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The highway between Mumbai, Delhi and Calcutta, Delhi, has a high incidence of accidents along its 100 kilometers.

Public safety officers say that the occurrence of accidents along the highway is randomly (uniformly) distributed, but the news media say otherwise. The Delhi Department of Public Safety published records for month of June. These records indicated the point at which 30 accidents involving an injury or death occured as follows ( the data points represent the distance from the city limits of Mumbai):

88.3 40.7 36.3 27.3 36.8 91.7 67.3 7.0 45.2 23.3
98.8 90.1 17.2 23.7 97.4 32.4 87.8 69.8 62.6 99.7
20.6 73.1 21.6 6.0 45.3 76.6 73.2 27.3 87.6 87.2

Use the Kolmogorov-Smimov goodness of fit test to determine whether the distribution of location of accidents is uniformly distributed for the month of June. Use a level of significane of α = 0.05.

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Solution:

$H_0$: The data points are uniformly distributed

$H_1$: Data points are not uniformly distributed

1. Normalize the data points to (0,1) for K-S Test. The results in following 30 data points are given.

2. Arrange data points in increasing order and compute $D^{-1}$ and $D^{-}$ (K-S Test)

Table:

$\text{i}$ $R(i)$ $\frac{i}{N}$ $\frac{i}{N}-R(i)$ $\frac{i-1}{N}$ $R_{i}-\frac{i-1}{n}$
1 0.06 0.033 - 0 0.06
2 0..07 0.066 - 0.033 0.037
3 0.172 0.100 - 0.066 0.106
4 0.206 0.133 - 0.100 0.106
5 0.2016 0.166 - 0.133 0.083
6 0.233 0.200 - 0.166 0.067
7 0.237 0.233 - 0.200 0.037
8 0.273 0.266 - 0.233 0.040
9 0.273 0.300 0.027 0.266 0.007
10 0.324 0.333 0.009 0.300 0.024
11 0.363 0.400 0.003 0.333 0.030
12 0.368 0.433 0.032 0.366 0.002
13 0.407 0.466 0.026 0.400 0.007
14 0.452 0.500 0.014 0.433 0.019
15 0.453 0.500 0.047 0.466 -
16 0.626 0.533 - 0.500 0.126
17 0.673 0.566 - 0.533 0.140
18 0.698 0.600 - 0.566 0.132
19 0.731 0.633 - 0.600 0.131
20 0.732 0.666 - 0.633 0.099
21 0.766 0.700 - 0.666 0.100
22 0.872 0.733 - 0.700 0.172
23 0.876 0.766 - 0.733 0.143
24 0.878 0.800 - 0.766 0.112
25 0.883 0.833 - 0.800 0.083
26 0.901 0.866 - 0.833 0.068
27 0.917 0.900 - 0.866 0.051
28 0.974 0.933 - 0.900 0.074
29 0.988 0.966 - 0.933 0.055
30 0.997 1.000 0.003 0.966 0.031

$D^+ = max_{1 \le i \le 30} \left\{\frac{i}{N}-R_{(i)}\right\}=0.047$

$D^- =\max _{1 \leq i \leq 30}\left\{R_{i}-\frac{i-1}{N}\right\} = 0.172$

\begin{aligned} \therefore D &= max(D^+, D^-) \\ &= max(0.047, 0.172) \\ &= 0.172 \end{aligned}

Critical value $D_\alpha$ for $\alpha = 0.05 \text{ of } N = 30$

$D_{0.05, 30} = 0.24$

Since $D = 0.172 \lt D_{0.05, 30} = 0.24$

$\therefore H_0 \text{ is not rejected }$