Question

The interarrival time of customers for 60 min was monitored and reported below:

0.56974538 6.25245930 4.38116270 2.73018725 0.01947573 2.05621644

0.95476822 0.20938206 0.36555334 13.31611807 1.84342165 3.85799384

0.78148032 0.06480221 1.01586982 1.54638918 1.65045201 8.62226018

1.41718532 3.75108604

What is p-value of the KS test testing the hypothesis that the interarrival times are exponentially distributed?

Answer 1

In order to solve this question I used R software.

R codes and output:

> y=scan('clipboard')
Read 20 items
> y
[1] 0.56974538 6.25245930 4.38116270 2.73018725 0.01947573 2.05621644
[7] 0.95476822 0.20938206 0.36555334 13.31611807 1.84342165 3.85799384
[13] 0.78148032 0.06480221 1.01586982 1.54638918 1.65045201 8.62226018
[19] 1.41718532 3.75108604
> mean(y)
[1] 2.7703
> ks.test(y,pexp(1/2.7703))

Two-sample Kolmogorov-Smirnov test

data: y and pexp(1/2.7703)
D = 0.85, p-value = 0.381
alternative hypothesis: two-sided

Hypothesis:

H₀ : Exponential distribution fits the given data.

H₁ : Exponential distribution does not fit the given data.

Test statistic, D = 0.85

P-value = 0.381

Since p-value is greater than 0.05, we accept null hypothesis. And conclude that exponential distribution fit the given data.