are assembled on four different production lines. Random samples of motors are taken from each line and inspected. The numbers that pass and that fail the inspection are counted for each line, with the following results:

State a conclusion.

To determine

To state : the conclusion

Answer to Problem 9CQ

The value of chi-square statistics ${\chi }^2=4.676$ is less thanthe critical value from table 7.815.

So, do not reject $H_0$ . There is not enough evidence to conclude that the failure probabilities are not all the same.

Explanation of Solution

Given information :

	Line
	1	2	3	4
Pass	482	467	458	404
Fail	57	59	37	47

Concept Involved:

In order to decide whether the presumed hypothesis for data sample stands accurate for the entire population or not we use the hypothesis testing.

  $H_0$ represents null hypothesis test and $H_a$ represents alternative hypothesis test.

  ${\chi }^2$ represent value of chi square test statistics found using the formula ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where O is observed frequency and E is expected frequency.

The value of test statistics and the critical value identified from the table help us to decide whether to reject or do not reject null hypothesis.

The critical value from Table A.4, using degrees of freedom of contingency table of any given study is provided.

If ${\chi }^2\ge CriticalValue$ , then reject $H_0$ otherwise reject $H_0$ .

The values of two qualitative variables are connected and denoted in a contingency table. This table consists of rows and column. The variables in each row and each column of the table represent a category. The number of rows of contingency table is represented by letter ‘r’ and number of column of contingency table is represented by letter ‘c’.

The formula to find the number of degree of freedom of contingency table is $\left(r-1\right)\left(c-1\right)$ .

Calculation:

Row totals reported	Column totals reported
Electric motors passed   $=482+467\text{+}458+404=1811$ Electric motors failed   $=57+59+37+47=200$	Electric motors in Line 1   $=482+57=539$ Electric motors in Line 2   $=467+59=526$ Electric motors in Line 3   $=458+37=495$ Electric motors in Line 4   $=404+47=451$
Grand total = 482 + 467 + 458 + 404 + 57 + 59 + 37 + 47 = 2011

From the results we have the below table:

	Line
	1	2	3	4	Row Total
Pass	482	467	458	404	1811
Fail	57	59	37	47	200
Column Total	539	526	495	451	2011

Finding the expected frequency for the cell corresponding to:	The expected frequency
Electric motor that pass and in production Line 1 The row total is 1811, the column total is 539, and the grand total is 2011.	$\frac{1811\cdot 539}{2011}\approx 485.395$
Electric motor that pass and in production Line 2 The row total is 1811, the column total is 526, and the grand total is 2011.	$\frac{1811\cdot 526}{2011}\approx 473.688$
Electric motor that pass and in production Line 3 The row total is 1811, the column total is 495, and the grand total is 2011.	$\frac{1811\cdot 495}{2011}\approx 445.771$
Electric motor that pass and in production Line 4 The row total is 1811, the column total is 451, and the grand total is 2011.	$\frac{1811\cdot 451}{2011}\approx 406.147$
Electric motor that fail and in production Line 1 The row total is 200, the column total is 539, and the grand total is 2011	$\frac{200\cdot 539}{2011}\approx 53.605$
Electric motor that fail and in production Line 2 The row total is 200, the column total is 526, and the grand total is 2011.	$\frac{200\cdot 526}{2011}\approx 52.312$
Electric motor that fail and in production Line 3 The row total is 200, the column total is 495, and the grand total is 2011.	$\frac{200\cdot 495}{2011}\approx 49.229$
Electric motor that fail and in production Line 4 The row total is 200, the column total is 451, and the grand total is 2011.	$\frac{200\cdot 451}{2011}\approx 44.853$

All the expected frequencies are at least 5. From the results of previous part we have the below table:

Observed Frequency	Expected frequency
Line1234Pass482 467 458 404 Fail57 59 37 47	Line1234Pass485.395 473.688 445.771 406.147 Fail53.605 52.312 49.229 44.853

Finding the value of the chi-square corresponding to:	$\frac{{\left(O-E\right)}^2}{E}$
Electric motor that pass and in production Line 1 Observed frequency is482 and expected frequency is 485.395	$\frac{{\left(482-485.395\right)}^2}{485.395}\approx 0.024$
Electric motor that pass and in production Line 2 Observed frequency is467 and expected frequency is 473.688	$\frac{{\left(467-473.688\right)}^2}{473.688}\approx 0.094$
Electric motor that pass and in production Line 3 Observed frequency is458 and expected frequency is 445.771	$\frac{{\left(458-445.771\right)}^2}{445.771}\approx 0.335$
Electric motor that pass and in production Line 4 Observed frequency is404 and expected frequency is 406.147	$\frac{{\left(404-406.147\right)}^2}{406.147}\approx 0.011$
Electric motor that fail and in production Line 1 Observed frequency is57 and expected frequency is 53.605	$\frac{{\left(57-53.605\right)}^2}{53.605}\approx 0.215$
Electric motor that fail and in production Line 2 Observed frequency is59 and expected frequency is 52.312	$\frac{{\left(59-52.312\right)}^2}{52.312}\approx 0.855$
Electric motor that fail and in production Line 3 Observed frequency is37 and expected frequency is 49.229	$\frac{{\left(37-49.229\right)}^2}{49.229}\approx 3.038$
Electric motor that fail and in production Line 4 Observed frequency is47 and expected frequency is 44.853	$\frac{{\left(47-44.853\right)}^2}{44.853}\approx 0.103$

To compute the test statistics, we use the observed frequencies and expected frequency:

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\}=0.024+0.094+0.335+0.011+0.215+0.855+3.038+0.103\\ {\chi }^2\approx 4.676\end{array}$

Here r represents the number of rows and c represents the number of columns.

Given $r=2$ , $c=4$ so the number of degree of freedom is $\left(2-1\right)\left(4-1\right)=\left(1\right)\left(3\right)=3$

Degrees of freedom	Table A.4 Critical Values for the chi-square Distribution
	0.995	0.99	0.975	0.95	0.90	0.10	0.05	0.025	0.01	0.005
1	0.000	0.000	0.001	0.004	0.016	2.706	3.841	5.024	6.635	7.879
2	0.010	0.020	0.051	0.103	0.211	4.605	5.991	7.378	9.210	10.597
3	0.072	0.115	0.216	0.352	0.584	6.251	7.815	9.348	11.345	12.838
4	0.207	0.297	0.484	0.711	1.064	7.779	9.488	11.143	13.277	14.860
5	0.412	0.554	0.831	1.145	1.610	9.236	11.070	12.833	15.086	16.750