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How to Perform a Log Rank Test in R, The most frequent technique to compare survival curves between two groups is to use a log-rank test.

The following hypotheses are used in this test

H0: There is no difference in survival between the two groups.

HA: There is a difference in survival between the two groups.

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We can reject the null hypothesis and conclude that there is enough evidence to claim there is a difference in survival between the two groups if the p-value of the test is less than some significance level (e.g. =0.05).

In R, we may use the survdiff() function from the survival package to do a log-rank test, which has the following syntax.

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## How to Perform a Log Rank Test in R

`survdiff(Surv(time, status) ~ predictors, data)`

The Chi-Squared test statistic and related p-value are returned by this function.

This function is used to execute a log-rank test in R, as seen in the example below.

### Example: Log Rank Test in R

We’ll use the ovarian dataset from the survival package for this example. The following information about 26 patients may be found in this dataset:

After being diagnosed with ovarian cancer, how long do you live (in months)?

Whether or whether the time spent surviving was censored

(rx = 1 or rx = 2) Type of treatment received

The following code demonstrates how to inspect the dataset’s first six rows.

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`library(survival)`

Let’s view the first six rows of the dataset

```head(ovarian)
futime fustat     age resid.ds rx ecog.ps
1     59      1 72.3315        2  1       1
2    115      1 74.4932        2  1       1
3    156      1 66.4658        2  1       2
4    421      0 53.3644        2  2       1
5    431      1 50.3397        2  1       1
6    448      0 56.4301        1  1       2```

The following code demonstrates how to use a log-rank test to see if patients who got various treatments had different survival rates.

make a log-rank test

```survdiff(Surv(futime, fustat) ~ rx, data=ovarian)
Call:
survdiff(formula = Surv(futime, fustat) ~ rx, data = ovarian)
N Observed Expected (O-E)^2/E (O-E)^2/V
rx=1 13        7     5.23     0.596      1.06
rx=2 13        5     6.77     0.461      1.06
Chisq= 1.1  on 1 degrees of freedom, p= 0.3```

With one degree of freedom, the Chi-Squared test statistic is 1.1, and the associated p-value is 0.3. We cannot reject the null hypothesis because the p-value is not smaller than 0.05.

To put it another way, we don’t have enough evidence to establish that the two treatments have a statistically significant difference in survival.

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The survival curves for each group can alternatively be plotted using the following syntax.

For each therapy group, draw a survival curve.

```plot(survfit(Surv(futime, fustat) ~ rx, data = ovarian),
xlab = "Time",
ylab = "Survival probability")```

In R, a plot of survival curves. Although the survival curves change somewhat, the difference is not statistically significant, according to the log-rank test.

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