Download Goodness-of-fit Testing Examples

IME 301
Goodness-of-fit Testing Examples
By: Dr. Parisay
This handout presents discussion on Goodness-of-fit testing examples. The detail
solutions were performed in class. Dr. Parisay’s comments are in red
Problem :
The grades in a statistics course for a particular semester were as follows:
Grade
F
A
14
B
18
C
32
D
20
F
16
Test the hypothesis at the 0.5 level of significance, that the distribution of grades is
uniform.
Solution:
H0 = data has uniform distribution
H1 = data has not uniform distribution
n = number of observations = 14+18+32+20+16 = 100
k = number of intervals = 5
for a uniform distribution ei = expected number of observations in each interval = n/k =
100/5 = 20
p = number of parameters of fitted distribution (uniform distribution parameters: min and
max) = 2
2,( k  p1)   02.05,2  5.991
 02  (14-20)2/20 + (18-20)2/20 +(32-20)2/20 +(20-20)2/20 +(16-20)2/20 = 200/20 = 10
 02  10 > 5.991 conclude cdf = uniform is not a good fit
Version 2: The grades in a statistics course for a particular semester were as follows:
Grade
A
B
C
D
F
F
14
18
32
20
16
Test the hypothesis at the 0.5 level of significance, that the distribution of grades follows
a normal distribution with 15% for A, 20% for B, 30% for C, 20% for D, and 15% for F.
In fact 15% for A means that the area under the normal curve for the range
of grades (above 90%) that will receive A is 0.15, or probability of receiving
an A is 15%. Using probability notation [F(max grade for A: 100%) – F(min
grade for A: 90%)], where F(X) is cumulative distribution function for
variable X.
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Solution:
n = number of observations = 14+18+32+20+16 = 100
Interval
Oi
ei = probability * n
A
B
C
D
F
14
18
32
20
16
15%(100) =15
20%(100) =20
30%(100) =30
20%(100) =20
15%(100) =15
Total
(Oi - ei )2 / ei
(14-15)2 /15
(18-20)2 /20
(32-30)2 /30
(20-20)2 /20
(16-15)2 /15
0.47
p = number of parameters of fitted distribution (normal distribution parameters: mean and
std. dev.) = 2
k = number of intervals = 5
2,( k  p1)   02.05,2  5.991
 02  0.47 < 5.991
conclude cdf = normal under the above condition is a good fit
Version 3: The grades in a statistics course for a particular semester were as follows:
Grade
A
B
C
D
F
F
14
18
32
20
16
Test the hypothesis at the 0.5 level of significance, that the distribution of grades follows
a normal distribution with 10% for A, 20% for B, 40% for C, 20% for D, and 10% for F.
Solution:
n = number of observations = 14+18+32+20+16 = 100
Interval
Oi
ei = probability * n
A
B
C
D
F
14
18
32
20
16
10%(100) =10
20%(100) =20
40%(100) =40
20%(100) =20
10%(100) =10
Total
(Oi - ei )2 / ei
(14-10)2 /10
(18-20)2 /20
(32-40)2 /40
(20-20)2 /20
(16-10)2 /10
7
p = number of parameters of fitted distribution (normal distribution parameters: mean and
std. dev.) = 2
k = number of intervals = 5
2,( k  p1)   02.05,2  5.991
 02  7 > 5.991 conclude cdf = normal under the above condition is not a good fit
Date prepared: November 29, 2003
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