A federal agency studied the records pertaining to the number of job-related injuries at an underground coal mine. The value for the past 100 months were as follows:

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teamques10 ★ 68k

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Injuries per Month	0	1	2	3	4	5	6
Frequency of Occurrence	35	40	13	6	4	1	1

i) Apply the Chi-Square test to these data to test the hypothesis that the underlying distribution is Poisson.

ii) Apply the Chi-Square test to these data to test the hypothesis that the underlying distribution is Poisson with mean 1.0. Use level of significance $\alpha$= 0.05 and $\chi ^2$= 5.99, $\chi ^2 _{0.05,3}$= 7.81

computer simulation and modelling

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written 5.6 years ago by

teamques10 ★ 68k

• modified 5.6 years ago

i) Define hypothesis

$H_0:$ data fits to Poisson distribution

$H_a:$ data does not fits to Poisson distribution

For Poisson distribution,

$P_{i}=\frac{e^{-\alpha} \alpha^x}{x !}$

$\alpha=\overline{x}=\frac{\sum f_{i} m_{i}}{n}$

$\therefore \alpha = \frac{0 \times 35 + 1 \times 40 + 2 \times 13 + 3 \times 6 + 4 \times 4 + 5 \times 1 + 6 \times 1}{100}$

$\therefore \alpha = 1.11$

$X_i$	$O_i$	$P_i$	$E_i = h * P_i$	$\frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$
0	35	0.3296	32.96	0.126
1	40	0.3658	36.58	0.320
2	13	0.2030	20.30	2.625
3	6 $\quad$ \|	0.0751	7.51 $\quad$ \|	$\quad$ \|
4	4 $\quad$ 12	0.0209	2.09 $\quad$ 10.16	0.333
5	1 $\quad$ \|	0.0046	0.46 $\quad$ \|	$\quad$ \|
6	1 $\quad$ \|	0.0010	0.10 $\quad$ \|	$\quad$ \|
-	$\sum = 100$	$\sum = 1.0000$	$\sum = 100$	$\chi_0^2 = 3.400$

To find $\alpha$:

We need to estimate value of $\alpha$

$\therefore \quad S=1$

$\therefore \quad d \cdot f=k-s-1=4-1-1=2$

$\because x_{0.05, 2}^{2}=5.99$

$\chi_{0}^{2}=3.404\lt\chi^{2}_{0.05, 2} = 5.99$

$\therefore H_0$ is accepted.

$\therefore$ given data fits to Poisson distribution.

ii) Define hypothesis:

$H_0:$ data fits to Poisson distribution

$H_a:$ data does not fits to Poisson distribution

$\because p_{i}=\frac{e^{-\alpha} \alpha^{x}}{x !} \quad \alpha_{i}=1$

$X_i$	$O_i$	$P_i$	$E_i = h * P_i$	$\frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$
0	35	0.3279	36.79	0.087
1	40	0.3679	36.79	0.280
2	13	0.1839	18.39	1.580
3	6 $\quad$ \|	0.0613	6.13 $\quad$ \|	$\quad$ \|
4	4 $\quad$ 12	0.0153	1.53 $\quad$ 10.16	1.963
5	1 $\quad$ \|	0.0031	0.31 $\quad$ \|	$\quad$ \|
6	1 $\quad$ \|	0.0006	0.06 $\quad$ \|	$\quad$ \|
-	$\sum = 100$	$\sum = 1.0$	$\sum = 100$	$\chi_0^2 = 3.91 $

To find $\alpha$:

We do not estimate value of $\alpha$

$\therefore s = 0$

$\therefore d.f = k - s - 1 = 4 - 0 - 1 = 3$

$\because \chi_{0.05, 3}^2 = 7.81$

$\because \chi_{0}^{2} = 3.91 \lt \chi_{0.05, 3}^{2}=7.81$

$\therefore H_0$ is accepted.

$\therefore$ given data fits to Poisson distribution.

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