Question

When the first proof of 392 pages of a book of 1200 pages were read, the distribution of printing mistakes were found to be as follows:
No. Of mistakes in page(x) : 0 1 2 3 4 5 6
No. Of pages (f) :275 72 30 7 5 2 1
Fit a Poisson distribution to the above data and test the of goodness of fit.

Answer 1

When the first proof of 392 pages of a book of 1200 pages were read, the distribution of printing mistakes were found to be as follows: 

$\begin{matrix} \small \text{No. of mistakes in a page}\ (x): & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \small \text{No. of pages}\ (f): & 275 & 72 & 30 & 7 & 5 & 2 & 1 \end{matrix}$

Fit a Poisson distribution to the above data and test the goodness of fit at 5 percent Level of significance. 

The null hypothesis is $H_0: X\sim Poisson$

The alternative hypothesis is $H_1:X$ does not follow a Poisson distribution. 

The mean of the (assumed) Poisson distribution is unknown so must be estimated from the data by the sample mean: 

$$\hat{\mu}={0(275)+1(72)+2(30)+3(7)+4(5)+5(2)+6(1) \over 392}\approx0.482$$

Using the Poisson distribution with $\mu=0.482$ we can compute $p_i,$ the hypothesised probabilities associated with each class. From these we can calculate the expected frequencies (under the null hypothesis): 

$$p_0=P(X=0)={e^{-0.482}\cdot(0.482)^0 \over 0!}\approx0.614547$$ $$0.614547\cdot392\approx240.90$$

$$p_1=P(X=1)={e^{-0.482}\cdot(0.482)^1 \over 1!}\approx0.297658$$ $$0.297658\cdot392\approx116.68$$

$$p_2=P(X=2)={e^{-0.482}\cdot(0.482)^2 \over 2!}\approx0.071736$$ $$0.071736\cdot392\approx28.12$$

$$p_3=P(X=3)={e^{-0.482}\cdot(0.482)^3 \over 3!}\approx0.011526$$ $$0.011526\cdot392\approx4.52$$

$$p_4=P(X=4)={e^{-0.482}\cdot(0.482)^4 \over 4!}\approx0.001389$$ $$0.001389\cdot392\approx0.54$$

$$p_5=P(X=5)={e^{-0.482}\cdot(0.482)^5 \over 5!}\approx0.000134$$ $$0.000134\cdot392\approx0.05$$

$$p_6=P(X=0)={e^{-0.482}\cdot(0.482)^6 \over 6!}\approx0.000011$$ $$0.000011\cdot392\approx0.004$$

$$\begin{matrix} x & Observed, f_O & Expected,f_E & {(f_O-f_E)^2 \over f_E} \\ 0 & 275 & 240.90 & 4.827 \\ 1 & 72 & 116.68 & 17.109 \\ 2 & 30 & 28.12 & 0.126 \\ 3 & 7 & 4.52 & 1.361 \\ 4 & 5 & 0.54 & 36.836 \\ 5 & 2 & 0.05 & 76.05 \\ 6 & 1 & 0.004 & 248.004 \\ Sum= & 392 & 392 & 384.313 \end{matrix}$$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0:p_0=0.614547,p_1=0.297658,p_2=0.071736,$

$p_3=0.011526,p_4=0.001389,p_5=0.000134,p_6=0.000011$

$H_1:$ Some of the population proportions differ from the values stated in the null hypothesis.

This corresponds to a Chi-Square test for Goodness of Fit.

Based on the information provided, the significance level is $\alpha=0.05,$ the number of degrees of freedom is equal to the number of levels (k) of the categorical variable minus 1: $df=k-1=7-1=6,$ so then the rejection region for this test is 

$R=\{\chi^2:\chi^2>12.592\}$

The Chi-Squared statistic is computed as follows:

$$\chi^2=\displaystyle\sum_{i=1}^n{(O_i-E_i)^2 \over E_i}=4.827+17.109+0.126+$$ $$+1.361+36.836+76.05+248.004=384.313$$

Since it is observed that  $\chi^2=384.313>\chi_c^2=12.592,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the $\alpha=0.05$ significance level.

We conclude that there is enough evidence to claim that the data do not follow a Poisson distribution.