Chapter 8.4, Problem 23E

The Environmental Protection Agency and the University of Florida recently cooperated in a large study of the possible effects of trace elements in drinking water on kidney-stone disease. The accompanying table presents data on age, amount of calcium in home drinking water (measured in parts per million), and smoking activity. These data were obtained from individuals with recurrent kidney-stone problems, all of whom lived in the Carolinas and the Rocky Mountain states.

1. a Estimate the average calcium concentration in drinking water for kidney-stone patients in the Carolinas. Place a bound on the error of estimation.
2. b Estimate the difference in mean ages for kidney-stone patients in the Carolinas and in the Rockies. Place a bound on the error of estimation.
3. c Estimate and place a 2-standard-deviation bound on the difference in proportions of kidney-stone patients from the Carolinas and Rockies who were smokers at the time of the study.

To determine

Evaluate the mean calcium concentration in drinking water for kidney-stone patients in Region C.

Obtain the bound on error of estimation.

Answer to Problem 23E

The point estimate for the mean calcium concentration in drinking water for kidney-stone patients in Region C is 11.3 ppm.

The bound on the error of estimation is 1.5.

Explanation of Solution

Based on the given information, the data for calcium concentration in drinking water for kidney-stone patient in Region C is observed as given below:

$\begin{array}{l}{n}_C=467\\ \overline{y}=11.3\\ s_C=16.6\end{array}$

The critical value has to be obtained for $t_{\frac{0.05}{2},\left(n-1\right)}=t_{\frac{0.05}{2},\left(467-1\right)}$.

Step-by-step procedure to obtain the critical value using Applet:

• Go to Applets/Simulations.
• Select Student’s t Probabilities and Quantiles.
• In df, enter 466.
• In Probability, enter 0.025.

The output obtained is as follows:

From the output, it is clear that the critical value is 1.965.

Substitute the values in two sided t-interval formula as given below:

$\begin{array}{c}\left(\overline{y}\pm t_{\frac{\alpha }{2},\left(n-1\right)}\frac{s}{\sqrt{n}}\right)=\left(11.3\pm t_{\frac{0.05}{2},466}\frac{16.6}{\sqrt{467}}\right)\\ =11.3\pm 1.965\left(\frac{16.6}{\sqrt{467}}\right)\\ =\left(11.3\pm 1.5\right)\end{array}$

Therefore, the point estimate for the mean calcium concentration in drinking water for kidney-stone patients in Region C is 11.3 ppm and the bound on the error of estimation is 1.5.

To determine

Evaluate the difference in mean ages for kidney-stone patients in Region C and Region R.

Obtain the bound on error of estimation.

Answer to Problem 23E

The point estimate for the difference in mean ages for kidney-stone patients in Region C and Region R is −1.3

The bound on the error of estimation is 1.67.

Explanation of Solution

Based on the given information, the data are observed as given below:

$\begin{array}{l}n=467\\ m=191\\ \overline{x}=45.1\\ \overline{y}=46.4\\ s_x=10.2\\ s_y=9.8\end{array}$

Critical value:

The critical value is obtained as follows:

 $\begin{array}{c}{t}_{\frac{0.05}{2},\left(n+m-2\right)}=t_{\frac{0.05}{2},\left(467+191-2\right)}\\ =t_{\frac{0.05}{2},\left(467+191-2\right)}\\ =t_{0.025,656}\end{array}$.

Step-by-step procedure to obtain the critical value using Applet:

• Go to Applets/Simulations.
• Select Student’s t Probabilities and Quantiles.
• In df, enter 656.
• In Probability, enter 0.025.

The output obtained is as follows:

From the output, it is clear that the critical value is 1.964.

Substitute the values in two sided t-interval formula as given below:

$\begin{array}{c}\left(\left(\overline{x}-\overline{y}\right)\pm t_{\frac{\alpha }{2},\left(n+m-2\right)}\left(\sqrt{\frac{s_x}{n}+\frac{s_y}{m}}\right)\right)=\left(\left(45.1-46.4\right)\pm t_{0.025,656}\left(\sqrt{\frac{10.2}{467}+\frac{9.8}{191}}\right)\right)\\ =\left(\left(45.1-46.4\right)\pm t_{0.025,656}\left(\sqrt{\frac{10.2}{467}+\frac{9.8}{191}}\right)\right)\\ =-1.3\pm 1.964\left(0.8518\right)\\ =-1.3\pm 1.67\end{array}$

Therefore, the point estimate for the difference in mean ages for kidney-stone patients in Region C and Region R is −1.3 and the bound on the error of estimation is 1.67.

To determine

Evaluate and place 2-standard deviations bound on the difference in proportions of kidney-stone patients from Regions C and R.

Obtain the bound on error of estimation.

Answer to Problem 23E

The point estimate for the difference in proportions of kidney-stone patients from Regions C and R is 0.17, that is, (−2.97, 0.37).

The 2-standard deviations bound on the error of estimation is 0.0079.

Explanation of Solution

Based on the given information, the data are observed as given below:

$\begin{array}{l}{n}_1=467\\ n_2=191\\ p_1=0.78\\ p_2=0.61\end{array}$

The formula for difference in proportion is as follows:

$\begin{array}{l}\left(p_1-p_2\right)\pm 2\sqrt{\widehat{P}\widehat{Q}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ \text{where, }\widehat{P}=\frac{p_1+p_2}{n_1+n_2}\\ \widehat{Q}=1-\widehat{P}\end{array}$

Substitute the values as given below:

$\begin{array}{c}\widehat{P}=\frac{p_1+p_2}{n_1+n_2}\\ =\frac{0.78+0.61}{467+191}\\ =0.0021\\ \widehat{Q}=1-0.0021\\ =0.9979\\ \left(p_1-p_2\right)\pm 2\sqrt{\widehat{P}\widehat{Q}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}=\left(0.78-0.61\right)\pm 2\sqrt{\left(0.0021\right)\left(0.9979\right)\left(\frac{1}{467}+\frac{1}{191}\right)}\\ =0.17\pm 0.0079\end{array}$

The point estimate for the difference in proportions of kidney-stone patients from Regions C and R is 0.17, and the 2-standard deviations bound on the error of estimation is 0.0079.