Chapter 15, Problem 15.80LM

The data for three independent random samples are shown in the table below. It is known that the sampled populations normally distributed. Use an appropriate test to determine whether the data provide sufficient evidence to indicate that at least two of the populations differ in location. Test using α = .05.

Sample from Population 1		Sample from Population 2		Sample from Population 3
18	15	12	34	87	50
32	63	33	18	53	64
43		10		65	77

To determine

Whether the data provide sufficient evidence to indicate that at least two of the populations differ in location or not.

Answer to Problem 15.80LM

There is sufficient evidence to reject the claim. This implies that the probability distributions of at least two of the populations differ in location.

Explanation of Solution

Given info: The data shows that the three independent random samples, and it is known that the sampled proportions are not normally distributed.

Calculation:

The claim that the probability distributions of at least two of the populations are differ in location were presented is alternative hypothesis. Thus, the null hypothesis is the all three populations have the same distribution.

The hypotheses are given below:

Null hypothesis:

$H_0:$ The probability distributions of the three populations have the same distribution.

Alternative hypothesis:

$H_a:$ At least two of the population’s distribution is different from the location.

The test statistic is compute by using the below formula.

$H=\frac{12}{n\left(n+1\right)}{\displaystyle \sum n_j{\left({\overline{R}}_j-\overline{R}\right)}^2}$

Where,

$\begin{array}{l}{\overline{R}}_j=\frac{{\displaystyle \sum R_j}}{n_j}\\ \overline{R}=\frac{\left({\displaystyle \sum n_j}\right)+1}{2}\end{array}$

The ranks are obtained below:

Population	Observation	Rank
1	18	4.5
1	32	6
1	43	9
1	15	3
1	63	12
2	12	2
2	33	7
2	10	1
2	34	8
2	18	4.5
3	87	16
3	53	11
3	65	14
3	50	10
3	64	13
3	77	15

The sum of ranks of population 1 is,

$\begin{array}{c}{R}_1=4.5+6+9+3+12\\ =34.5\end{array}$

The sum of ranks of population 2 is,

$\begin{array}{c}{R}_2=2+7+1+8+4.5\\ =22.5\end{array}$

The sum of ranks of population 3 is,

$\begin{array}{c}{R}_3=16+11+14+10+13+15\\ =79\end{array}$

The value of ${\overline{R}}_1$ is,

$\begin{array}{c}{\overline{R}}_1=\frac{R_1}{n_1}\\ =\frac{34.5}{5}\\ =6.9\end{array}$

Thus, the value of ${\overline{R}}_1$ is 6.9.

The value of ${\overline{R}}_2$ is,

$\begin{array}{c}{\overline{R}}_2=\frac{R_2}{n_2}\\ =\frac{22.5}{5}\\ =4.5\end{array}$

Thus, the value of ${\overline{R}}_2$ is 4.5.

The value of ${\overline{R}}_3$ is,

$\begin{array}{c}{\overline{R}}_3=\frac{R_3}{n_3}\\ =\frac{79}{6}\\ =13.167\end{array}$

Thus, the value of ${\overline{R}}_3$ is 13.167.

The value of $\overline{R}$ is,

$\begin{array}{c}\overline{R}=\frac{\left(n_1+n_2+n_3\right)+1}{2}\\ =\frac{5+5+6+1}{2}\\ =\frac{17}{2}\\ =8.5\end{array}$

Thus, the value of $\overline{R}$ is 8.5.

The test statistic is,

$\begin{array}{c}H=\frac{12}{n\left(n+1\right)}{\displaystyle \sum n_j{\left({\overline{R}}_j-\overline{R}\right)}^2}\\ =\frac{12}{16\left(17\right)}\left[5{\left(6.9-8.5\right)}^2+5{\left(4.5-8.5\right)}^2+6{\left(13.167-8.5\right)}^2\right]\\ =\frac{12}{272}\left[\left(5\times 2.56\right)+\left(5\times 16\right)+\left(6\times 21.7809\right)\right]\\ =0.0441\left[12.8+80+130.6853\right]\end{array}$

    $\begin{array}{c}=0.0441\times 223.4853\\ =9.856\end{array}$

The test value is $H=9.856$.

Critical value:

Here, the level of significance $\left(\alpha \right)$ is 0.05, and the number of variables k is 3.

The degrees of freedom is,

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

The level of significance is,

$\alpha =0.05$

From “Appendix D Table IV: Critical Values of ${\chi }^2$” the critical region is obtained by following the below procedure:

Procedure:

• Locate the degrees of freedom as 2 in the first column.
• Locate the level of significance as 0.05 in the first row.
• The intersection value 5.99147 is the critical region for the degrees of freedom as 2 with $\alpha =0.05$.

Thus, the rejection region for the test is $H>5.99147$.

Rejection region:

If $H>5.99147$, then reject the null hypothesis $H_0$.

Decision:

It is clear that the test statistic value is 9.856, and critical value is 5.99147.

Here, test statistic is greater than critical value.

That is, $H\left(=9.856\right)>\text{critical value}\left(=5.99147\right)$.

Therefore, by the rejection region, the null hypothesis $H_0$ is rejected.

There is sufficient evidence to reject the claim. Hence, the probability distributions of at least two of the populations differ in location.