Question

The following table presents the observed and expected data on the number of plants found in each of 50 sampling quadrants. # of plants 0 Observed Frequency (0) 5 18 10 1 Expected Frequency (E;) 6.767 13.534 13.534 9.022 7.144 2 3 >4 12 5 Ho: The distribution is Poisson H7: The distribution is not Poisson A) Justify why the assumption of the Poisson distribution seem appropriate as a probability model for this data? Find the value of the only parameter Poisson distribution has. B) Using a = 0.01, find the test statistic and critical region to make a conclusion about this goodness of fit test. (Hint: we did not estimate the parameter) C) Calculate the P-value for this test. State the conclusion for this test.

Answer 1

a) Since the random variable is the number of plants found in each of the sampling quadrants, and the number of plants will always be a discrete value, hence, the Poisson's distribution will be an appropriate model for this data.

From the data, we can see that the expected number of sampling quadrants with no plants is = 6.767.

The hypothesized probability that a sampling quadrant will have no plants is = 6.767/50 = 0.135

If $\lambda$ is our parameter for the Poisson's distribution, then the probability of having no plant in sampling quadrant will be

$P(X = 0) = \frac{\lambda^x.e^{-\lambda}}{x!} = \frac{\lambda^0.e^{-\lambda}}{0!} = e^{-\lambda}$

Now, as calculated above, we have

e = 0.135

$\lambda = 2$

Hence, the value of the Poisson distribution parameter is = 2.

b) The value of the chi-squared test statistics is

$\chi^2 = \frac{\sum (O_{i} - E_{i})^2}{E_{i}}$

$\chi^2 = \frac{(5 -6.767 )^2}{6.767} + \frac{(18 -13.534 )^2}{13.534} + \frac{(10 -13.534 )^2}{13.534} + \frac{(12 -9.022 )^2}{9.022}+ \frac{(5 -7.144 )^2}{7.144}$

$\chi^2 = 0.461 +1.474+ 0.923 +0.983+0.643 = 4.484$

The critical chi-sqaured value for a degree of freedom of 4 and a significance level of 0.01 is = 13.277

Hence, the critical region will be the region to the right of the chi-sqaure value of 13.277, that is, if the chi-squared value is less than the critical value then we will accept the null hypothesis.

C. The p-value corresponding to a chi-sqaured statistics of 4.484 and a degree of freedom of 4 is = 0.344.

The significance level is = 0.01.

Since the p-value is greater than the significance level, hence we will fail to reject the null hypothesis.

Conclusion: The null hypothesis is true and there is not enough evidence to conclude that the data doesn't follow a Poisson distribution.

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