91影视

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.

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 :

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

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

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

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.

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

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鈥榮 core temperature reduced during surgery.