Transcript Goodness of Fit Test

Multinomial Experiments
Goodness of Fit Tests
We have just seen an example of comparing two
proportions. For that analysis, we used the normal
distribution as a sampling distribution.
We will now look an problems where we compare more
than two proportions. We will not be able to use the
normal distribution, but will use a different distribution
called the Chi-Square or Χ2 Distribution.
Consider the problem of testing a die to see if it is unfair.
The die has six numbers, all equally likely. If die is fair, then
each number should have a probability of 1/6. In the long
run, each number will come up 1/6 of the number of rolls.
Suppose weI take a sample of 60 rolls. Theoretically, each
number should come up 1/6*60 = 10 times. If the
numbers are not all 10, either the die is not fair, or, the die
is fair, and the numbers different from 10 are explained by
sampling variation.
To sort this out, I need a hypotheses test, a sampling
distribution, and a p-value.
Section 11.2, Page 236
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Goodness of Fit Test
Fair Die Example
Following is the distribution of the observed
frequencies of results from rolling a die 60 times. Is
the die fair?
The hypotheses are as follows:
Clearly, the observed frequencies are not all equal to
the theoretical frequencies of 10. We need a way to
measure how big the “miss” is to see if it likely to be
due to sampling variation, or if it is so large as to not be
explained by sampling variation.
Section 11.2, Page 239
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Chi Square Statistic
Fair Die Example
We calculate the “miss” called the chi-square statistic
similarly to the way we calculate the variance.
Note that the expected frequencies always equal the
total number of observations × Ho true proportion for
each cell or proportion. Also note that the total
expected frequencies always equals the total observed
frequencies.
The χ2 Statistic is the Total, 2.2. Also, note that
minimum value of the χ2 is zero. If we took another
sample, we would likely get a different value for the
chi-square statistic.
Section 11.1, Page 239
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Chi-Square Distribution
Fair Die Example
Now we need a sampling distribution for the Χ2 statistic
= 2.2, so we can calculate the probability of getting a Χ2
≥ 2.2 when the true proportions are all equal to 1/6.
Χ2 Distribution for 5 df
This is a distribution of all possible Χ2 statistics
calculated from all possible samples of 60
observations when there are 6 proportions or cells.
Note that the degree of freedom equals the number
of proportions – 1.
Finding the p-value on the TI-83, Given Χ2 Stat, df
PRGM – CHI2DIST
LOWER BOUND: 2.2
UPPER BOUND: 2ND E99
df: 5
Output: P-VALUE = 0.8208
The null hypothesis cannot be rejected.
Section 11.2, Page 240
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Chi-Square Distribution
Conditions
The sample is random and the observed
data represents counts of individuals in
individual categories of a categorical
variable
Each expected count is 5 or greater
Section 11.1, Page 241
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Goodness of Fit Test
Fair Die Example – TI-83 Add-In
Following is the distribution of the observed
frequencies of results from rolling a die 60 times. Is
the die fair?
The hypotheses are as follows:
Each expected cell = 1/6*60 = 10.
STAT-EDIT – LI: Enter the observed frequency
numbers
L2: Enter the expected values, 10 in each of 6
cells.
PRGM – GOODFIT
OBSERVED LIST = 2ND L1
EXPECTED LIST = 2ND L2
Answer: p-value = .8208, Chi-Square Stat = 2.2
Since p-value > 0.05, Ho cannot be rejected.
Section 11.2, Page 240
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Goodness of Fit Test
Mendelian Theory Problem
Mendel’s genetic theory of inheritance claims that the
frequencies of round and yellow, wrinkled and yellow, round
and green, and wrinkled and green peas will occur in the ratio of
9:3:3:1. In testing the theory, Mendel obtained frequencies of
315, 101, 108, and 32 respectively. Does the data contradict the
theory. Do a hypotheses test.
Ho: The data fits the theory
Ha: The data does not fit the theory.
Calculation of Expected Values
Observed
Expected
Expected Count
Proportions
315
9/16
9/16 *556 = 312.75
101
3/16
3/16 * 556 = 104.25
108
3/16
3/16 *556 = 104.25
32
1/16
1/16 *556= 34.75
Total = 556
Total = 1
Total = 556
Section 11.2, Page 245
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Goodness of Fit Test
Mendelian Theory Problem
Observed
Expected
Expected Count
Proportions
315
9/16
9/16 *556 = 312.75
101
3/16
3/16 * 556 = 104.25
108
3/16
3/16 *556 = 104.25
32
1/16
1/16 * 556 = 34.75
Total = 556
Total = 1
Total = 556
STAT – EDIT: Enter observed data in L1 and expected in L2
PRGM – GOODFIT
OBSERVED LIST = 2ND L1
EXPECTED LIST = 2ND L2
Answer: p-value = .9254, Chi-Square Stat = .47
The null hypothesis cannot be rejected. The observed
data does not contradict the theory
Section 11.2, Page 243
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Problems
a. Perform a hypotheses test to see if the colors are not
equally likely. State the hypotheses.
b. Find the p-value and state your conclusion
c. What is the name of the model used for the sampling
distribution?
d. What conditions must be satisfied? Are they?
Problems, Page 252
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Problems
a. Perform a hypotheses test to see if the
preferences are not all the same. State the
hypotheses.
b. Find the p-value and state your conclusion
c. What is the name of the model used for the
sampling distribution?
d. What conditions must be satisfied? Are they?
Problems, Page 252
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Problems
a. Perform a hypotheses test to see of the
observed data is inconsistent with the stated
ratios. State the appropriate hypotheses.
b. Find the expected counts for each color.
c. What are the necessary conditions for the
sampling distribution?
d. What is the name of the model used for the
sampling distribution?
e. Find the p-value and state your conclusion.
Problems, Page 252
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Test for Independence
Following is a two way table. In this case, two categorical
variables are measured on one group of college students. For
each student, their Gender and Favorite Subject Area are
recorded.
Independence of Two Variables
Consider the Social Science category. 113/300 or 38% of all
students chose Social Science. However, 41/122 or 34% of
males chose the category and 72/178 or 40% of Females
chose the category.
Considering this a probability distribution, if I pick a person
at random, there is a 38% chance the person chose Social
Science. However, it you tell me the person is a female, then
the probability is 40% they chose the category.
This is an indication that the two variables are not
independent, but related.
Two variables are independent, if knowing the outcome of
one variable does not change the probability of the
outcome of the other variable.
Section 11.3, Page 244
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Tests for Independence
The sample data gives an indication the
variables are not independent, but this
indication may be due to sampling variation.
To test for independence, we will use Chi-Square
methods. The appropriate hypotheses are:
Ho: The variables are independent
Ha: The variables are not independent
Next, we need to calculate the expected values
for each cell of the data matrix under the
assumption that the variables are independent.
For example, if the variables are independent,
then the the overall proportion of of students in
the Social science category is 113/300 = .3767.
Both the proportions for the category have to be
the same. The expected value for Males is
0.3767*122= 45.95 and the expected values for
Females is 0.3767*178 = 67.05.
Section 11.3, Page 244
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Test for Independence
Shown above in the parentheses are all the
expected values. Next we need to calculate the
χ2 statistic for each data cell. For example, for
the first cell: (Observed-Expected)2/Expected =
(37-29.28)2/29.28 = 2.0355.
Adding up the cell calculations for the 6 cells
gives total χ2 statistic of 4.604. The formula for
df =(#rows – 1)*(#columns – 1) = (2-1)*(3-1) = 2.
The area under the curve to the right of 4.604
= .1001 > .05. The null hypotheses cannot be
rejected.
Section 11.3, Page 246
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Test for Independence
Black Box Program
Ho: The variables are independent
Ha: The variables are not independent
2nd MATRIX – EDIT
2 ENTER 3 (The data table is 2 rows and 3
columns. Ignore total row and
total column)
Enter the data in matrix [A] left to right
STAT-TESTS-C:χ2-TEST
Observed: [A]
Expected: [B]
Calculate
Answer: p-value = .0999, χ2-Stat = 4.6063
2nd MATRIX – EDIT – [B] – ENTER
Displays the Expected Values Matrix
All cells ≥ 5; conditions satisfied
Section 11.3, Page 246
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Problems
a. Test the hypotheses that the size of
community reared in is independent of the
size of community residing in. State the
appropriate hypotheses.
b. Find the p-value and state your conclusion
c. What is the name of the sampling
distribution?
d. What are the necessary conditions, and are
they satisfied? What is the value of the
smallest expected cell?
Section 11.3, Page 254
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Problems
a. Test the hypotheses that years of employment
and knowing what supervisor expects are
independent. State the appropriate
hypotheses.
b. Find the p-value and state your conclusion
c. What is the name of the sampling
distribution?
d. What are the necessary conditions, and are
they satisfied? What is the value of the
smallest expected cell?
Section 11.3, Page 254
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Problems
a. Test the hypotheses that the survival rate and
the treatment are independent. State the
appropriate hypotheses.
b. Find the p-value and state your conclusion
c. What is the name of the sampling distribution?
d. What are the necessary conditions, and are
they satisfied? What is the value of the
smallest expected cell?
Problems, Page 253
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Tests for Homogeneity
Another application of Chi-Square procedures is test
for homogeneity, or essentially, a test whether
different groups have the same distribution for a
given variable.
Consider the table below that gives voter’s opinion
on a proposal broken down by separate locations.
In the case of a test for independence, we had one
group of individuals and measure two categorical
variables in that group.
In the case of a test for homogeneity, we have one
categorical variable, Opinion on Proposal, and three
separately located groups of voters. The hypothesis
are:
Ho: The distributions are homogeneous
Ha: The distributions are not homogeneous
Section 11.3, Page 247
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Tests for Homogeneity
The mechanics for a test of homogeneity are exactly
the same as for a test of independence. We calculate
the expected values under the assumption Ho is true.
The proportion favor are all assumed to be
254/500 = .5080. The expected value for
urban is .5080*200 = 101.6. The χ2 Stat for
cell 1 = (143-101.6)2/101.6 = 16.8897. The
total χ2 statistic for all cells is 91.72.
The df = 2 and the p-value = 1.21E-20 ≅ 0
Ho is rejected, the distributions are not the same.
Section 11.3, Page 248
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Problems
a. State the hypotheses.
b. Find the p-value and state your
conclusion.
c. What is the name of the model used
for the sampling distribution.
d. What is the value of the smallest
expected cell?
Section 11.3, Page 253
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Problems
a. State the hypotheses.
b. Find the p-value and state your
conclusion.
c. What is the name of the model used for
the sampling distribution.
d. What is the value of the smallest
expected cell?
Section 11.3, Page 253
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Summary of Chi-Square Applications
Goodness of Fit Test
Given one categorical variable with a fixed set of
proportions for the categories.
Ha: The observed data does not fit the proportions.
Calculate expected values (Ho true proportion * total
observations)
Observed and Expected data in List Editor
PRGM: GOODFIT
Test for Independence
Given two categorical variables measured on the
same population.
Ha: The variables are not independent (They are
related)
Observed data in Matrix Editor
Stat-Tests-χ2 Test
Test for Homogeneity
Given one categorical variable and two or more
populations.
Ha: The proportions for the categories are not the
same for for all populations.
Observed data in Matrix Editor
Stat-Tests-χ2 Test
Chapter 11 Summary
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Problems
a. Is this a goodness of fit test, a test for
independence, or a test or homogeneity?
b. State the hypotheses.
c. Find the p-value and state your
conclusion.
d. What is the name of the model used for
the sampling distribution.
e. What is the value of the smallest
expected cell?
Problems, Page 252
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