Chapter PVI, Problem 25RE

(a)

To determine

To write a $95\%$ confidence interval and interpret it in context.

Answer to Problem 25RE

We are $95\%$ confident that between $19.77\%\text{ and 38}\text{.66\%}$ of children with bipolar symptoms will be helped with medication and psychotherapy.

Explanation of Solution

It is given in the question that the June American Journal of Psychiatry reported that researchers use medication and psychotherapy to treat children aged $\text{7 to 16}$ who exhibit bipolar symptoms. Thus, to find the confidence interval for proportion we will use the calculator $\text{TI}-89$ . To calculate a confidence interval for a population proportion: Go to the Intsmenu and select $5:1-$ PropZInt. Enter the number of successes observed and the sample size. Specify a confidence level. Calculate the interval. So, we have,

  $\begin{array}{l}x=26\\ n=89\\ \alpha =0.05\end{array}$

Thus, by using the calculator $\text{TI}-89$ , the confidence interval will be as:

  $(0.1977,0.3866)$ .

So, we are $95\%$ confident that between $19.77\%\text{ and 38}\text{.66\%}$ of children with bipolar symptoms will be helped with medication and psychotherapy, based on this study.

(b)

To determine

To find out how many patients should they study.

Answer to Problem 25RE

They should study $221$ patients.

Explanation of Solution

It is given in the question that the June American Journal of Psychiatry reported that researchers use medication and psychotherapy to treat children aged $\text{7 to 16}$ who exhibit bipolar symptoms. Thus, the margin of error given is $6\%=0.06$ . So, the patients they should studyis calculated as:

  $\begin{array}{l}ME=z*\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ \Rightarrow 0.06=1.96\times \sqrt{\frac{\frac{26}{89}(1-\frac{26}{89})}{n}}\\ \Rightarrow \frac{0.06}{1.96}=\sqrt{\frac{\frac{26}{89}(1-\frac{26}{89})}{n}}\\ $ \Rightarrow {(\frac{0.06}{1.96})}^2=\frac{\frac{26}{89}(1-\frac{26}{89})}{n}$\Rightarrow n=220.6\\ \approx 221\end{array}$

Thus, they should study $221$ patients.