Chapter 10, Problem 11PT

(a)

To determine

To construct: and interpret a 95% confidence interval for the difference in the true mean length.

Answer to Problem 11PT

(-4.175, -1.025)

Explanation of Solution

Given:

Group	$n$	$\overline{X}$	$S_x$
Normothermic	104	12.1	4.4
Hypothermic	96	14.7	6.5

C=95%

Formula used:

For lower $\left(x_1-x_2\right)-t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}$

For upper $\left(x_1-x_2\right)+t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}$

Calculation:

Finding the degree of freedom

  $\begin{array}{c}df=\min (n_1-1,n_2-1)\\ =\min (104-1,96-1)\\ =95>80\end{array}$

  $t_c$ with df = 80 and c=95% using the table

  $t_c$ =1.990

Endpoints of the ${\mu }_1-{\mu }_2$ confidence interval

  $\begin{array}{c}\left(x_1-x_2\right)-t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}=\left(12.1-14.7\right)-1.990\times \sqrt{\frac{{4.4}^2}{104}+\frac{{6.5}^2}{96}}\\ =-4.175\end{array}$

  $\begin{array}{c}\left(x_1-x_2\right)+t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}=\left(12.1-14.7\right)-1.990\times \sqrt{\frac{{4.4}^2}{104}+\frac{{6.5}^2}{96}}\\ =-1.025\end{array}$

There are 95% confidence that the mean population difference is between -4.175 and -1.025

(b)

To determine

To find: that in part (a) suggest that keeping patients warm during surgery affects the average length of patients’ hospital stays yes or not.

Answer to Problem 11PT

Yes

Explanation of Solution

Given:

Group	$n$	$\overline{X}$	$S_x$
Normothermic	104	12.1	4.4
Hypothermic	96	14.7	6.5

From the result of part (a), it is observed that the confidence interval is not having 0 and therefore suggests that having patients warm during surgery affects the average length of the patients’ hospital stay.