Question

Let \({X_L},....,{X_m}\)be a random sample from a Poisson distribution with parameter\({\mu _1}\), and let \({Y_1},....,{Y_n}\)be a random sample from another Poisson distribution with parameter\({\mu _2}\). We wish to test \({H_0}:{\mu _1} - {\mu _2} = 0\) against one of the three standard alternatives. When m and n are large, the large-sample z test of Section 9.1 can be used. However, the fact that \(V(\bar X) = \mu /n\) suggests that a different denominator should be used in standardizing\(\bar X - \bar Y\). Develop a large sample test procedure appropriate to this problem, and then apply it to the following data to test whether the plant densities for a particular species are equal in two different regions (where each observation is the number of plants found in a randomly located square sampling quadrate having area\(1\;{m^2}\), so for region 1 there were 40 quadrates in which one plant was observed, etc.):

Short answer

Reject null hypothesis

Step-by-step solution

Step 1: To obtain a large sample test procedure

Proposition: For a random variable X with Poisson distribution with parameter\(\mu > 0\), the following is true

\(E(X) = V(X) = \mu \)

As mentioned,

\(V(\bar X) = \frac{\mu }{n}\)

The standard deviation for\(\bar X - \bar Y\)can be computed as

\(\begin{array}{l}{\sigma _{\bar X - \bar Y}} = \sqrt {V(\bar X - \bar Y)} = \sqrt {V(\bar X) + V(\bar Y)} \\ = \sqrt {\frac{{{\mu _1}}}{m} + \frac{{{\mu _2}}}{n}} .\end{array}\)

To form a test statistic, the standard deviation needs to be estimated. First estimate\({\mu _1}\)and\({\mu _2}\), as usual, with

\(\begin{array}{l}{{\hat \mu }_1} = \bar X\\{{\hat \mu }_2} = \bar Y.\end{array}\)

This yields test statistic with standard normal distribution (for large m and n)

\(Z = \frac{{\bar X - \bar Y}}{{\sqrt {\frac{{{{\hat \mu }_1}}}{m} + \frac{{{{\hat \mu }_2}}}{n}} }}\)

The accompanying table summarize the data to perform this test.

Step 2: Final proof

The sample mean for the number of plants in the region 1 is

\(\bar x = {\hat \lambda _1} = \frac{1}{{125}} \cdot (0 + 40 + 56 + \ldots + 7) = 1.616\)

and for the region is

\(\bar y = {\hat \lambda _2} = \frac{1}{{140}} \cdot (0 + 25 + 60 + \ldots + 7) = 2.557\)

The z statistic value is

\(z = \frac{{1.611 - 2.557}}{{\sqrt {1.616/125 + 2.557/140} }} = - 5.3\)

Depending on the alternative hypothesis the P value would differ. The alternative hypothesis of interest is\({H_a}:{\mu _1} \ne {\mu _2}\); thus the P value is two times the area under the z curve to the right of value |z|

\(P = 2 \cdot P(Z > 5.3) = 2 \cdot 0 = 0\)

so

Reject null hypothesis

At any reasonable significance level. The value was computed using software (you could use the table in the appendix.