### Introduction

*Korean Journal of Anesthesiology*has thus far published several papers using survival analysis for clinical outcomes: a comparison of 90-day survival rate for pressure ulcer development or non-development after surgery under general anesthesia [3], a comparison of postoperative 5-year recurrence rate in patients with breast cancer depending on the type of anesthetic agent used [4], and a comparison of airway intubation success rate and intubation time between two types of video laryngoscopy in difficult intubation cases [5]. It is not hard to consider the application of survival analysis for the data of first two articles, using survival analysis for the intubation time comparison in the third paper is not easy to think. For such topics, the main obstacle to the widespread use of survival analysis is the word “survival,” which leads to the misunderstanding that can only be used for data related to death or failure. However, as is the case with the third paper, survival analysis can be applied not only to data of patients’ death, but also to any data related to an “event of interest” that may or may not occur during the observation period.

^{1)}The following are several examples of questions for which survival analysis may be applied:

*t*-test or the Mann-Whitney

*U*test may be used. However, even in such cases, non-occurrence or termination of participation (due to improvement or exacerbation, moving, death, etc.) often cannot be ruled out. Also, the observation period may vary due to a limited study period or different outcome time frames. For these reasons, some of data are subject to incompleteness, and these data should be analyzed using another statistical method or excluded. Survival analysis is necessary to analyze incomplete data by setting the time-to-event as a primary outcome. In summary, survival analysis can be applied in a range of situations in which any event of interest (not just mortality) is analyzed in terms of its occurrence or non-occurrence during a specified observation period [1].

### Censored Data

^{2)}and are excluded from analysis as missing values in other statistical analysis methods; in survival analysis, however, they are processed as important data influencing the outcome [6,7].

### Survival Function, Hazard Function

^{3)}t = 0 corresponds to a probability of 1.0 (i.e., 100% survival at the onset), and the point in time with 50% survival probability is the median survival time.

*t*). First, the incidence rate for the period between a specific time t and the next measurement time

*t*+ α can be calculated by dividing the number of events occurring between t and

*t*+ α by the total number of observations at time t. By α approaches 0, i.e., by taking the limit as the interval between

*t*and

*t*+ α closes to 0, the instantaneous incidence rate at t, which constitutes the hazard, can be calculated.

^{4)}The hazard function is a function for calculating the instantaneous incidence rate at any given point in time, and is denoted by h (

*t*) [9].

### Survival Analysis Using Kaplan-Meier Curves (Estimates)

### Multiple comparison method for Kaplan-Meier survival analysis

^{5)}The null hypothesis tested with the log-rank method is that “there is no overall difference between the two survival curves to be compared.” The observed count and the expected count are compared using the chi-square test. The log-rank method is known to have good statistical power when there are considerable differences in the incidence rate among the groups being compared. If three or more groups are compared, pairwise multiple comparisons can also be performed using post-hoc tests such as the Bonferroni test and Holm-Sidak method [12].

^{6)}For example, the incidence rate of postoperative nausea/vomiting is highest immediately after surgery and decreases with time, regardless of whether the type of general anesthesia is inhalational or intravenous. That is, the risk of postoperative nausea/vomiting varies (decreases) with time independently of the type of anesthesia. If the proportional hazards assumption is valid, then a finding that the risk of postoperative nausea/vomiting in the patient group receiving an inhalational anesthetic agent is twice that of the patient group receiving an intravenous anesthetic means that there is a twofold difference in hazard at all points in time. With the hazard ratio greater than 1, there is a greater risk of postoperative nausea/vomiting in the inhalational anesthesia group than in the intravenous anesthesia group.

^{7)}As illustrated by this example, the hazard ratio offers information that may allow understanding hazard information at a glance [13] (Equation 2).

### An example of Kaplan-Meier survival analysis

^{2}(1) = 5.597, P = 0.018, Cohen’s Φ [phi] = 0.23). The relative risk for Group A compared to Group B is 0.7 (95% CI: 0.5–0.9). Similar results are yielded by incidence rate estimation, although it is an overlapping method, in which the nausea/vomiting symptoms manifested during the first 24 postoperative hours account for 49.0% (95% CI: 35.3–62.7%) of patients in Group A and 71.7% (95% CI: 59.6–83.8%) in Group B, showing a significant intergroup difference (z-test, P = 0.018, Cohen’s

*h*= 0.47). These statistics reflect the results including all censored data without manifestation of nausea/vomiting during the first 24 hours. Given that nausea/vomiting symptoms are potentially observable among the patients with censored data during the study period (24 hours), the accuracy of the above results may not be high. Excluding all censored cases without nausea/vomiting before the first 24 postoperative hours, the number of patients included in the analysis is 73, and the incidence rate recalculated accordingly is 89.3% (95% CI: 77.8–100.0%) for Group A and 84.4% (95% CI: 73.9–95.0%) for Group B. These increased incidence rates reflect the omission of patients who dropped out of the study before completing it (24-hour observation) without manifesting symptoms of nausea/vomiting. Regardless of the researcher’s intention, the actual incidence rates are overestimated due to the omission of patients without symptom manifestation.

^{2}(1) = 6.802, P = 0.009). Examination of the Kaplan-Meier curves for the two groups shows that the cumulative survival rate of Group B rapidly decreases with time compared with that of Group A (Fig. 4).

### Cox Proportional Hazards Regression Model

^{−λt}). The null hypothesis for comparison of two survival curves is that “the hazard ratio for the two groups is 1.”

^{8)}

### An example of regression analysis with the Cox proportional hazards regression model

### Sample Size for Survival Analysis

^{9)}In survival analysis based on the proportional hazards assumption, the hazard ratio (presumed to have a constant value throughout the observation period) is used as the effect size [23].

^{10)}and setting the hazard ratios of the two groups to be compared. For example, if the incidence rate for the experimental group is 20% lower than that of the control group, the hazard ratio is 0.8 (alternative hypothesis). The proportional hazards assumption presupposes the maintenance of the hazard ratio at a constant value throughout the observation period, and the null hypothesis is that “there is no difference in the risk of the two groups for experiencing the event,” i.e., hazard ratio = 1 [24].

^{11)}If the literature does not provide an incidence rate including the time factor, a pilot test should be designed for incidence rate estimation. An incidence rate estimate including the time factor is the event count per uniform time segment, such as the person-day or person-year, and represents the hazard ratio in the time segment concerned. Using the hazard ratio, it is possible to estimate the survival function based on the proportional hazards assumption, as well as the survival functions of the groups to be compared, using the hazard ratio set in the hypothesis formulation stage.

_{A}(4) = exp (−0.1 × 4) = 0.018. Since new drug B decreases recurrence by 30%, the hazard ratio is 0.7, and the value of the 4-week survival function for new drug B is S

_{B}(4) = exp ((−0.1 × 0.7) × 4) = 0.061. If both groups have the same sample size, π

_{1}= π

_{2}= 0.5, the probability of an event, which is the denominator of the sample size calculation formula, is 1 − (π

_{1}S1 (t) + π

_{2}S2 (

*t*)) = 1 − (0.5 × 0.018 + 0.5 × 0.061) = 0.960. The total event count, which is the numerator of the sample size calculation formula, can be obtained from Equation 3. z

_{α/2}and z

_{β}, which represent the values of probability in a standard normal distribution, are 1.96 and 0.842, respectively, for a significance level of 0.05 and statistical power of 80%. With the values of π

_{1}and π

_{2}set to 0.5 each and the hazard ratio set at 0.7, the total event count required is (1.96 + 0.842)

^{2}/{0.5 × 0.5 × (log0.7)

^{2}} = 246.9, i.e., 247 events. Substituting this value and the incidence rate into Equation 4, 247/0.960 = 257.3, i.e., 258 is obtained. Applying the generally assumed withdrawal rate of 10% to the value obtained, 258/(1 − 0.1) = 286.7, i.e., a total of 287 subjects, is set as the required sample size. With the group size ratio set at 0.5, 144 subjects are to be assigned to each group.

^{12)}