Chapter 5, Problem 19SE

a.

To determine

Show that for a proportion $1-\alpha$ of all possible samples, $X-z_{\frac{\alpha }{2}}{\sigma }_X<n\lambda <X+z_{\frac{\alpha }{2}}{\sigma }_X$.

Explanation of Solution

Given info:

Consider X represents the number of events that are observed to occur in n units of time or space. That is, $X\sim \text{Poisson}\left(n\lambda \right)$. if X is large, $X\sim N\left(n\lambda ,n\lambda \right)$

Calculation:

Here, X follows normally distributed with mean $n\lambda$. Hence, proportion $1-\alpha$ of all possible samples is, $-z_{\frac{\alpha }{2}}{\sigma }_X<X-n\lambda <z_{\frac{\alpha }{2}}{\sigma }_X$.

Multiply by −1 on both sides in the above inequalities.

$\begin{array}{c}-1\left(-z_{\frac{\alpha }{2}}{\sigma }_X\right)<-1\left(X-n\lambda \right)<-1\left(z_{\frac{\alpha }{2}}{\sigma }_X\right)\\ z_{\frac{\alpha }{2}}{\sigma }_X<-X+n\lambda <-z_{\frac{\alpha }{2}}{\sigma }_X\end{array}$

Add X on both sides in the above inequalities.

$\begin{array}{c}X+z_{\frac{\alpha }{2}}{\sigma }_X<-X+n\lambda +X<-z_{\frac{\alpha }{2}}{\sigma }_X+X\\ X-z_{\frac{\alpha }{2}}{\sigma }_X<n\lambda <z_{\frac{\alpha }{2}}{\sigma }_X+X\end{array}$

Hence, a proportion $1-\alpha$ of all possible samples is $X-z_{\frac{\alpha }{2}}{\sigma }_X<n\lambda <X+z_{\frac{\alpha }{2}}{\sigma }_X$.

b.

To determine

Show that ${\sigma }_{\widehat{\lambda }}=\frac{{\sigma }_X}{n}$.

Explanation of Solution

Given info:

$\widehat{\lambda }=\frac{X}{n}$

Calculation:

Here, n is constant, then

$\begin{array}{c}{\sigma }_{\frac{X}{n}}=\frac{{\sigma }_X}{n}\\ =\frac{\sqrt{n\lambda }}{n}\\ =\frac{\sqrt{\lambda }}{\sqrt{n}}\\ =\sqrt{\frac{\lambda }{n}}\\ {\sigma }_{\widehat{\lambda }}=\frac{{\sigma }_X}{n}\end{array}$

Thus, ${\sigma }_{\widehat{\lambda }}=\frac{{\sigma }_X}{n}$.

c.

To determine

Conclude that for a proportion $1-\alpha$ of all possible samples, $\widehat{\lambda }-z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}<\lambda <\widehat{\lambda }+z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}$.

Explanation of Solution

Calculation:

From part (a), $X-z_{\frac{\alpha }{2}}{\sigma }_X<n\lambda <X+z_{\frac{\alpha }{2}}{\sigma }_X$.

Divide the above inequality by n.

$\begin{array}{c}\frac{X-z_{\frac{\alpha }{2}}{\sigma }_X}{n}<\frac{n\lambda }{n}<\frac{X+z_{\frac{\alpha }{2}}{\sigma }_X}{n}\\ \frac{X}{n}-z_{\frac{\alpha }{2}}{\sigma }_{\frac{X}{n}}<\frac{n\lambda }{n}<\frac{X}{n}+z_{\frac{\alpha }{2}}{\sigma }_{\frac{X}{n}}\\ \widehat{\lambda }-z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}<\lambda <\widehat{\lambda }+z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}\left(\text{Since }\frac{X}{n}=\widehat{\lambda }\right)\end{array}$

Thus, a proportion $1-\alpha$ of all possible samples is $\widehat{\lambda }-z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}<\lambda <\widehat{\lambda }+z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}$.

d.

To determine

Derive the expression for the level $100\left(1-\alpha \right)\%$ confidence interval for $\lambda$.

Explanation of Solution

Given info:

${\sigma }_{\widehat{\lambda }}\approx \sqrt{\frac{\widehat{\lambda }}{n}}$

Calculation:

From part (c), $\widehat{\lambda }-z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}<\lambda <\widehat{\lambda }+z_{\frac{\alpha }{2}}{\sigma }_{\widehat{\lambda }}$.

Substitute ${\sigma }_{\widehat{\lambda }}=\sqrt{\frac{\widehat{\lambda }}{n}}$ in the above inequality.

$\widehat{\lambda }-z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{\lambda }}{n}}<\lambda <\widehat{\lambda }+z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{\lambda }}{n}}$

Thus, the expression for the level $100\left(1-\alpha \right)\%$ confidence interval for $\lambda$ is $\underset{\_}{\widehat{\lambda }\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{\lambda }}{n}}}$.

e.

To determine

Find the 95% confidence interval for $\lambda$.

Answer to Problem 19SE

The 95% confidence interval for $\lambda$ is $\underset{\_}{\left(\text{53}\text{.2104},\text{66}\text{.7896}\right)}$.

Explanation of Solution

Given info:

$n=5$ and $X=300$.

Calculation:

The value of $\widehat{\lambda }$ is,

$\begin{array}{c}\widehat{\lambda }=\frac{X}{n}\\ =\frac{300}{5}\\ =60\end{array}$

The value of ${\sigma }_{\widehat{\lambda }}$ is,

$\begin{array}{c}{\sigma }_{\widehat{\lambda }}=\sqrt{\frac{\widehat{\lambda }}{n}}\\ =\sqrt{\frac{60}{5}}\\ =3.4641\end{array}$

From Table A.2 of the Cumulative Normal distribution, the value of $z_{\frac{\alpha }{2}}$ with 95% level of confidence is 1.96.

The 95% confidence interval for $\lambda$ is,

$\begin{array}{c}\widehat{\lambda }\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{\lambda }}{n}}=60\pm 1.96\left(3.4641\right)\\ =60\pm \text{6}\text{.7896}\\ \text{=}\left(\text{53}\text{.2104},\text{66}\text{.7896}\right)\end{array}$

Thus, the 95% confidence interval for $\lambda$ is $\underset{\_}{\left(\text{53}\text{.2104},\text{66}\text{.7896}\right)}$.