1. ENGINEERING STATISTICS I
INDE 2333
ENGINEERING STATISTICS I
GOODNESS OF FIT
University of Houston
Dept. of Industrial Engineering
Houston, TX
(713)

2. AGENDA
Chi-square goodness of fit test

3. GOODNESS OF FIT TESTS
Used to determine if a sample could have come from a distribution with the specified parameters
Commonly used to determine if data is normally distributed
Many tests such as the ones that we have been using require normally distributed data.
If data is not normally distributed, non-parametric tests must be used (next subject in the course)
Also used for input distributions in system modeling
Customers or jobs arrive exponentially distributed?
Service times follow what distribution?
Failures occur according to what distribution?

4. GOODNESS OF FIT TESTS Based on a comparison of observations between
Observed data
Theoretical data
The comparison utilizes a set of intervals or cells
Each cell has a lower and upper boundary values
The determination of the boundaries are a function of
Theoretical distribution
Number of observations in the sample
2 different approaches…

5. TWO DIFFERENT APPROACHES
Used in the book
Equal interval approach
No cell grouping can have less than 5 expected observations
Approach 2
Used in other books
Equiprobable approach
Maximum number of cells not to exceed 100 such that the expected number of observations is at least 5 = Int ( obs/5 )
Expected number of obs in each cell = obs / cells
More statistically robust

6. HYPOTHESES TEST PROCEDURE
Identify Ho and Ha
Determine level of significance (generally 0.05 or 0.01)
Determine “critical value” criterion from level of significance
Calculate “test statistic”
Make decision
Fail to reject Ho
Reject Ho

7. HYPOTHESES Ho
The sample could have come from a distribution with the specified parameters Ha
The sample could not have come from a distribution with the specified parameters

8. CRITICAL VALUE Chi-square distribution chart One sided test
Alpha typically 0.05
Degrees of freedom
# of cells - # of parameters used from sample -1
The -1 is always used due to the known sample size n
Note, if the parameters are specified not sampled then they do not reduce the number of degrees of freedom in the above equation

9. CHI-SQUARE for a particular number of degrees of freedom
f(X^2)
Right tail probability, alpha, typically 0.05
X^2
X^2 Critical value

10. TEST STATISTIC

11. DECISION Cannot reject Reject
Test statistic is less than the critical value
Sample could have come from a distribution with the specified parameters
Reject
Test statistic is greater than the critical value
Sample could not have come from a distribution with the specified parameters

12. EXAMPLE 1 EQUAL INTERVAL APPROACH
400 5 minute intervals were observed for air traffic control messages
At alpha=0.01, is the distribution of the number of messages able to be considered as having a poisson distribution with a mean of 4.6?
Approach
Lamba parameter of 4.6 is given
Use the poisson table probability table for 4.6
Multiply the probability by 400 to obtain the expected observations
Compare the actual observations to the expected observations

13. HYPOTHESES Ho: Ha: Poisson distribution with mean of 4.6
Not poisson distribution with a mean of 4.6

14. Messages
Observed
Probability
Expected 3
Combine
0.010
4.0 1 15
for 18
0.046
18.4
For 22.4 2 47
0.107
42.8 76
0.163
65.2 4 68
0.187
74.8 5 74
0.173
69.2 6 46
0.132
52.8 7 39
0.087
34.8 8
0.050
20.0 9
0.025
10.0 10
0.012
4.8 11
these 4
0.005
2.0
These 4 12
cells for a
0.002
0.8
Cells for a 13
total of 8
0.001
0.4

15. CHI-SQUARE for 10-1 degrees of freedom
f(X^2)
Right tail probability, alpha = 0.01
X^2
Critical value

16. TEST STATISTIC

17. DECISION
Test statistic of is less than the critical value of
Cannot reject Ho of distribution being poisson with a mean of 4.6
There is evidence to support the claim that the data came from a poisson distribution with a mean of 4.6 at an alpha level of 0.01

18. EXAMPLE 2 EQUIPROBABLE APPROACH
Were the scores from an INDE 2333 exam normally distributed?
Sample statistics
Mean=71.95
Std=11.93
N=43

19. HYPOTHESES Ho
The sample could have come from a normally distributed population with a mean of and a std of 11.93 Ha
The sample could not have come from a normally distributed population with a mean of and a std of 11.93

20. CRITICAL VALUE Chi-square distribution chart One sided test 0.05
Degrees of freedom
The sample size is 43
Want the maximum number of cells not to exceed 100 with a minimum expected number of observation of 5
43/5=8.6 cells
With 8 cells, the expected number of observations is 5.375
Degrees of freedom is number of cells – number of parameters used from sample-1
Degrees of freedom=8-2-1=5

21. CHI-SQUARE for 5 of degrees of freedom
f(X^2)
0.05
X^2
11.070

22. TEST STATISTIC

23. CELL BOUNDARIES
To calculate observed values in each cell, we must determine the actual x cell boundaries from the 8 equiprobable cells
For normal distributions
Look up z value corresponding to probability
Boundaries =mean+std * Z

24. CALCULATING OBSERVATIONS
Cell
Lower%
Upper%
Lower Z
Upper
lowerx
upperx
obs
exp 1
0.125
-inf
58.227 5
5.375 2
0.250
63.905 6 3
0.375
68.151 4
0.500
71.953
0.625
75.756
0.750
80.002 7
0.875
85.680 8
1.000
+inf
100.00

25. CALCULATING TEST STATISTIC
Cell
obs
exp
((O-E)^2)/E 1 5
5.375
0.026 2 6
0.072 3 4
1.049 7 8
1.282
Total
2.534

26. DECISION 2.581 < 11.070 Cannot reject the Ho
Evidence to support the claim that the test scores are normally distributed with a mean of and std of 11.93

27. IN EXCEL Frequency Range operation Norminv function Chiinv function
Data_array, bins_array
Range operation
CTRL-SHIFT-ENTER
Norminv function
Probability, mean, std
Chiinv function
Probability, df