Chapter 9, Problem 94SE

To determine

Develop the appropriate large sample test procedure.

Check whether the plant densities are equal for region 1 and region 2.

Answer to Problem 94SE

The data do not suggest the sufficient result that the plant densities are equal for region 1 and region 2.

Explanation of Solution

Given info:

The data represents the samples taken from the randomly located square sampling quadrate having the area of 1 m_­².

Here, $X_1,X_2,\mathrm{...}{X}_m$ and $Y_1,Y_2,\mathrm{...}{Y}_n$ follows Poisson distribution with parameter ${\mu }_1$ and ${\mu }_2$.

Calculation:

Large sample test procedure:

The estimate of ${\widehat{\lambda }}_1$ is obtained as given below:

Here $E\left({\mu }_1\right)=\overline{X}$

$\begin{array}{c}E\left({\widehat{\lambda }}_1\right)=E\left({\mu }_1\right)\\ E\left({\widehat{\lambda }}_1\right)=\overline{X}\end{array}$

The estimate of ${\widehat{\lambda }}_2$ is obtained as given below:

Here $E\left({\mu }_2\right)=\overline{Y}$

$\begin{array}{c}E\left({\widehat{\lambda }}_2\right)=E\left({\mu }_2\right)\\ E\left({\widehat{\lambda }}_2\right)=\overline{Y}\end{array}$

Then it is assumed that $\lambda ={\lambda }_1={\lambda }_2$.

The variance of $\left(\overline{X}-\overline{Y}\right)$ is obtained as shown below:

$\begin{array}{c}V\left(\overline{X}-\overline{Y}\right)=\frac{{\lambda }_1}{m}+\frac{{\lambda }_2}{n}\\ =\frac{\lambda }{m}+\frac{\lambda }{n}\\ =\lambda \left(\frac{1}{m}+\frac{1}{n}\right)\end{array}$

The pooled estimate of  ${\lambda }_{\text{Pooled}}$ is given by:

$\begin{array}{c}{\widehat{\lambda }}_{\text{Pooled}}=\frac{{\widehat{\lambda }}_1}{m}+\frac{{\widehat{\lambda }}_2}{n}\\ =\frac{m\overline{X}-n\overline{Y}}{m+n}\end{array}$

Test statistic:

Here, the sample size is large. That is, $m>40$ and $n>40$. Hence, the test statistic for the large samples is obtained as:

$\begin{array}{c}Z=\frac{\left(\overline{X}-\overline{Y}\right)-{\Delta }_0}{\sqrt{{\lambda }_{\text{Pooled}}\left(\frac{1}{m}+\frac{1}{n}\right)}}\\ =\frac{\left(\overline{X}-\overline{Y}\right)-0}{\sqrt{\frac{m\overline{X}-n\overline{Y}}{m+n}\left(\frac{1}{m}+\frac{1}{n}\right)}}\\ =\frac{\left(\overline{X}-\overline{Y}\right)}{\sqrt{\left(\frac{\overline{Y}}{m}+\frac{\overline{X}}{n}\right)}}\end{array}$

Test procedure:

The aim of the problem is to test the hypotheses that the plant densities are equal for region 1 and region 2.

Assume that the alternative hypothesis as $H_a:{\lambda }_1\ne {\lambda }_2$ and level of significance $\alpha =0.05$.

The mean of the region1 is ${\lambda }_1$ and mean of the region 2 is ${\lambda }_2$.

The test hypotheses are,

Null hypothesis:

$H_0:{\lambda }_1={\lambda }_2$

That is, the plant density of region 1 is equal to the density of region 2.

Alternative hypothesis:

$H_a:{\lambda }_1\ne {\lambda }_2$

That is, the plant density of region 1 is different from the density of region 2.

The value of the $\overline{X}$ is obtained as given below:

$\begin{array}{c}\overline{X}=\frac{\left\{\begin{array}{l}\left(0\times 28\right)+\left(1\times 40\right)+\left(2\times 28\right)+\left(3\times 17\right)+\left(4\times 8\right)+\\ \left(5\times 2\right)+\left(6\times 1\right)+\left(7\times 1\right)\end{array}\right\}}{125}\\ =\frac{0+40+56+51+32+10+6+7}{125}\\ =\frac{202}{125}\\ =1.616\end{array}$

The value of the $\overline{Y}$ is obtained as given below:

$\begin{array}{c}\overline{Y}=\frac{\left\{\begin{array}{l}\left(0\times 14\right)+\left(1\times 25\right)+\left(2\times 30\right)+\left(3\times 18\right)+\left(4\times 49\right)+\\ \left(5\times 2\right)+\left(6\times 1\right)+\left(7\times 1\right)\end{array}\right\}}{125}\\ =\frac{0+25+60+54+196+10+6+7}{140}\\ =\frac{358}{140}\\ =2.557\end{array}$

The test statistic value is obtained below:

Substitute $\overline{X}=1.616$,$\overline{Y}=2.557$,$m=125$ and $n=140$

$\begin{array}{c}Z=\frac{\left(1.616-2.557\right)}{\sqrt{\frac{1.616}{125}+\frac{2.557}{140}}}\\ =\frac{-0.9411}{\sqrt{0.012928+0.01826}}\\ =\frac{-0.9411}{0.1766}\\ =-5.33\end{array}$

Thus, the test statistic value is –5.33.

From Appendix, “Table A.3 Standard Normal Curves”, the standard normal value of –3.40 at 5% level of significance is 0.000.

So, the $P\left(Z\le -5.3\right)=0.000$

The P-value is obtained as given below:

$\begin{array}{c}P\left(\left|Z\right|\ge \left|-5.3\right|\right)=2\times \left[\Phi \left(-5.3\right)\right]\\ =0\end{array}$

Thus, the P-value is 0.

Decision rule based on P-value approach:

Rejection region for a right-tailed test:

If $P-\text{value}<\alpha$ , then reject the null hypothesis $\left(H_0\right)$.

If $P-\text{value>}\alpha$ , then fail to reject the null hypothesis $\left(H_0\right)$

Conclusion:

Here, the P-value is less than the level of significance.

That is,$P\text{-value}\left(=0\right)<\alpha \left(=0.5\right)$.

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

Thus, it can be concluded that there is no enough evidence to infer that the plant densities are equal for region 1 and region 2.