91影视

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 confidence interval for \({\mu _1} - {\mu _2}\)

can be obtained from

\(P\left( { - {z_{\alpha /2}} \le Z \le {z_{\alpha /2}}} \right) = 1 - \alpha \)

which with transformations becomes

\({\hat \mu _1} - {\hat \mu _2} \pm {z_{\alpha /2}} \cdot \sqrt {\frac{{{{\hat \mu }_1}}}{m} + \frac{{{{\hat \mu }_2}}}{n}} \)

Remember that \(\bar x = {\hat \mu _1}\) and\(\bar y = {\hat \mu _2}\).

The accompanying table summarize the data to perform this test.

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

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

Thus, the 95 % confidence interval \(\left( {\alpha = 0.05,{z_{\alpha /2}} = 1.96} \right)\) is

\(\begin{array}{l}1.616 - 2.557 - 1.96 \cdot \sqrt {\frac{{1.616}}{{30}} + \frac{{2.557}}{{30}}} ,1.616 - 2.557 + 1.96 \cdot \sqrt {\frac{{1.616}}{{30}} + \frac{{2.557}}{{30}}} \\ = ( - 0.94 - 1.96 \cdot 0.177, - 0.94 + 1.96 \cdot 0.177)\\ = ( - 1.29, - 0.59).\end{array}\)

Finally we get,

\(\begin{array}{l}{{\hat \mu }_1} - {{\hat \mu }_2} \pm {z_{\alpha /2}} \cdot \sqrt {\frac{{{{\hat \mu }_1}}}{m} + \frac{{{{\hat \mu }_2}}}{n}} \\( - 1.29, - 0.59)\end{array}\)