Chapter 11, Problem 3CRP

(a)

To determine

Identify the test to be used.

Answer to Problem 3CRP

The test to be used is rank-sum test.

Explanation of Solution

The rank-sum test is the nonparametric test which is used when the data is independent. It is used for comparing the sample distributions that are taken from two populations which are independent. The rank-sum test is used for testing the difference between sample means and when the assumption of normal distribution is not satisfied. It is also referred as Mann–Whitney or Wilcoxon's rank-sum test.

In the scenario, the samples of ‘with catalyst’ and ‘without catalyst’ are independent of each other. This shows that, rank-sum test is appropriate to apply.

Hence, the test to be used is rank-sum test.

(b)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 3CRP

The level of significance is 0.05.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.05, and to test the claim that the viscosity index has changed.

Hence, the level of significance is 0.05.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The distributions of viscosity index for with catalyst and without catalyst are same.

Alternative hypothesis:

$H_1:$ The distributions of viscosity index for with catalyst and without catalyst are different.

(c)

To determine

Find the value of the sample test statistic.

Answer to Problem 3CRP

The value of the sample test statistic is 0.12.

Explanation of Solution

Calculation:

Test statistic:

The z value for the sample test statistic R is,

$z=\frac{R-{\mu }_R}{{\sigma }_R}$

In the formula R is the sum of ranks from the sample of size $n_1$ (smaller sample),

$\begin{array}{l}{\mu }_R=\frac{n_1\left(n_1+n_2+1\right)}{2}\\ {\sigma }_R=\sqrt{\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}}\end{array}$

$n_1$ be the sample size of the smaller sample and $n_2$ be the sample size of the larger sample

Procedure for assigned rank to data values:

• First combine both the samples.
• Arrange the data values in ascending order.
• Rank each of the data value in sequential order.

The rank for each of tied data is computed as,

• First assign the sequential position ranks for all the values that are same.
• Take the mean of all the position ranks that are assigned for same data values.
• Assign the mean rank as the rank position for all the tied data.
• The assigned mean rank is used in calculating the test statistic.

The ranks are,

Index	Rank	Group
1.1	1	with catalyst
1.5	2	without catalyst
1.6	3	with catalyst
1.8	4	with catalyst
1.9	5	without catalyst
2.2	6	without catalyst
2.4	7	without catalyst
2.5	8	with catalyst
2.8	9	without catalyst
2.9	10	with catalyst
3.1	11	without catalyst
3.2	12	with catalyst
3.3	13	without catalyst
3.5	14	without catalyst
3.6	15	without catalyst
3.7	16	with catalyst
3.8	17	with catalyst
3.9	18	without catalyst
4.0	19	without catalyst
4.1	20	with catalyst
4.2	21	with catalyst
4.4	22	with catalyst
4.6	23	without catalyst

The smaller sample is 11 which corresponds the ‘with catalyst’, and without catalyst sample size is 12. That is, $n_1=11,n_2=12$.

The value of R is,

$\begin{array}{c}R=1+3+4+8+10+12+16+17+20+21+22\\ =134\end{array}$

The value of R is 134.

The mean of R is,

$\begin{array}{c}{\mu }_R=\frac{11\left(11+12+1\right)}{2}\\ =\frac{11\left(24\right)}{2}\\ =\frac{264}{2}\\ =132\end{array}$

The standard deviation of R is,

$\begin{array}{c}{\sigma }_R=\sqrt{\frac{\left(11\times 12\right)\left(11+12+1\right)}{12}}\\ =\sqrt{\frac{132\left(24\right)}{12}}\\ =\sqrt{\frac{3,168}{12}}\\ =\sqrt{264}\\ =16.25\end{array}$

Test statistic:

Substitute R as 134, ${\mu }_R$ as 132 and ${\sigma }_R$ as 16.25 in the test statistic formula

$\begin{array}{c}z=\frac{134-132}{16.25}\\ =\frac{2}{16.25}\\ =0.12\end{array}$

Hence, the z value is 0.12.

(d)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 3CRP

The P-value is 0.9044.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tail, for the region of the curve to shade.
• Enter the X value as 0.12.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.4522 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.4522\\ =0.9044\end{array}$

Hence, the P-value is 0.9044.

(e)

To determine

Mention the conclusion of the test.

Interpret the conclusion in the context of the application.

Answer to Problem 3CRP

The null hypothesis is failed to be rejected.

Explanation of Solution

Calculation:

From part (d), the P-value is 0.9044.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.9044 and the level of significance is 0.05.

The P-value is greater than the level of significance.

That is, $0.9044\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.05.

There is no sufficient evidence that the distributions of viscosity index for with catalyst and without catalyst are different at level of significance 0.05.