Page 345 - Fundamentals of Probability and Statistics for Engineers

328 Fundamentals of Probability and Statistics for Engineers

distributions. We also remark that the values of c n, given in Table A.6 are based

on a completely specified hypothesized distribution. When the parameter values
must be estimated, no rigorous method of adjustment is available. In these cases,
it can be stated only that the values of c n, should be somewhat reduced.
The step-by-step procedure for executing the K–S test is now outlined as
follows:
. Step 1: rearrange sample values x 1 , x 2 ,..., x n in increasing order of magni-
tude and label them x (1) , x (2) ,..., x (n) .
0
. Step 2: determine observed distribution function F (x) at each x (i) by using
0
F [x (i) ] ˆ i/n.
. Step 3: determine the theoretical distribution function F X (x) at each x (i) by
using the hypothesized distribution. Parameters of the distribution are esti-
mated from the data if necessary.
0
. Step 4: form the differences jF -x -i) ) F X -x -i) )j for i ˆ 1, 2, ... , n.
. Step 5: calculate
n
0
d 2 ˆ maxfjF ‰x i† Š F X ‰x i† Šjg:
iˆ1
The determination of this maximum value requires enumeration of n quan-
0
tities. This labor can be somewhat reduced by plotting F (x) and F X (x) as
functions of x and noting the location of the maximum by inspection.
. Step 6: choose a value of and determine from Table A.6 the value of c n, .
. Step 7: reject hypothesis H if d 2 > c n, : Otherwise, accept H.

Example 10.5. Problem: 10 measurements of the tensile strength of one type
of engineering material are made. In dimensionless forms, they are 30.1, 30.5,
28.7, 31.6, 32.5, 29.0, 27.4, 29.1, 33.5, and 31.0. On the basis of this data set, test
the hypothesis that the tensile strength follows a normal distribution at the 5%
significance level.
Answer: a reordering of the data yields x -1) ˆ 27:4, x -2) ˆ 28:7, ... , x -10) ˆ
0
.
33 5. The determination of F (x (i) ) is straightforward. We have, for example,
0
0
0
F 27:4†ˆ 0:1; F 28:7†ˆ 0:2; .. . ; F 33:5†ˆ 1:
With regard to the theoretical distribution function, estimates of the mean and
variance are first obtained from

10
1 X
^ m ˆ x ˆ x j ˆ 30:3;
10
jˆ1
10

n 1 2 1 X 2
b 2
ˆ s ˆ x j 30:3† ˆ 3:14:
n 10
jˆ1

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