Chapter 15.2, Problem 15.8E

a.

To determine

Check whetherthe river community was releasing semi treated sewage into the river.

Answer to Problem 15.8E

There is no sufficient evidence to indicate that the river community was releasing semi treated sewage into the river.

Explanation of Solution

Given:

The given data set is

  

Calculation:

The sample size $n_1=5$ and $n_2=5$

  $\alpha =$ significance level $=0.05$

The null hypothesis tells that there is no difference in the populations. The alternative hypothesis is opposite of null hypothesis.

  $h_0\to$ There is no difference between the two population distributions.

  $h_a\to$ first distribution is shifted to the right of second distribution.

Find the rank of every data.

The rank of smallest value is one. The rank of second smallest value is two and so on.

The multiple data have the same value. The rank is the average of the corresponding ranks.

Sample 1	Rank	Sample 2	Rank
$\begin{array}{l}4.8\\ 4.9\\ 5\\ 5.1\\ 5.2\end{array}$	$\begin{array}{l}2.5\\ 5\\ 7.5\\ 9\\ 10\end{array}$	$\begin{array}{l}4.7\\ 4.8\\ 4.9\\ 4.9\\ 8\end{array}$	$\begin{array}{l}1\\ 2.5\\ 5\\ 5\\ 7.5\end{array}$

Find a distribution one is shifted either to the left or to the right of distribution two.

The test is left tailed $\to$ $T_1$

The test is right tailed $\to$ $T_1{}^{*}$

The test is two tailed $\to$ min $\left(T_1,T_1{}^{*}\right)$

Find $T_1$ .

  $T_1=2.5+5+7.5+9+10=34$

The given test is right tailed. Find the value of $T_1{}^{*}$ .

  $\begin{array}{l}{T}_1{}^{*}=n_1\left(n_1+n_2+1\right)-T_1\\ =5\left(5+5+1\right)-34\\ =55-34\\ =21\end{array}$

  $\alpha =$ significance level $=.05$

The critical value for the Wilcoxon rank sum test

Row $n_2=5$

Column $n_1=5$

  $\alpha =0.05=5\%$

From the table of the Wilcoxon rank sum test

  $T=19$

Hence the rejection region that contains all values $\le 19$ .

If the value of the test statistic is in the rejection region, then reject the null hypothesis.

  $21>19$

The test statistic is notin the rejection region.

Hence there is no sufficient evidence to indicate thatthe river community was releasing semi treated sewage into the river.

b.

To determine

Check whether the river community was releasing semi treated sewage into the river.

Answer to Problem 15.8E

It is the same conclusion as part ‘a’.

Explanation of Solution

Given:

The given data set is

  

Calculation:

The sample size $n_1=5$ and $n_2=5$

  $\alpha =$ significance level $=0.05$

The null hypothesis tells that there is no difference in the populations. The alternative hypothesis is opposite of null hypothesis.

  $h_0\to$ There is no difference between the two population distributions.

  $h_a\to$ first distribution is shifted to the right of second distribution.

Find the rank of every data.

The rank of smallest value is one. The rank of second smallest value is two and so on.

The multiple data have the same value. The rank is the average of the corresponding ranks.

Sample 1	Rank	Sample 2	Rank
$\begin{array}{l}4.8\\ 4.9\\ 5\\ 5.1\\ 5.2\end{array}$	$\begin{array}{l}2.5\\ 5\\ 7.5\\ 9\\ 10\end{array}$	$\begin{array}{l}4.7\\ 4.8\\ 4.9\\ 4.9\\ 8\end{array}$	$\begin{array}{l}1\\ 2.5\\ 5\\ 5\\ 7.5\end{array}$

Find mean of both the sample.

  $\begin{array}{l}\overline{m_1}=\frac{{\displaystyle \sum _{i=1}^n{m}_i}}{n}=\frac{4.8+5.2+5.0+4.9+5.1}{5}=5\\ \overline{m_2}=\frac{{\displaystyle \sum _{i=1}^n{m}_2}}{n}=\frac{5+4.7+4.9+4.8+4.9}{5}=4.86\end{array}$

Find the sample standard deviation.

  $\begin{array}{l}{s}_1=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}=\sqrt{\frac{{\left(4.8-5\right)}^2+{\left(5.2-5\right)}^2+{\left(5.0-5\right)}^2+{\left(4.9-5\right)}^2+{\left(5.1-5\right)}^2}{5-1}}=0.158\\ s_2=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}=\sqrt{\frac{{\left(5-4.86\right)}^2+{\left(4.7-4.86\right)}^2+{\left(4.9-4.86\right)}^2+{\left(4.8-4.86\right)}^2+{\left(4.9-4.86\right)}^2}{5-1}}=0.114\end{array}$

There is no one deviation is more than three times the other deviation.

Use pooled t-test.

The means of above town is higher.

Find the pooled standard deviation.

  $s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right)s_2{}^2}{n_1+n_2-2}}=\sqrt{\frac{\left(5-1\right){0.158}^2+\left(5-1\right){0.114}^2}{5+5-2}}\approx 0.138$

Find the test statistic.

  $t=\frac{\overline{x_1}-\overline{x_2}}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{5-4.86}{0.138\sqrt{\frac{1}{5}+\frac{1}{5}}}\approx 1.6$

The p-value is the probability of obtaining the value of the test static.

Use student t distribution table and find the value of p.

Row $df=5+5-2=8$

  $0.05<p<0.10$

If the p-value is less than or equal to the significance level, then the null hypothesis is rejected.

  $p>0.05$

There is no sufficient evidence to indicate that the river community was releasing semi treated sewage into the river.

Hence it is the same conclusion as part ‘a’.