Chapter 15, Problem 15.59SE

To determine

To find:the data present sufficient efficient to indicate a difference in brightness measurements for two process. Use both parametric and non-parametric test and compare the results

Answer to Problem 15.59SE

Non-parametric:There is sufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.

Parametric: There is not sufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.

Explanation of Solution

Given:

The coded values for measures of brightness in paper, prepared by two different process, are given in the table.

  

Calculation:

  $\begin{array}{l}{n}_1=\text{sample size}=9\\ n_2=\text{sample size}=9\\ \alpha =\text{significant level}=0.05\end{array}$

The null hypothesis states that there is no difference in the populations. The alternatives hypothesis states the opposite of the null hypothesis.

  $\begin{array}{l}{H}_0:\text{ There is no difference between the two population distributions}\\ H_a:\text{ There is a difference between the two population distributions}\end{array}$

Determine the difference in the data in the data values of each pair and determine the sign of the difference. The smallest value receives the rank 1, the second smallest value receives the rank 2, the third smallest value receives the rank 3 and so on.

If the multiple data value has the same, then their rank is the average of the corresponding rank.

Now, to determine the difference in the ranks.

First sample	Rank	Second sample	Rank
6.1	1	6.9	2
7.1	3	7.5	4
7.6	5	7.8	6
8.3	9.5	7.9	7
8.7	12	8.2	8
8.9	13.5	8.3	9.5
9	15	8.6	11
9.2	17	8.9	13.5
9.5	18	9.1	16

When the test is left-tailed, $T_1^{\ast }$ when the test is right-tailed and min $\left(T_1,T_1^{\ast }\right)$ when the test is two tailed. Since, the test is right-tailed, it is need to use $T_1^{\ast }$ .

  $T_1$ is the sum of the ranks in the first sample:

  $T_1=1+3+9.5+12+13.5+15+17+18=94$

Next, it is determining the value of $T_1^{\ast }$ :

  $\begin{array}{l}{T}_1^{\ast }=n\left(n_1+n_2+1\right)-T_1\\ =9\left(9+9+1\right)-94=59\end{array}$

The value of the test statistics is equal to min $\left(T_1,T_1^{\ast }\right)$ .

  $T=\min \left(T_1,T_1^{\ast }\right)=\min \left(94,59\right)=59$

The critical value of the Wilcoxon signed is given in the row $n_2=9$ and in the column $n_1=9$ of the in the ranked table of the table of the Wilcoxon rank sum test with $\alpha =\raisebox{1ex}{$0.05$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=2.5\%$ .

  $T=62$

The rejection region contains all values smaller than or equal to $T=62$ .

If the value of the test statistics is in the rejection region, then it rejects the null hypothesis $H_0$

  $59\le 62\Rightarrow \text{ Reject }{H}_0$

There issufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.

Descriptive measures:

The mean is the sum of all values divided by the number of values:

Now, let us determine the squares of the difference of the rank

  $\begin{array}{l}\overline{x_1}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}=\frac{6.1+9.2+8.7+8.9+7.6+7.1+9.5+8.3+9.0}{9}\\ \overline{x_1}=8.2667\\ \overline{x_2}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}=\frac{9.1+8.2+8.6+6.9+7.5+7.9+8.3+7.8+8.9}{9}\\ \overline{x_2}=8.1222\end{array}$

The sample variance is the sum of the squared deviation from the mean divided by $n-1$ . The sample standard derivation is the square root of the population sample.

  $\begin{array}{l}{s}_1=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\left(n-1\right)}}\\ =\sqrt{\frac{\begin{array}{l}{\left(6.1-8.2667\right)}^2+{\left(9.2-8.2667\right)}^2+{\left(8.7-8.2667\right)}^2+{\left(8.9-8.2667\right)}^2\\ +{\left(7.6-8.2667\right)}^2+{\left(7.1-8.2667\right)}^2+{\left(9.5-8.2667\right)}^2+{\left(8.3-8.2667\right)}^2+{\left(9.0-8.2667\right)}^2\end{array}}{\left(9-1\right)}}\\ =1.1192\end{array}$

  $\begin{array}{l}{s}_2=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\left(n-1\right)}}\\ =\sqrt{\frac{\begin{array}{l}{\left(9.1-8.1222\right)}^2+{\left(8.2-8.1222\right)}^2+{\left(8.6-8.1222\right)}^2+{\left(8.6-8.1222\right)}^2+{\left(6.9-8.1222\right)}^2\\ +{\left(7.5-8.1222\right)}^2+{\left(7.9-8.1222\right)}^2+{\left(8.3-8.1222\right)}^2+{\left(7.8-8.1222\right)}^2+{\left(8.9-8.1222\right)}^2\end{array}}{\left(9-1\right)}}\\ =0.6778\end{array}$

It is noted that neither of the variance is more than the other variance (as the same is true for the sample standard deviations) thus it is then appropriate to use the pooled test.

Parametric test (pooled $t\text{-test}$ )

Given claim: The means differ

The claim is either null hypothesis or the alternatives hypothesis. The null hypothesis or the alternatives hypothesis states that these are opposite to each other. The null hypothesis needs to contain an equality.$\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1\ne {\mu }_2\end{array}$

Now, let us determine the pooled standard deviation:

  $\begin{array}{l}{s}_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{\left(9-1\right){1.1192}^2+\left(9-1\right){0.6678}^2}{9+9-2}}\approx 0.9252\end{array}$

Determine the test statistic:

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{8.2667-8.1222}{0.9252\sqrt{\frac{1}{9}+\frac{1}{9}}}\approx 0.331\end{array}$

The $P\text{-value}$ is the probability of obtaining a value of the test statistics, or a value more extreme. The $P\text{-value}$ is the number in the common title of student distribution table in the appendix containing the $t$ value in the row $df=n_1+n_2-2=9+9-2=16$ :

  $P>2\times 0.10=0.20$

If the $P\text{-value}$ is less than the significance level $\alpha$ , then rejected the null hypothesis:

  $P>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is not sufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.

Hence, it is concluded that both Parametric and Non-parametric:

Non-parametric: There is sufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.

Parametric: There is not sufficient evidence to support the claim that there is a difference in the brightness measurements for the two process.