Question

Refer to Exercise 10.83 and find a 90 % confidence interval for the difference between the mean numbers of acute postoperative days in the hospital with the dynamic and static systems.

Short answer

The interval is $0.00264$ to $0.01299$

Step-by-step solution

:Given information

With the dynamic and static systems, calculate a $90\%$confidence interval for the difference in the mean numbers of acute postoperative days in the hospital.

calculation

The degree of freedom is computed using the formula:

$\begin{array}{rl}& \mathrm{df}=\frac{{[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)]}^2}{\frac{{\left(\frac{s_1}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}\\ & =\frac{{[\left(\frac{{0.00514}^2}{10}\right)+\left(\frac{{0.00470}^2}{15}\right)]}^2}{\frac{{\left(\frac{{0.00614}^2}{10}\right)}^2}{10-1}+\frac{{\left(\frac{{0.00470}^2}{15}\right)}^2}{15-1}}\\ & \approx 18\end{array}$

The interval's end point is,

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\frac{a}{2}}\cdot \sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)}=(0.24246-0.01643)\pm 2.552.\sqrt{\left(\frac{{0.00514}^2}{10}+\frac{{0.00470}^2}{15}\right)}$

$=0.00264\text{to}0.01299$