Chapter 12, Problem 27SE

a.

To determine

Perform a hypothesis test to determine if the population proportion of good parts is the same for all three shifts at $\alpha =0.05$ level significance.

Find the p-value and draw conclusion.

Answer to Problem 27SE

All population proportions for three shifts are not equal.

The p-value is 0.0174.

Explanation of Solution

Calculation:

The given data related to the quality (good and defective) of goods for first, second and third shifts.

Let $p_1$ be the sample proportion of first shift, $p_2$ be the sample proportion of second shift and $p_3$ be the sample proportion of third shift.

State the test hypotheses:

Null hypothesis:

$H_0:p_1=p_2=p_3$

That is, all population proportions for three shifts are equal.

Alternative hypothesis:

$H_a:\text{Not all population proportions are equal}$.

That is, not all population proportions for three shifts are equal.

The row and column total is tabulated below:

Quality	First Shift	Second Shift	Third Shift	Total
Good	285	368	176	829
Defective	15	32	24	71
Total	300	400	200	900

The formula for expected frequency is given below:

$e_{ij}=\frac{\left(\text{Row }i\text{ Total}\right)\left(\text{Coloumn }j\text{ Total}\right)}{\text{Total Sample Size}}$.

The expected frequency for each category is calculated as follows:

Quality	First Shift	Second Shift	Third Shift
Good	$\frac{\left(300\right)\left(829\right)}{900}=276.33$	$\frac{\left(829\right)\left(400\right)}{\text{900}}=368.44$	$\frac{\left(829\right)\left(200\right)}{\text{900}}=184.22$
Defective	$\frac{\left(300\right)\left(71\right)}{\text{900}}=23.67$	$\frac{\left(400\right)\left(71\right)}{\text{900}}=31.56$	$\frac{\left(200\right)\left(71\right)}{\text{900}}=15.78$

The formula for chi-square test statistic is given as,

${\chi }^2={\displaystyle \sum _i{\displaystyle \sum _j\frac{{\left(f_{ij}-e_{ij}\right)}^2}{e_{ij}}}}$.

Therefore, the value of chi-square test statistic is,

$\begin{array}{c}{\chi }^2=\left\{\begin{array}{l}\frac{{\left(285-276.33\right)}^2}{276.33}+\frac{{\left(368-368.44\right)}^2}{368.44}+\frac{{\left(176-184.22\right)}^2}{184.22}+\frac{{\left(15-23.67\right)}^2}{23.67}\\ +\frac{{\left(32-31.56\right)}^2}{31.56}+\frac{{\left(24-15.78\right)}^2}{15.78}\end{array}\right\}\\ =\frac{75.1689}{276.33}+\frac{0.1936}{368.44}+\frac{67.5684}{184.22}+\frac{75.1689}{23.67}+\frac{0.1936}{31.56}+\frac{67.5684}{15.78}\\ =0.272+0.000525+0.36678+3.175+0.006134+4.282\\ \approx 8.10\end{array}$

Thus, the chi-square test statistic is 8.10.

Degrees of freedom:

$\begin{array}{c}df=k-1\\ =3-1\\ =2\end{array}$

Thus, the degree of freedom is 2.

Level of significance:

The given level of significance is $\alpha =0.05$.

p-value:

Software procedure:

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

• Select Graph > Probability distribution plot > view probability
• Select chi-square under distribution and enter 2 in degrees of freedom.
• Enter the X-value as 8.10 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

From the MINITAB output, the p-value is 0.0174.

Rejection rule:

If the $p\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the p-value less than the level of significance.

That is, $p\text{-value}\left(=0.0174\right)\le \alpha \left(=0.05\right)$.

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

Therefore, the data provide sufficient evidence to conclude that not all population proportions for three shifts are equal. That is, the shifts differ in terms of production quality.

b.

To determine

Perform a multiple comparison test to determine how the shifts differ in terms of quality.

Answer to Problem 27SE

Supplier A and B can be used as they are not significantly different in terms of the proportion defective components and supplier C can be eliminated due to higher proportion of defective components.

Explanation of Solution

Calculation:

The critical values for pairwise comparison procedure of k population proportions are given as,

$C{V}_{ij}=\sqrt{{\chi }_{\alpha }^2}\sqrt{\frac{\overline{p_i}\left(1-\overline{p_i}\right)}{n_i}+\frac{\overline{p_j}\left(1-\overline{p_j}\right)}{n_j}}$.

Where, ${\chi }_{\alpha }^2$ be the chi square with level of significance $\alpha$ and degrees of freedom $k-1$; $\overline{p_i}$ and $\overline{p_j}$ be the sample proportion for populations i and j with sample size $n_i$ and $n_j$, respectively.

The sample proportion of good item for first shift is,

$\begin{array}{c}\overline{p_1}=\frac{285}{300}\\ =0.95\end{array}$

The sample proportion of good item for second shift is,

$\begin{array}{c}\overline{p_2}=\frac{368}{400}\\ =0.92\end{array}$

The sample proportion of good item for third shift is,

$\begin{array}{c}\overline{p_3}=\frac{176}{200}\\ =0.88\end{array}$

Multiple comparisons for first and second shift:

Degrees of freedom:

For a population of size k, the degrees of freedom is given as $k-1$.

In this given problem, for three populations the degrees of freedom is,

$\begin{array}{c}df=k-1\\ =3-1\\ =2\end{array}$

Thus, the degree of freedom is 2.

Level of significance:

The given level of significance is $\alpha =0.05$.

From the table “Area in Upper Tail” table it is found that the highest ${\chi }^2$ table value for degrees of freedom 2 and level of significance 0.05 is 5.991.

Thus, the critical values for pairwise comparison procedure of first and second shift is,

$\begin{array}{c}C{V}_{12}=\sqrt{{\chi }_{\alpha }^2}\sqrt{\frac{\overline{p_1}\left(1-\overline{p_1}\right)}{n_1}+\frac{\overline{p_2}\left(1-\overline{p_2}\right)}{n_2}}\\ =\sqrt{5.991}\sqrt{\frac{0.95\left(1-0.95\right)}{300}+\frac{0.92\left(1-0.92\right)}{400}}\\ =\left(2.447652\right)\sqrt{\left(\frac{0.0475}{300}+\frac{0.0736}{400}\right)}\end{array}$

       $\begin{array}{c}=\left(2.447652\right)\sqrt{\left(0.000158+0.000184\right)}\\ =\left(2.447652\right)\sqrt{\left(0.000342\right)}\\ =\left(2.447652\right)\left(0.0185\right)\\ \approx 0.0453\end{array}$

Multiple comparisons for first and third shift:

Degrees of freedom:

For a population of size k, the degrees of freedom is given as $k-1$.

In this given problem, for three populations the degrees of freedom is $3-1=2$.

Level of significance:

The given level of significance is $\alpha =0.05$.

From the table “Area in Upper Tail” table it is found that the highest ${\chi }^2$ table value for degrees of freedom 2 and level of significance 0.05 is 5.991.

Thus, the critical values for pairwise comparison procedure of first and third shift is,

$\begin{array}{c}C{V}_{13}=\sqrt{{\chi }_{\alpha }^2}\sqrt{\frac{\overline{p_1}\left(1-\overline{p_1}\right)}{n_1}+\frac{\overline{p_3}\left(1-\overline{p_3}\right)}{n_3}}\\ =\sqrt{5.991}\sqrt{\frac{0.95\left(1-0.95\right)}{300}+\frac{0.88\left(1-0.88\right)}{200}}\\ =\left(2.447652\right)\sqrt{\left(\frac{0.0475}{300}+\frac{0.1056}{200}\right)}\end{array}$

        $\begin{array}{c}=\left(2.447652\right)\sqrt{\left(0.000158+0.000528\right)}\\ =\left(2.447652\right)\sqrt{\left(0.000686\right)}\\ =\left(2.447652\right)\left(0.02619\right)\\ \approx 0.0641\end{array}$

Multiple comparisons for second and third shift:

Degrees of freedom:

For a population of size k, the degrees of freedom is given as $k-1$.

In this given problem, for three populations the degrees of freedom is $3-1=2$.

Level of significance:

The given level of significance is $\alpha =0.05$.

From the table “Area in Upper Tail” table it is found that the highest ${\chi }^2$ table value for degrees of freedom 2 and level of significance 0.05 is 5.991.

Thus, the critical values for pairwise comparison procedure of supplier B and C is,

$\begin{array}{c}C{V}_{23}=\sqrt{{\chi }_{\alpha }^2}\sqrt{\frac{\overline{p_2}\left(1-\overline{p_2}\right)}{n_2}+\frac{\overline{p_3}\left(1-\overline{p_3}\right)}{n_3}}\\ =\sqrt{5.991}\sqrt{\frac{0.92\left(1-0.92\right)}{400}+\frac{0.88\left(1-0.88\right)}{200}}\\ =\left(2.447652\right)\sqrt{\left(\frac{0.0736}{400}+\frac{0.1056}{200}\right)}\end{array}$

         $\begin{array}{c}=\left(2.447652\right)\sqrt{\left(0.000184+0.000528\right)}\\ =\left(2.447652\right)\sqrt{\left(0.000712\right)}\\ =\left(2.447652\right)\left(0.02668\right)\\ \approx 0.0653\end{array}$

Now,

Comparison	$p_i$	$p_j$	Difference $\left(p_i\sim p_j\right)$	$n_i$	$n_j$	Critical Value $\left(C{V}_{ij}\right)$	Significant $\left(\text{Diff}\text{>}\text{CV}\right)$
First vs. second	0.95	0.92	0.03	300	400	0.0453	No
First vs. Third	0.95	0.88	0.07	300	200	0.0641	Yes
Second vs. Third	0.92	0.88	0.04	400	200	0.0653	No

Conclusion:

It can be concluded that first and third shift, both are significantly different from second shift. Due to proportion of defective goods than other shifts, third shift can be eliminated.

Thus, the shift first and third can be used as they are not significantly different in terms of the proportion good components.