Biostats SPSS Kaplan-Meier Curve

Part3

Step-by-Step Guide to Assignment 8.3

Problem 3. Kaplan-Meier Survival Analysis (With Factor)

a. Run a Kaplan-Meier analysis in SPSS, using Time as the Time variable and Event as the Status variable (Be sure to define the event). Add Tx as the factor.

Step 1. Open the SPSS dataset. Go to Analyze (Survival ( Kaplan-Meier. The information from 8.2 should be saved. Select Treatment Status [tx] and place it in the Factor box.

(Note: Time to event/censor [time] should be in the Time box and event(“0” “1”) should be in the Status box from Problem 8.2; if not, you need to repeat steps to place them there and define Event again).

b. Test the difference between Tx groups (Compare Factor) using Log-Rank, Breslow, and Tarone-Ware

Step 2. Click on the Compare Factor button.

Step 3. In the open window, check Log-Rank, Breslow, and Tarone-Ware.. Click Continue.

Step 4. Click the Options button. (Make sure Survival table(s) and Mean and median survival are checked in Statistics and that Survival is checked in Plots.)

Click Continue.

Step 5. Click OK in the Kaplan-Meier window.

SPSS output:

c. Produce a plot of the survival function

d. Describe what the Overall Comparisons mean in terms of treatment groups and survival times.

Overall Comparisons
	Chi-Square	df	Sig.
Log Rank (Mantel-Cox)	4.308	1	.038
Breslow (Generalized Wilcoxon)	.926	1	.336
Tarone-Ware	2.074	1	.150
Test of equality of survival distributions for the different levels of Treatment Status.

Means and Medians for Survival Time
Treatment Status	Meana	Median
	Estimate	Std. Error	95% Confidence Interval	Estimate	Std. Error	95% Confidence Interval
			Lower Bound	Upper Bound			Lower Bound	Upper Bound
Placebo	12.422	1.155	10.157	14.686	8.000	1.007	6.027	9.973
Tx Thiotepa	16.278	1.839	12.674	19.881	11.000	2.121	6.842	15.158
Overall	14.017	1.026	12.007	16.027	9.000	1.097	6.849	11.151
a. Estimation is limited to the largest survival time if it is censored.

Notes:

Recall that the log-rank test is based on a 2 x 2 table looking at the number of deaths and expected number of deaths for each group. The log-rank chi square statistic is 4.31. For 1 df, the chi square statistic must be greater than the 3.84 (table p. 466 in the text) to reject the null hypothesis of no difference between the two treatment groups (placebo and Thiotepa) at the 95% probability level. Since 4.31 > 3.84, we can reject the null. Additionally, since the significant value of the Log Rank test is less than 0.05 and the Breslow and Tarone-Ware tests are greater than 0.05, there is a difference in survival times between the two groups. This is confirmed by the log-rank p-value = 0.038, which is p < 0.05.