Transcript Summary Table of Influence Procedures for a Single Sample (II)

Summary Table of Influence Procedures
&4-8 (&8-6)
for a Single Sample (I)
Horng-Chyi Horng
Statistics II
127
Summary Table of Influence Procedures
for a Single Sample (II)
Horng-Chyi Horng
Statistics II
128
Testing for Goodness of Fit
&4-9 (&8-7)

In general, we do not know the underlying distribution of
the population, and we wish to test the hypothesis that a
particular distribution will be satisfactory as a population
model.

Probability Plotting can only be used for examining
whether a population is normal distributed.

Histogram Plotting and others can only be used to guess
the possible underlying distribution type.
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Goodness-of-Fit Test (I)



A random sample of size n from a population whose
probability distribution is unknown.
These n observations are arranged in a frequency
histogram, having k bins or class intervals.
Let Oi be the observed frequency in the ith class interval,
and Ei be the expected frequency in the ith class interval
from the hypothesized probability distribution, the test
statistics is
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Goodness-of-Fit Test (II)


If the population follows the hypothesized distribution, X02
has approximately a chi-square distribution with k-p-1 d.f.,
where p represents the number of parameters of the
hypothesized distribution estimated by sample statistics.
That is,
2
k
 
2
0

Oi  Ei 
Ei
i 1
~  k2 p 1
Reject the hypothesis if
   ,k  p1
2
0
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Goodness-of-Fit Test (III)

Class intervals are not required to be equal width.

The minimum value of expected frequency can not be to
small. 3, 4, and 5 are ideal minimum values.

When the minimum value of expected frequency is too
small, we can combine this class interval with its
neighborhood class intervals. In this case, k would be
reduced by one.
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Statistics II
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Example 8-18
The number of defects in printed circuit boards is
hypothesized to follow a Poisson distribution. A random sample of size 60 printed
boards has been collected, and the number of defects observed as the table below:

The only parameter in Poisson distribution is l, can be estimated by the
sample mean = {0(32) + 1(15) + 2(19) + 3(4)}/60 = 0.75. Therefore, the
expected frequency is:
e 0.75 (0.75) 0
p1  P( X  0) 
 0.472
0!
E1  0.472  60  28.32
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Statistics II
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Example 8-18 (Cont.)

Since the expected frequency in the last cell is less than 3, we combine the last
two cells:
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Example 8-18 (Cont.)
1. The variable of interest is the form of distribution of defects in printed circuit
boards.
2. H0: The form of distribution of defects is Poisson
H1: The form of distribution of defects is not Poisson
3. k = 3, p = 1, k-p-1 = 1 d.f.
4. At  = 0.05, we reject H0 if X20 > X20.05, 1 = 3.84
5. The test statistics is:
(Oi  Ei ) 2 (32  28.32) 2 (15  21.24) 2 (13  10.44) 2
 



 2.94
Ei
28.32
21.24
10.44
i 1
k
2
0
6. Since X20 = 2.94 < X20.05, 1 = 3.84, we are unable to reject the null hypothesis
that the distribution of defects in printed circuit boards is Poisson.
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Statistics II
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Contingency Table Tests

(&8-8)
Example 8-20
A company has to choose among three pension plans. Management wishes to
know whether the preference for plans is independent of job classification and
wants to use  = 0.05. The opinions of a random sample of 500 employees are
shown in Table 8-4.
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Contingency Table Test
- The Problem Formulation (I)

There are two classifications, one has r levels and the other has c
levels. (3 pension plans and 2 type of workers)
Want to know whether two methods of classification are statistically
independent. (whether the preference of pension plans is independent
of job classification)

The table:

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Statistics II
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Contingency Table Test
- The Problem Formulation (II)


Let pij be the probability that a random selected element falls in the ijth
cell, given that the two classifications are independent. Then pij = uivj,
where the estimator for ui and vj are


1 c
1 r
 i   Oij
v j   Oij
n j 1
n i 1
Therefore, the expected frequency of each cell is



r
1 c
Eij  n  i v j   Oij  Oij
n j 1 i 1
Then, for large n, the statistic
r
c
 02  
i 1 j 1
(Oij  Eij ) 2
Eij
has an approximate chi-square distribution with (r-1)(c-1) d.f.
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Statistics II
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Example 8-20
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