Chapter 22, Problem 21E

(a)

To determine

To Explain: that this is an experiment or not.

Answer to Problem 21E

No

Explanation of Solution

This is not an experiment, as subjects have not been arbitrary assigned to treatment groups. This is an observational study

(b)

To determine

To conduct: a suitable hypothesis and give conclusion.

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=157\\ y_1=42\\ n_2=89\\ y_2=7\end{array}$

Formula used:

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\end{array}$

Calculation:

hypotheses would be

  $\begin{array}{l}{H}_0:p_1-p_2=0\\ H_1:p_1-p_2\ne 0\end{array}$

Normal model conducts a two-proportion z-test

  $\begin{array}{l}{\widehat{p}}_1=\frac{42}{157}=0.2675\\ {\widehat{p}}_2=\frac{7}{89}=0.0787\end{array}$

Pool the sample data

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_1+y_2}{n_1+n_2}\\ =\frac{42+7}{157+89}\\ =\frac{49}{246}\\ =0.1992\end{array}$

SE to estimate $SD\left(P_1-P_2\right)$ .thus,

  $\begin{array}{c}S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_1}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_2}}\\ =\sqrt{\frac{0.1992\left(1-0.1992\right)}{157}+\frac{0.1992\left(1-0.1992\right)}{89}}\\ =\sqrt{\frac{0.1992\left(0.8008\right)}{157}+\frac{0.1992\left(0.8008\right)}{89}}\\ =0.0530\end{array}$

The observed proportions is

  $\begin{array}{c}{\widehat{p}}_1-{\widehat{p}}_2=0.2675-0.0787\\ =0.1889\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-0}{S{E}_{Pooled}\left({\widehat{p}}_1-{\widehat{p}}_2\right)}\\ =\frac{0.1889-0}{0.0530}\\ =3.56\end{array}$

The P-value is

  $\begin{array}{c}\text{P-value =2}P\left(z>3.56\right)\\ =2\left(0.0002\right)\\ =0.0004\end{array}$

P-value = 0.0004 is very small, we do not consider the null hypothesis, we do conclude that the strategies of clinics are not efficient for older women.

(c)

To determine

To find: the difference between a confidence interval and an interpretation interval, if there was a difference when concluded.

Answer to Problem 21E

(0.0999, 0.2779)

Explanation of Solution

Formula used:

  $\begin{array}{c}SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_1{\widehat{q}}_1}{n_1}+\frac{{\widehat{p}}_2{\widehat{q}}_2}{n_2}}\\ ME=z^{\ast }\times SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)\end{array}$

For the confidence interval

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

Calculation:

  $\begin{array}{c}SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)=\sqrt{\frac{{\widehat{p}}_1{\widehat{q}}_1}{n_1}+\frac{{\widehat{p}}_2{\widehat{q}}_2}{n_2}}\\ =\sqrt{\frac{0.2675\left(1-0.2675\right)}{157}+\frac{0.0787\left(1-0.0787\right)}{89}}\\ =\sqrt{\frac{0.2675\left(0.7325\right)}{157}+\frac{0.0787\left(0.9213\right)}{89}}\\ =0.0454\end{array}$

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

  $\begin{array}{c}ME=z^{\ast }\times SE\left({\widehat{p}}_1-{\widehat{p}}_2\right)\\ =1.96\times 0.0454\\ =0.0890\end{array}$

So, the 95% confidence interval is

  $\begin{array}{c}\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm ME=\left(0.2675-0.0787\right)\pm 0.0890\\ =0.1889\pm 0.0890\\ =\left(0.0999,0.2779\right)\end{array}$

A 95 % confidence Interval for the difference Between me, women under the age of 38 and women age 38 and older are living births (0.0999, 0.2779). On the basis of these samples, we are 95 % sure that live births for women under the age of 38 are 10% to 27.8% higher than live births for women 38 and older. The success / failure condition is not fulfilled, because success in women aged 38 and older is 7, which is less than 10. We cannot depend on this confidence Interval.