Chapter 10: Q. 11 (page 666)

T10.11. Researchers wondered whether maintaining a patient's body temperature close to normal by heating the patient during surgery would affect wound infection rates. Patients were assigned at random to two groups: the normothermic group (patients' core temperatures were maintained at near normal, $36.5 ° C$ , with heating blankets) and the hypothermic group (patients" core temperatures were allowed to decrease to about $34.5 ° C$ ). If keeping patients warm during surgery alters the chance of infection, patients in the two groups should have hospital stays of very different lengths. Here are summary statistics on hospital stay (in number of days) for the two groups:
$\begin{array}{lccc} Group & n & \bar{x} & s_{x} \\ Normothemic & 104 & 12.1 & 4.4 \\ Hypothermic & 96 & 14.7 & 6.5 \end{array}$
(a) Construct and interpret a $95 %$ confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic patients.
(b) Does your interval in part (a) suggest that keeping patients warm during surgery affects the average length of patients' hospital staves? Justify your answer.

Short Answer

Expert verified

a) The $95 %$ confidence interval is between $(- 4.175, - 1.025)$

b)Yes

Step by step solution

Part (a) Step1: Given Information

${\bar{x}}_{1} = 12.1 {\bar{x}}_{2} = 14.7 s_{1} = 4.4 s_{2} = 6.5 n_{1} = 104 n_{2} = 96 c = 95 %$

Part (a) Step 2: Explanation

Determine the degrees of freedom:

$d f = \min (n_{1} - 1, n_{2} - 1) = \min (104 - 1, 96 - 1) = 95 > 80$

Determine $t_{c}$ with $d f = 80$ and $c = 95 %$ using table B:

$t_{c} = 1.990$

The endpoints of the confidence interval for $μ_{1} - μ_{2}$ are:

localid="1650383792637" $({\bar{x}}_{1} - {\bar{x}}_{2}) - t_{α / 2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} = (12.1 - 14.7) - 1.990 \cdot \sqrt{\frac{4 . 4^{2}}{104} + \frac{6 . 5^{2}}{96}} \approx - 4.175$

localid="1650383809654" $({\bar{x}}_{1} - {\bar{x}}_{2}) + t_{α / 2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} = (12.1 - 14.7) - 1.990 \cdot \sqrt{\frac{4 . 4^{2}}{104} + \frac{6 . 5^{2}}{96}} \approx - 1.025$

Part (b) Step 1: Given Information

Need to find interval in part (a) suggests that keeping patients warm during surgery affects the average length of patients’ hospital stays.

Part (b) Step 2: Explanation

According to the information from part (a), the $95 %$ of confidence interval is between: $(- 4.175, - 1.025)$

The confidence interval does not include $0$ , implying that keeping patients warm during surgery has an impact on the average length of stay in the hospital.