Transcript Lecture 43 - Test of Goodness of Fit

Test of Goodness of
Fit
Lecture 43
Section 14.1 – 14.3
Mon, Apr 17, 2006
Count Data

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
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Count data – Data that counts the number of
observations that fall into each of several
categories.
The data may be univariate or bivariate.
Univariate example – Observe a student’s final
grade: A – F.
Bivariate example – Observe a student’s final
grade and year in college: A – F and freshman –
senior.
Univariate Example

Observe students’ final grades in statistics: A, B,
C, D, or F.
A
5
B
12
C
8
D
3
F
2
Bivariate Example

Observe students’ final grade in statistics and
year in college.
A
B
C
D
F
Fresh
3
6
3
2
1
Soph
1
4
4
1
0
Junior
1
2
0
0
1
Senior
0
1
1
0
0
Observed and Expected Counts
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
Observed counts – The counts that were
actually observed in the sample.
Expected counts – The counts that would be
expected if the null hypothesis were true.
Tests of Goodness of Fit
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The goodness-of-fit test applies only to
univariate data.
The null hypothesis specifies a discrete
distribution for the population.
We want to determine whether a sample from
that population supports this hypothesis.
The Chi-Square Statistic
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
Denote the observed counts by O and the
expected counts by E.
Define the chi-square (2) statistic to be
2
(
O

E
)
2  
E
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Clearly, if the observed counts are close to the
expected counts, then 2 will be small.
But if even a few observed counts are far from
the expected counts, then 2 will be large.
Think About It

Think About It, p. 923.
Chi-Square Degrees of Freedom
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
The chi-square distribution has an associated
degrees of freedom, just like the t distribution.
Each chi-square distribution has a slightly
different shape, depending on the number of
degrees of freedom.
Chi-Square Degrees of Freedom
0.5
0.4
2(2)
0.3
2(5)
0.2
2(10)
0.1
5
10
15
20
Properties of 2

The chi-square distribution with df degrees of
freedom has the following properties.
2  0.
 It is unimodal.
 It is skewed right (not symmetric!)
 2 = df.
 2 = (2df).
 If df is large, then 2(df) is approximately N(df, (2df)).

Chi-Square vs. Normal
0.05
0.04
N(30,60)
N(32, 8)
2(30)2
 (32)
0.03
0.02
0.01
10
20
30
40
50
Chi-Square vs. Normal
0.025
0.02
N(128, 16)
2(128)
0.015
0.01
0.005
100
120
140
160
The Chi-Square Table
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
See page A-11.
The left column is degrees of freedom: 1, 2, 3,
…, 15, 16, 18, 20, 24, 30, 40, 60, 120.
The column headings represent areas of lower
tails:
0.005, 0.01, 0.025, 0.05, 0.10,
 0.90, 0.95, 0.975, 0.99, 0.995.
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Of course, the lower tails 0.90, 0.95, 0.975, 0.99,
0.995 are the same as the upper tails 0.10, 0.05,
0.025, 0.01, 0.005.
Example
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If df = 10, what value of 2 cuts off an lower
tail of 0.05?
If df = 10, what value of 2 cuts off a upper tail
of 0.05?
TI-83 – Chi-Square Probabilities

To find a chi-square probability (p-value) on the
TI-83,
Press DISTR.
 Select 2cdf (item #7).
 Press ENTER.
 Enter the lower endpoint, the upper endpoint, and
the degrees of freedom.
 Press ENTER.
 The probability appears.
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Example
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If df = 8, what is the probability that 2 will fall
between 4 and 12?
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
If df = 32, what is the probability that 2 will fall
between 24 and 40?
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Compute 2cdf(4, 12, 8).
Compute 2cdf(24, 40, 32).
If df = 128, what is the probability that 2 will
fall between 96 and 160?

Compute 2cdf(96, 160, 128).
Tests of Goodness of Fit



The goodness-of-fit test applies only to
univariate data.
The null hypothesis specifies a discrete
distribution for the population.
We want to determine whether a sample from
that population supports this hypothesis.
Examples

If we rolled a die 60 times, we expect 10 of each
number.
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If we got frequencies 8, 10, 14, 12, 9, 7, does that
indicate that the die is not fair?
If we toss a fair coin, we should get two heads
¼ of the time, two tails ¼ of the time, and one
of each ½ of the time.

Suppose we toss a coin 100 times and get two heads
16 times, two tails 36 times, and one of each 48
times. Is the coin fair?
Examples

If we selected 20 people from a group that was
60% male and 40% female, we would expect to
get 12 males and 8 females.

If we got 15 males and 5 females, would that
indicate that our selection procedure was not
random (i.e., discriminatory)?
Null Hypothesis
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The null hypothesis specifies the probability (or
proportion) for each category.
Each probability is the probability that a random
observation would fall into that category.
Null Hypothesis
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To test a die for fairness, the null hypothesis
would be
H0: p1 = 1/6, p2 = 1/6, …, p6 = 1/6.
The alternative hypothesis will always be a
simple negation of H0:
H1: At least one of the probabilities is not 1/6.
or more simply,
H1: H0 is false.
Expected Counts
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To find the expected counts, we apply the
hypothetical (H0) probabilities to the sample
size.
For example, if the hypothetical probabilities are
1/6 and the sample size is 60, then the expected
counts are
(1/6)  60 = 10.
Example
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We will use the sample data given for 60 rolls of
a die to calculate the 2 statistic.
Make a chart showing both the observed and
expected counts (in parentheses).
1
8
(10)
2
10
(10)
3
14
(10)
4
12
(10)
5
9
(10)
6
7
(10)
Example
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Now calculate 2.
2
2
2
2
2
2
(
8

10
)
(
10

10
)
(
14

10
)
(
12

10
)
(
9

10
)
(
7

10
)
2 





10
10
10
10
10
10
 0.4  0.0  1.6  0.4  0.1  0.9
 3.4
Computing the p-value
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The number of degrees of freedom is 1 less
than the number of categories in the table.
In this example, df = 5.
To find the p-value, use the TI-83 to calculate
the probability that 2(5) would be at least as
large as 3.4.
p-value = 2cdf(3.4, E99, 5) = 0.6386.
Therefore, p-value = 0.6386 (accept H0).
The Effect of the Sample Size
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What if the previous sample distribution
persisted in a much larger sample, say n = 6000?
Would it be significant?
1
800
(1000)
2
1000
(1000)
3
1400
(1000)
4
1200
(1000)
5
900
(1000)
6
700
(1000)
TI-83 – Goodness of Fit Test
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The TI-83 will not automatically do a goodnessof-fit test.
The following procedure will compute 2.
Enter the observed counts into list L1.
 Enter the expected counts into list L2.
 Evaluate the expression (L1 – L2)2/L2.
 Select LIST > MATH > sum and apply the sum
function to the previous result, i.e., sum(Ans).
 The result is the value of 2.
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The List of Expected Counts
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To get the list of expected counts, you may
Store the list of hypothetical probabilities in L3.
 Multiply L3 by the sample size and store in L2.
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For example, if the probabilities for 4 categories
are p1 = 0.25, p2 = 0.15, p3 = 0.40, and p4 = 0.20,
and the sample size is n = 225, then
Store {0.25,0.25,0.40,0.20} in L3.
 Compute L3*225 and store in L2.
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Example
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To test whether the coin is fair, the null
hypothesis would be
H0: pHH = 1/4, pTT = 1/4, pHT = 1/2.
The alternative hypothesis would be
H1: H0 is false.
Let  = 0.05.
Expected Counts

To find the expected counts, we apply the
hypothetical probabilities to the sample size.
Expected HH = (1/4) 100 = 25.
 Expected TT = (1/4)  100 = 25.
 Expected HT = (1/2)  100 = 50.
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Example
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We will use the sample data given for 60 rolls of
a die to calculate the 2 statistic.
Make a chart showing both the observed and
expected counts (in parentheses).
HH
16
(25)
TT
36
(25)
HT
48
(50)
Example

Now calculate 2.
2
2
2
(
16

25
)
(
36

25
)
(
48

50
)
2 


25
25
50
 3.24  4.84  0.08
 8.16
Compute the p-value
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In this example, df = 2.
To find the p-value, use the TI-83 to calculate
the probability that 2(2) would be at least as
large as 8.16.
2cdf(8.16, E99, 2) = 0.0169.
Therefore, p-value = 0.0169 (reject H0).
The coin appears to be unfair.
Example
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Suppose we select 20 people from a group that
is 60% male and 40% female and we get 15
males and 5 females.
Is it reasonable to believe that we selected the 20
people at random?
The Hypotheses
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To test that the process was random, the null
hypothesis would be
H0: pM = 0.60, pF = 0.40.
The alternative hypothesis would be
H1: H0 is false.
Let  = 0.05.
Calculate the Expected Counts
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To find the expected counts, multiply the
hypothetical probabilities to the sample size.
Expected no. of males = 0.60  20 = 12.
 Expected no. of females = 0.40  20 = 8.

Make the Chart

Make a chart showing both the observed and
expected counts (in parentheses).
M
15
(12)
F
5
(8)
Compute 2

Now calculate 2.
2
2
(
15

12
)
(
5

8
)
2 

12
8
 0.75  1.125
 1.875.
Compute the p-value
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In this example, df = 1.
To find the p-value, use the TI-83 to calculate
the probability that 2(1) would be at least as
large as 1.875.
2cdf(1.875, E99, 1) = 0.1709.
Therefore, p-value = 0.1709 (accept H0).
There is no evidence that the people were not
selected at random.