Chapter 22, Problem 11E

(a)

To determine

To find: the standard error of the two-proportion difference.

Answer to Problem 11E

0.0354

Explanation of Solution

Given:

  $n_1=827$

  $n_2=130$

Formula used:

  $SE\left({\widehat{p}}_F-{\widehat{p}}_M\right)=\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}$

Calculation:

The proportion of dogs with lymphoma observed is

  $\begin{array}{c}{\widehat{p}}_1=\frac{473}{827}=0.5719\\ {\widehat{p}}_2=\frac{19}{130}=0.1462\end{array}$

The standard error is

  $\begin{array}{c}SE\left({\widehat{p}}_F-{\widehat{p}}_M\right)=\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}\\ =\sqrt{\frac{0.5719\left(1-0.5719\right)}{827}+\frac{0.1462\left(1-0.1462\right)}{130}}\\ =\sqrt{\frac{0.5719\left(0.4281\right)}{827}+\frac{0.1462\left(0.8538\right)}{130}}\\ =0.0354\end{array}$

The standard difference error in both proportions is 0.0354.

(b)

To determine

To construct: a confidence interval of 95 percent for this distinction.

Answer to Problem 11E

(0.3563, 0.4953)

Explanation of Solution

Given:

From part (a), $SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=0.0354$

Formula used:

  $ME=z^{\ast }\times SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)$

For the confidence interval

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm ME$

Calculation:

for a 95% confidence level, $z^{\ast }=1.96$ , therefore

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)\\ =1.96\times 0.0354\\ =0.0695\end{array}$

Therefore, the confidence interval for the 95 percent

  $\begin{array}{c}\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm ME=\left(0.5719-0.1462\right)\pm 0.0695\\ =0.4258\pm 0.0695\\ =\left(0.3563,0.4953\right)\end{array}$

The difference in proportions of dogs with lymphoma from homes where a herbicide was used from homes in which no herbicides were used is a 95 percent confidence interval (0.3563, 0.4953).

(c)

To determine

To State: a suitable conclusion.

Explanation of Solution

Based on these samples, there are 95 percent confident that the real difference in the proportions of dogs with lymphoma is between 0.3563 and 0.4953 from homes in which a herbicide has been used and from homes where no herbicides have been used.