Question

Researchers wondered whether maintaining a patient’s body temperature close to normal by heating the patient during surgery would affect rates of infection of wounds. Patients were assigned at random to two groups: the normothermic group (core temperatures were maintained at near normal, $36.5°\mathrm{C}$, using heating blankets) and the hypothermic group (core temperatures were allowed to decrease to about $34.5°\mathrm{C}$). If keeping patients warm during surgery alters the chance of infection, patients in the two groups should show a difference in the average length of their hospital stays. Here are summary statistics on hospital stay (in number of days) for the two groups:

a. Construct and interpret a $95\%$confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients like these.

b. Does your interval in part (a) suggest that keeping patients warm during surgery affects the average length of patients’ hospital stays? Justify your answer.

c. Interpret the meaning of “$95\%$confidence” in the context of this study.

Short answer

Part(a) $95\%$confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients like these is $(-4.175,-1.025)$

Part(b) Yes, keeping patients warm during surgery affects the average length of patients’ hospital stays

Part(c) "$95\%$confidence" means $95\%$ of the intervals would grasp the true difference in mean hospital stay.

Step-by-step solution

Part(a) Step 1 : Given information

We need to interpret a confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients like these.

Group	n	${\mathrm{x}}^{-\frac{305}{1526}=0.200=20.0\%}\overline{\mathrm{x}}$	$S\times \frac{305}{1526}=0.200=20.0\%S_x$
Normothermic	$104$	$12.1$	$4.4$
Hypothermic	$96$	$14.7$	$6.5$

Part(a) Step 2 : Simplify

As given :

$$\begin{gathered} \overline{{\mathrm{x}}_1}=12.1 \\ \overline{{\mathrm{x}}_2}=14.7 \\ {\mathrm{s}}_1=4.4 \\ {\mathrm{s}}_2=6.5 \\ {\mathrm{n}}_1=104 \\ {\mathrm{n}}_2=96 \\ \mathrm{c}=95\% \end{gathered}$$

Now, degree of freedom :

$$\begin{gathered} \mathrm{df}=\min ({\mathrm{n}}_1-1,{\mathrm{n}}_2-1) \\ =\min (104-1,96-1) \\ =95>80 \end{gathered}$$

Now, $t_c=1.990$using table B.

The endpoints of confidence interval are :

$$\begin{gathered} (\overline{x_1}-\overline{x_2})-t_{\frac{\alpha }{2}}\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(12.1-14.7)-1.990\times \sqrt{\frac{4.4^2}{104}+\frac{6.5^2}{96}} \\ =-4.175 \\ (\overline{x_1}-\overline{x_2})+t_{\frac{\alpha }{2}}\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(12.1-14.7)+1.990\times \sqrt{\frac{4.4^2}{104}+\frac{6.5^2}{96}} \\ =-1.025 \end{gathered}$$

Therefore, $95\%$ confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients like these.

Part(b) Step 1 : Given information

We need to find that keeping patients warm during surgery affects the average length of patients’ hospital stays.

Part(b) Step 2 : Simplify

Yes, The confidence interval $(-4.175,-1.025)$in component (a) does not contain zero, implying that keeping patients warm during surgery influences the average length of their hospital stay.

Part(c) Step 1 : Given information

We need to interpret meaning of " $95\%$confidence".

Part(c) Step 2 : Simplify

In the context of this study, "$95$ percent confidence" means that if we repeated the experiment many times, about $95$ percent of the intervals would grasp the true difference in mean hospital stay between patients who are receiving heating blankets during surgery and patients who have one‘s core temperature reduced during surgery.