Download Overview Section 11-1 & 11-2 Overview and Multinomial Experiments: Goodness of Fit

Section 11-1 & 11-2
Overview and Multinomial
Experiments: Goodness of Fit
Key Concept
Overview
We focus on analysis of categorical (qualitative
or attribute) data that can be separated into
different categories (often called cells).
Use the χ2 (chi-square) test statistic (Table A- 4).
The goodness-of-fit test uses a one-way
frequency table (single row or column).
Given data separated into different
categories, we will test the hypothesis that
the distribution of the data agrees with or
“fits” some claimed distribution.
The hypothesis test will use the chi-square
distribution with the observed frequency
counts and the frequency counts that we
would expect with the claimed distribution.
The contingency table uses a two-way frequency
table (two or more rows and columns).
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Definition
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Example: Last Digits of Weights
Multinomial Experiment
This is an experiment that meets the following
conditions:
When asked, people often provide
weights that are somewhat lower
than their actual weights. So how
can researchers verify that weights
were obtained through actual
measurements instead of asking
subjects?
1. The number of trials is fixed.
2. The trials are independent.
3. All outcomes of each trial must be classified
into exactly one of several different
categories.
4. The probabilities for the different categories
remain constant for each trial.
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Example: Last Digits of Weights
Example: Last Digits of Weights
Test the claim that the digits in Table 11-2 do not
occur with the same frequency.
Table 11-2
summarizes the last
digit of weights of 80
randomly selected
students.
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Verify that the four conditions of a
multinomial experiment are satisfied.
1. The number of trials (last digits) is the fixed number 80.
2. The trials are independent, because the last digit of
any individual’s weight does not affect the last digit of any
other weight.
3. Each outcome (last digit) is classified into exactly 1 of
10 different categories. The categories are 0, 1, … , 9.
4. Finally, in testing the claim that the 10 digits are
equally likely, each possible digit has a probability of 1/10,
and by assumption, that probability remains constant for
each subject.
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Definition
Goodness-of-Fit Test
Notation
Goodness-of-fit Test
A goodness-of-fit test is used to
test the hypothesis that an
observed frequency distribution
fits (or conforms to) some claimed
distribution.
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Expected Frequencies
represents the expected frequency of an outcome.
k
represents the number of different categories or
outcomes.
n
represents the total number of trials.
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E=np
the sum of all observed frequencies
divided by the number of categories
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Each expected frequency is found by
multiplying the sum of all observed
frequencies by the probability for the
category.
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Goodness-of-Fit Test in
Multinomial Experiments
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Goodness-of-Fit Test in
Multinomial Experiments
Requirements
Test Statistics
1. The data have been randomly selected.
2. The sample data consist of frequency counts
for each of the different categories.
3. For each category, the expected frequency is at
least 5. (The expected frequency for a category
is the frequency that would occur if the data
actually have the distribution that is being
claimed. There is no requirement that the
observed frequency for each category must be
at least 5.)
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E
If expected frequencies are not all
equal:
n
k
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represents the observed frequency of an outcome.
Expected Frequencies
If all expected frequencies are equal:
E=
O
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χ2 = Σ
(O – E)2
E
Critical Values
1. Found in Table A-4 using k – 1 degrees of
freedom, where k = number of categories.
2. Goodness-of-fit hypothesis tests are always
right-tailed.
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Relationships Among the χ2 Test
Statistic, P-Value, and Goodness-of-Fit
Goodness-of-Fit Test in
Multinomial Experiments
Figure 11-3
A close agreement between observed and
expected values will lead to a small value of χ2
and a large P-value.
A large disagreement between observed and
expected values will lead to a large value of χ2
and a small P-value.
A significantly large value of χ2 will cause a
rejection of the null hypothesis of no difference
between the observed and the expected.
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Example: Last Digit Analysis
Example: Last Digit Analysis
Test the claim that the digits in Table 11-2
do not occur with the same frequency.
Test the claim that the digits in Table 11-2
do not occur with the same frequency.
H0: p0 = p1 = … = p9
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Because the 80 digits
would be uniformly
distributed through
the 10 categories, each
expected frequency
should be 8.
H1: At least one of the
probabilities is different
from the others.
α = 0.05
k–1=9
χ2.05, 9 = 16.919
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Example: Last Digit Analysis
Example: Last Digit Analysis
Test the claim that the digits in Table 11-2
do not occur with the same frequency.
Test the claim that the digits in Table 11-2
do not occur with the same frequency.
From Table 11-3, the test statistic is χ2 = 156.500.
Since the critical value is 16.919, we reject the
null hypothesis of equal probabilities.
There is sufficient evidence to support the claim
that the last digits do not occur with the same
relative frequency.
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Example: Detecting Fraud
Unequal Expected Frequencies
Example: Detecting Fraud
In the Chapter Problem, it was noted that statistics can
be used to detect fraud. Table 11-1 lists the percentages
for leading digits from Benford’s Law.
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Example: Detecting Fraud
Test the claim that there is a significant discrepancy
between the leading digits expected from Benford’s
Law and the leading digits from the 784 checks.
Observed Frequencies and Frequencies Expected with
Benford’s Law
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Example: Detecting Fraud
Test the claim that there is a significant discrepancy
between the leading digits expected from Benford’s
Law and the leading digits from the 784 checks.
Test the claim that there is a significant discrepancy
between the leading digits expected from Benford’s
Law and the leading digits from the 784 checks.
H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, p4 = 0.097, p5 = 0.079,
p6 = 0.067, p7 = 0.058, p8 = 0.051 and p9 = 0.046
The test statistic is χ2 = 3650.251.
H1: At least one of the proportions is different from the
claimed values.
Since the critical value is 20.090, we reject the null
hypothesis.
α = 0.01
There is sufficient evidence to reject the null
hypothesis. At least one of the proportions is
different than expected.
k–1=8
χ2.01,8 = 20.090
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Example: Detecting Fraud
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Example: Detecting Fraud
Test the claim that there is a significant discrepancy
between the leading digits expected from Benford’s
Law and the leading digits from the 784 checks.
Figure 11-5
Figure 11-6 Comparison of Observed Frequencies and
Frequencies Expected with Benford’s Law
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