Chapter 10.2, Problem 18P

(a)

To determine

Find the probability that the number of bacteria colonies per field.

Answer to Problem 18P

The probability that the number of bacteria colonies per field is,

r	0	1	2	3	4	5 or more
$P\left(r\right)$	0.061	0.170	0.238	0.222	0.156	0.153

Explanation of Solution

Calculation:

From the given information the probability distribution of r is,

$P\left(r\right)=\frac{e^{-\lambda }{\lambda }^r}{r!}$

Where, $\lambda =2.80$ is the average number of bacteria colonies per field, r is the number of bacteria colonies per field.

For $r=0$,

The probability that there are no bacteria colonies is,

$\begin{array}{c}P\left(0\right)=\frac{e^{-2.80}{\left(2.80\right)}^0}{0!}\\ =0.061\end{array}$

For $r=1$,

The probability that there is one bacteria colony is,

$\begin{array}{c}P\left(1\right)=\frac{e^{-2.80}{\left(2.80\right)}^1}{1!}\\ =0.170\end{array}$

For $r=2$,

The probability that there are two bacteria colonies is,

$\begin{array}{c}P\left(1\right)=\frac{e^{-2.80}{\left(2.80\right)}^2}{2!}\\ =\frac{0.477}{2}\\ =0.238\end{array}$

For $r=3$,

The probability that there are three bacteria colonies is,

$\begin{array}{c}P\left(3\right)=\frac{e^{-2.80}{\left(2.80\right)}^3}{3!}\\ =\frac{1.335}{6}\\ =0.222\end{array}$

For $r=4$,

The probability that there are four bacteria colonies is,

$\begin{array}{c}P\left(4\right)=\frac{e^{-2.80}{\left(2.80\right)}^4}{4!}\\ =\frac{3.737}{24}\\ =0.156\end{array}$

For $r=5\text{ or more}$,

The probability that there are five or more accidents is,

$\begin{array}{c}P\left(5\text{ or more}\right)=1-P\left(4\text{or less}\right)\\ =1-\left[P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)\right]\\ =1-\left[0.061+0.170+0.238+0.222+0.156\right]\\ =1-0.847\\ =0.153\end{array}$

Hence, the probability that the number of bacteria colonies per field is,

r	0	1	2	3	4	5 or more
$P\left(r\right)$	0.061	0.170	0.238	0.222	0.156	0.153

(b)

To determine

Find the expected number of colonies $E=100P\left(r\right)$ for $r=0,1,2,3,4,5\text{ or more}$.

Answer to Problem 18P

The expected number of colonies is,

r	Expected value
0	6.1
1	17.0
2	23.8
3	22.2
4	15.6
5 or more	15.3

Explanation of Solution

Calculation:

For $r=0$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(0\right)\\ =100\times 0.061\\ =6.1\end{array}$

For $r=1$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(1\right)\\ =100\times 0.170\\ =17.0\end{array}$

For $r=2$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(2\right)\\ =100\times 0.238\\ =23.8\end{array}$

For $r=3$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(3\right)\\ =100\times 0.222\\ =22.2\end{array}$

For $r=4$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(4\right)\\ =100\times 0.156\\ =15.6\end{array}$

For $r=5\text{ or more}$,

The expected number of colonies is,

$\begin{array}{c}E=100P\left(5\right)\\ =100\times 0.153\\ =15.3\end{array}$

Hence, the expected number of colonies is,

r	Expected value
0	6.1
1	17.0
2	23.8
3	22.2
4	15.6
5 or more	15.3

(c)

To determine

Find the sample statistic ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$.

Find the degrees of freedom.

Answer to Problem 18P

The sample statistic is 13.116.

The degrees of freedom are 5.

Explanation of Solution

Calculation:

Test statistic:

The sample chi-square test statistic is,

${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

In the formula O is the observed frequency, E is the expected frequency, with degrees of freedom $d.f.=k-1$, k is the number of categories,

$E=\text{(sample size }n\text{)(probability assigned to category)}$

The value of the chi-square statistic for the sample is,

$\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\left[\frac{{\left(12-6.1\right)}^2}{6.1}+\frac{{\left(15-17.0\right)}^2}{17.0}+\frac{{\left(29-23.8\right)}^2}{23.8}+\frac{{\left(18-22.2\right)}^2}{22.2}+\frac{{\left(19-15.6\right)}^2}{15.6}+\frac{{\left(7-15.3\right)}^2}{15.3}\right]\\ =5.706+0.235+1.136+0.795+0.741+4.503\\ =13.116\end{array}$

Hence, the sample statistic is 13.116.

Substitute 6 for k in the degrees of freedom formula.

$\begin{array}{c}d.f.=6-1\\ =5\end{array}$

Hence, the degrees of freedom are 5.

(d)

To determine

Check whether the Poisson distribution fits the sample data or not.

Answer to Problem 18P

There is sufficient evidence that the Poisson distribution does not fit the sample data.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.05, and to test the statement that the Poisson distribution fits the sample data.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The Poisson distribution fits the sample data.

Alternative hypothesis:

$H_1:$ The Poisson distribution does not fit the sample data.

P-value:

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 ‘Chi-Square’ distribution.
• In Degrees of freedom, enter the value as 5.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 13.116.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0223.

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 0.0223 and the level of significance is 0.05.

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

That is, $0.0223\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, there is sufficient evidence that the Poisson distribution does not fit the sample data at level of significance 0.05.