Chapter 22, Problem 12E

(a)

To determine

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

Answer to Problem 12E

0.0682

Explanation of Solution

Given:

  $n_1=88$

  $n_2=88$

  ${\widehat{p}}_1=0.80$

  ${\widehat{p}}_2=0.48$

Formula used:

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

Calculation:

The standard error is

  $\begin{array}{c}SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_1{\widehat{q}}_1}{n_1}+\frac{{\widehat{p}}_2{\widehat{q}}_2}{n_2}}\\ =\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}\\ =\sqrt{\frac{0.80\left(1-0.80\right)}{88}+\frac{0.48\left(1-0.48\right)}{88}}\\ =0.0682\end{array}$

Therefore, the standard error in sample proportions is 0.0682

(b)

To determine

To construct: a 95 percent confidence interval.

Answer to Problem 12E

(0.1863, 0.4537) or (18.6%, 45.4%)

Explanation of Solution

Given:

From part (a) is $SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=0.0682$ .

Formula used:

For 95 percent confidence interval

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

Calculation:

For 95 percent confidence interval

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

the critical value obtained is $z^{\ast }=1.96$ .

Therefore,

  $\begin{array}{c}CI=\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm z^{\ast }SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)\\ =\left(0.80-0.48\right)\pm 1.96\times 0.0682\\ =0.32\pm 0.1337\\ =\left(0.1863,0.4537\right)\end{array}$

Therefore, the 95 percent confidence interval for the difference in proportions occurred is (0.1863, 0.4537) or (18.6%, 45.4%)

(c)

To determine

To Explain: a suitable conclusion.

Explanation of Solution

The acquired outcome is as follows: After surgery, the 95 percent confidence interval for the difference in proportions of patients progressed and patients progressed after using wrist splints (18.6,45.4). So, among the recorded patients, 95 percent are positive. The number of patients who have surgery is between 18.6 percent and 45.4 percentage points higher than the percentage of patients who have wrist splints. And it is more likely that the surgery will be successful than wrist splints.