Question

Time between failures (in months) of a particular bearing is assumed to follow normal distribution. The data collected over 50 failures are 

11.394 10.728 6.680 8.050 8.382 

8.740 8.287 7.979 5.857 13.521 

12.000 9.496 9.248 6.529 12.137 

11.383 8.135 11.752 10.040 8.615 

8.686 6.416 9.987 11.282 4.732 

9.344 7.019 6.735 12.176 4.247 

10.099 6.254 5.557 9.376 5.780 

7.129 7.835 9.648 4.381 5.801 

8.334 9.454 8.486 7.256 10.963 

10.544 10.433 10.425 10.078 7.709 

Using Kolmogorov-Smirnov test, check whether the distribution follows normal.

Answer 1

Let the observations in a random sample of size n from the distribution of the random variable x be arranged in icreasing order of magnitude and let x_(i) be the ith value in this arrangement. Now suppose F_n is defined by

Thus nF_n(x) is nothing but the number of sample observations that are less than or equal to x.

Define

The computational part is done by R.

x=c(11.394,8.740,12.000,11.383,8.686,9.344,10.099,7.129,8.334,10.544,10.728,8.287,9.496,
8.135,6.416,7.019,6.254,7.835,9.454,10.433,6.680,7.979,9.248,11.752,9.987,6.735,5.557,
9.648,8.486,10.425,8.050,5.857,6.529,10.040,11.282,12.176,9.376,4.381,7.256,10.078,
8.382,13.521,12.137,8.615,4.732,4.247,5.780,5.801,10.963,7.709)
ks.test(x,"pnorm", mean(x), sd(x))

Output:

One-sample Kolmogorov-Smirnov test

data: x
D_n = 0.058601, p-value = 0.9915
alternative hypothesis: two-sided

Since p-value is very large hence the distribution follows normal.