Transcript Chapter 11 Chi-Square Procedures

Chapter 11
Chi-Square Procedures
11.1
Chi-Square Goodness of Fit
Characteristics of the Chi-Square Distribution
1. It is not symmetric.
Characteristics of the Chi-Square Distribution
1. It is not symmetric.
2. The shape of the chi-square distribution
depends upon the degrees of freedom, just like
Student’s t-distribution.
Characteristics of the Chi-Square Distribution
1. It is not symmetric.
2. The shape of the chi-square distribution
depends upon the degrees of freedom, just like
Student’s t-distribution.
3. As the number of degrees of freedom
increases, the chi-square distribution becomes
more symmetric as is illustrated in Figure 1.
Characteristics of the Chi-Square Distribution
1. It is not symmetric.
2. The shape of the chi-square distribution
depends upon the degrees of freedom, just like
Student’s t-distribution.
3. As the number of degrees of freedom
increases, the chi-square distribution becomes
more symmetric as is illustrated in Figure 1.
4. The values are non-negative. That is, the
values of are greater than or equal to 0.
The Chi-Square Distribution
A goodness-of-fit test is an inferential
procedure used to determine whether a
frequency distribution follows a claimed
distribution.
Expected Counts
Suppose there are n independent trials an
experiment with k > 3 mutually exclusive possible
outcomes. Let p1 represent the probability of
observing the first outcome and E1 represent the
expected count of the first outcome, p2 represent
the probability of observing the second outcome
and E2 represent the expected count of the second
outcome, and so on. The expected counts for
each possible outcome is given by
Ei = i = npi
for
i = 1, 2, …, k
EXAMPLE
Finding Expected Counts
A sociologist wishes to determine whether the distribution for
the number of years grandparents who are responsible for
their grandchildren is different today than it was in 2000.
According to the United States Census Bureau, in 2000,
22.8% of grandparents have been responsible for their
grandchildren less than 1 year; 23.9% of grandparents have
been responsible for their grandchildren 1or 2 years; 17.6%
of grandparents have been responsible for their
grandchildren 3 or 4 years; and 35.7% of grandparents have
been responsible for their grandchildren for 5 or more years.
If the sociologist randomly selects 1,000 grandparents that
are responsible for their grandchildren, compute the
expected number within each category assuming the
distribution has not changed from 2000.
Test Statistic for Goodness-of-Fit Tests
Let Oi represent the observed counts of category i,
Ei represent the expected counts of an category i, k
represent the number of categories, and n represent
the number of independent trials of an experiment.
Then,
i = 1, 2, …, k
approximately follows the chi-square distribution
with k – 1 degrees of freedom provided (1) all
expected frequencies are greater than or equal to 1
(all Ei > 1) and (2) no more than 20% of the
expected frequencies are less than 5. NOTE: Ei =
npi for i = 1,2, ..., k.
The Chi-Square Goodness-of-Fit Test
If a claim is made regarding a distribution, we
can use the following steps to test the
claim provided
1. the data is randomly selected
The Chi-Square Goodness-of-Fit Test
If a claim is made regarding a distribution, we
can use the following steps to test the
claim provided
1. the data is randomly selected
2. all expected frequencies are greater than
or equal to 1.
The Chi-Square Goodness-of-Fit Test
If a claim is made regarding a distribution, we
can use the following steps to test the
claim provided
1. the data is randomly selected
2. all expected frequencies are greater than
or equal to 1.
3. no more than 20% of the expected
frequencies are less than 5.
Step 1: A claim is made regarding a
distribution. The claim is used to
determine the null and alternative
hypothesis.
Ho: the random variable follows the
claimed distribution
H1: the random variable does not follow
the claimed distribution
Step 2: Calculate the expected frequencies for
each of the k categories. The expected
frequencies are npi for i = 1, 2, …, k where n is
the number of trials and pi is the probability of the
ith category assuming the null hypothesis is true.
Step 3: Verify the requirements fort he
goodness-of-fit test are satisfied.
(1) all expected frequencies are greater
than or equal to 1 (all Ei > 1)
(2) no more than 20% of the expected
frequencies are less than 5.
EXAMPLE Testing a Claim Using the Goodness-of-Fit
Test
A sociologist wishes to determine whether the distribution
for the number of years grandparents who are
responsible for their grandchildren is different today than
it was in 2000. According to the United States Census
Bureau, in 2000, 22.8% of grandparents have been
responsible for their grandchildren less than 1 year;
23.9% of grandparents have been responsible for their
grandchildren 1or 2 years; 17.6% of grandparents have
been responsible for their grandchildren 3 or 4 years; and
35.7% of grandparents have been responsible for their
grandchildren for 5 or more years. The sociologist
randomly selects 1,000 grandparents that are responsible
for their grandchildren and obtains the following data.
Solution:
• Step 1. Construct the Hypothesis
•
H0 : The distribution for the number of years
grandparents who are responsible for their
grandchildren is the same today as it was in 2000.
H1 : The distribution for the number of years
grandparents who are responsible for their
grandchildren is different today from what it was
in 2000.
• Step 2. Compute the expected counts for each
category, assuming that the null hypothesis is true.
Number of Years
Frequency(Oi)
(observed count)
Expected Frequency(Ei)
(expected count)
Less than 1 year
252
228
1 or 2 years
255
239
3 or 4 years
5 or more years
162
331
176
357
Solution(cont’d):
• Step 3. Verify that the requirements for the
goodness-of-fit test are satisfied.
1. All expected frequencies( or expected
counts ) are bigger than or equal to 1?
2. No more than 20% of the expected
frequencies are less than 5.
Step 4. Find the critical values, determine the
critical region.
α=0.05, k = 4, degree of freedom = k-1 =3
Look in table IV,
χα2 =7.815
C:=(7.815, infinity)
•
Step 5. Compute the test statistic
χ2 = (252-228)^2/228+(255-239)^2/239
+( 162-176)^2/176+(331-357)^2/357
=6.605
• Step 6. Compare the test statistics with the critical values
the test statistic < the critical value
or the test statistic does not lie in th critical region.
• Step 7. Conclusion?
•
There is no sufficient evidence at the α=0.05 level of significance to reject
the null hypothesis, i.e., the claim of the distribution for the number of years
grandparents who are responsible for their grandchildren is the same today as
it was in 2000
Or ….