Chapter 22, Problem 27E

(a)

To determine

To explain: the 95 percent confidence interval for the percent of individuals taking medicine A that can get relief from this form of joint pain.

Answer to Problem 27E

(0.6698, 0.8302)

Explanation of Solution

Given:

Of the 112 individuals who received treatment A in the community, 84 said that this pain reliever was successful. With $n=112$

Formula used:

  $SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}$

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

For the confidence interval

  $\widehat{p}\pm ME$

Calculation:

  $\widehat{p}=\frac{84}{112}=0.75$

The standard error is

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ =\sqrt{\frac{0.75\left(1-0.75\right)}{112}}\\ =0.0409\end{array}$

For a 95% confidence level, $z^{\ast }=1.96$ , so

  $\begin{array}{c}ME=z^{\ast }\times SE\left(\widehat{p}\right)\\ =1.96\times 0.0409\\ =0.0802\end{array}$

therefore, the 95 percent confidence interval is

  $\begin{array}{c}\widehat{p}\pm ME=0.75\pm 0.0802\\ =\left(0.6698,0.8302\right)\end{array}$

If the observed proportion is 75 percent, then there are 95 percent sure that treatment A will be successful for between 67 percent and 83 percent of the population proportion of individuals who have complained of a certain form of joint pain.

(b)

To determine

To explain: the 95 percent confidence interval for the percent of individuals taking medicine B that can get relief from this form of joint pain.

Answer to Problem 27E

(0.5192, 0.7031)

Explanation of Solution

Given:

Of the 108 individuals in the group receiving treatment B, 66 indicated that this pain relief was successful with $n=108$

Formula used:

  $SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}$

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

For the confidence interval

  $\widehat{p}\pm ME$

Calculation:

Standard error is

  $\begin{array}{c}SE\left(\widehat{p}\right)=\sqrt{\frac{\widehat{p}\widehat{q}}{n}}\\ =\sqrt{\frac{0.61\left(1-0.61\right)}{108}}\\ =0.0469\end{array}$

For a 95% confidence level, $z^{\ast }=1.96$ , so

  $\begin{array}{c}ME=z^{\ast }\times SE\left(\widehat{p}\right)\\ =1.96\times 0.0469\\ =0.0919\end{array}$

Therefore, the 95 percent confidence interval is

  $\begin{array}{c}\widehat{p}\pm ME=0.61\pm 0.0919\\ =\left(0.5192,0.7031\right)\end{array}$

If the observed proportion is 61 percent, then there are 95 percent sure that treatment B will be successful between 51.9 percent and 70.3 percent of the population proportion of individuals who have complained of a certain type of joint pain.

(c)

To determine

To Explain: about the overlap of interval A and B and about the comparative effectiveness of this medications.

Explanation of Solution

Yes, the A and B intervals overlap. This may mean that the proportion of people who have complained of a certain type of joint pain in the population would find treatment A successful and medication B successful.

(d)

To determine

To explain: the 95 % confidence interval for the proportion difference of individuals who will find this drug effective.

Answer to Problem 27E

(0.0169, 0.2609)

Explanation of Solution

Given:

  $\begin{array}{l}{n}_A=112\\ y_A=84\\ n_B=108\\ y_B=66\end{array}$

Formula used:

  $\begin{array}{c}SE\left({\widehat{p}}_A-{\widehat{p}}_B\right)=\sqrt{\frac{p_A{q}_A}{n_A}+\frac{p_B{q}_B}{n_B}}\\ ME=z^{\ast }\times SE\left({\widehat{p}}_A-{\widehat{p}}_B\right)\end{array}$

For the confidence interval

  $\left({\widehat{p}}_A-{\widehat{p}}_B\right)\pm ME$

Calculation:

Observed proportions are

  $\begin{array}{c}{\widehat{p}}_A=\frac{84}{112}=0.75\\ {\widehat{p}}_B=\frac{66}{108}=0.61\end{array}$

Estimating the standard deviation

  $\begin{array}{c}SE\left({\widehat{p}}_A-{\widehat{p}}_B\right)=\sqrt{\frac{p_A{q}_A}{n_A}+\frac{p_B{q}_B}{n_B}}\\ =\sqrt{\frac{0.75\left(1-0.75\right)}{112}+\frac{0.61\left(1-0.61\right)}{108}}\\ =\sqrt{\frac{0.75\left(0.25\right)}{112}+\frac{0.61\left(0.39\right)}{108}}\\ =0.0622\end{array}$

For a 95% confidence level, $z^{\ast }=1.96$ , so

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_A-{\widehat{p}}_B\right)\\ =1.96\times 0.0622\\ =0.1220\end{array}$

So, the 95% confidence interval is

  $\begin{array}{c}\left({\widehat{p}}_A-{\widehat{p}}_B\right)\pm ME=\left(0.75-0.61\right)\pm 0.1220\\ =0.14\pm 0.1220\\ =\left(0.0169,0.2609\right)\end{array}$

95 percent confidence interval for the proportion difference of people who received treatment A and said that this pain reliever was beneficial and that this pain reliever was beneficial for people who received treatment B and said it was effective (0.0169. 0.2609). This indicates that 1.7% to 26% more users explore that treatment A is safe.

(e)

To determine

To Explain: about the including the zero in the interval and the meaning of this.

Explanation of Solution

No, 0.0 is not included in this interval. This means 0 is not the difference 's probable value. it might assume that the efficacy of treatment A and treatment B for joint pain differ.

(f)

To determine

To Explain: the reason that the results in part (c) and (e) looks contradictory, if want to compare the effectiveness of these two pain relievers and the find the correct approach.

Explanation of Solution

There is required to need to estimate the standard deviation of the difference if there is required to estimate the difference in the proportions of individuals that will find these treatments successful. The difference cannot be explained by two separate confidence intervals. If there is required to evaluate the efficacy of these two pain relievers, the two-sample method is the right strategy.