Transcript IB Math HL - William H. Peacock, LCDR USN, Ret.

Chi-Squared Significance
Tests
Chapters 26/27
Objectives:
•
•
•
•
Chi-Squared Distribution
Chi-Squared Test Statistic
Chi-Squared Goodness of Fit Test
Chi-Squared Test of Independence
A variable has a chi-square distribution if its distribution
has the shape of a special type of right-skewed curve, called
2

a chi-square ( ) curve. Actually, there are infinitely many
chi-square distributions, and we identify the chi-square
distribution in question by its number of degrees of freedom,
2

just as we did for t-distributions. Figure 12.1 shows three 2
curves and illustrates some basic properties of  -curves.
Chi-Square Table
Chi-Square (2) Test
for Goodness of Fit
Definition
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.
Expected Frequencies
If all expected frequencies are not all equal:
E=np
each expected frequency is found by
multiplying the sum of all observed
frequencies by the probability for the
category
Goodness-of-fit Test
Test Statistic
 =
2
(O – E)2
E
Critical Values
1.
Found in Table using k – 1 degrees of
freedom where k = number of categories
2.
Goodness-of-fit hypothesis tests are
always right-tailed.
A close agreement between observed
and expected values will lead to a small
value of 2 and a large P-value.
A large disagreement between
observed and expected values will
lead to a large value of 2 and a small
P-value.
A significantly large value of 2 will
cause a rejection of the null
hypothesis of no difference between
the observed and the expected.
Example
The U.S. Federal Bureau of Investigation (FBI) compiles data on
crimes and crime rates and publishes the information in Crime in
the United States. A violent crime is classified by the FBI as
murder, forcible rape, robbery, or aggravated assault. Table 12.1
gives a relative-frequency distribution for (reported) violent crimes
in 2000. For instance, in 2000, 28.6% of violent crimes were
robberies.
A random sample of 500 violent-crime reports from last year
yielded the frequency distribution shown in Table 12.2. Suppose
that we want to use the data in Tables 12.1 and 12.2 to decide
whether last year’s distribution of violent crimes has changed from
the 2000 distribution.
Solution Example
a. Formulate the problem statistically by posing it as a hypothesis
test.
b. Explain the basic idea for carrying out the hypothesis test.
c. Discuss the details for making a decision concerning the
hypothesis test.
Table 12.1
Table 12.2
Solution Example
a. The population is last year’s (reported) violent crimes.
The variable is “type of violent crime,” and its possible
values are murder, forcible rape, robbery, and
aggravated assault. We want to perform the
hypothesis test
H0 : Last year’s violent-crime distribution is the
same as the 2000 distribution.
Ha : Last year’s violent-crime distribution is different
from the 2000 distribution.
Solution Example
b. The idea behind the chi-square goodness-of-fit test is
to compare the observed frequencies in the second
column of Table 12.2 to the frequencies that would be
expected – the expected frequencies – if last year’s
violent-crime distribution is the same as the 2000
distribution. If the observed and expected frequencies
match fairly well, (i.e., each observed frequency is
roughly equal to its corresponding expected
frequency), we do not reject the null hypothesis;
otherwise, we reject the null hypothesis.
Solution Example
c.
To formulate a precise procedure for carrying out the
hypothesis test, we need to answer two questions:
1. What frequencies should we expect from a random
sample of 500 violent-crime reports from last year if
last year’s violent-crime distribution is the same as the
2000 distribution?
2. How do we decide whether the observed and
expected frequencies match fairly well?
The first question is easy to answer, which we
illustrate with robberies. If last year’s violent-crime
distribution is the same as the 2000 distribution, then,
according to Table 12.1, 28.6% of last year’s violent
crimes would have been robberies. Therefore, in a
random sample of 500 violent-crime reports from last
year, we would expect about 28.6% of the 500, or 143,
to be robberies.
Solution Example
c.
Compute each expected frequency using the formula E = np,
where n is the sample size and p is the relative frequency.
Calculations of the expected frequencies for all four types of
violent crime are shown in Table 12.3. The third column of Table
12.3 answers the first question. It gives the frequencies that we
would expect if last year’s violent-crime distribution is the same as
the 2000 distribution.
Table 12.3
Solution Example
c.
The second question – whether the observed and expected
frequencies match fairly well – is harder to answer. We need to
calculate a number that measures the goodness of fit. In Table
12.4, the second column repeats the observed frequencies from
the second column of Table 12.2. The third column of Table 12.4
repeats the expected frequencies from the third column of Table
12.3.
Table 12.4
Solution Example
c.
To measure the goodness of fit of the observed and expected
frequencies, we look at the differences, O − E, shown in the fourth
column of Table 12.4. Summing these differences to obtain a
measure of goodness of fit isn’t very useful because the sum is 0.
Instead, we square each difference (shown in the fifth column)
and then divide by the corresponding expected frequency. Doing
so gives the values (O − E)2/E, called chi-square subtotals, shown
in the sixth column. The sum of the chi-square subtotals,
O  E 
2
E  3.555
is the statistic used to measure the goodness of fit of the
observed and expected frequencies.
Solution Example
c.
If the null hypothesis is true, the observed and expected
frequencies should be roughly equal, resulting in a small value of
the test statistic,
(O − E)2/E. In other words, large values of
(O − E)2/E provide evidence against the null hypothesis.
As we have seen, (O − E)2/E = 3.555. Can this value be
reasonably attributed to sampling error, or is it large enough to
suggest that the null hypothesis is false? To answer this question,
we need to know the distribution of the test statistic
(O − E)2/E.
Key Fact
Procedure
Procedure (cont.)
Quick Tip
• The chi-square goodness of fit test is
always a right-tailed test.
• For the chi-square goodness-of-fit test,
the expected frequencies should be at
least 5.
• When the expected frequency of a class
or category is less than 5, this class or
category can be combined with another
class or category so that the expected
frequency is at least 5.
EXAMPLE
• There are 4 TV sets that are located in
the student center of a large university.
At a particular time each day, four
different soap operas (1, 2, 3, and 4) are
viewed on these TV sets. The
percentages of the audience captured by
these shows during one semester were
25 percent, 30 percent, 25 percent, and
20 percent, respectively. During the
first week of the following semester,
300 students are surveyed.
EXAMPLE (Continued)
• (a) If the viewing pattern has not
changed, what number of students is
expected to watch each soap opera?
• Solution: Based on the information, the
expected values will be: 0.25300 = 75,
0.30300 = 90, 0.25300 = 75, and
0.20300 = 60.
EXAMPLE (Continued)
• (b) Suppose that the actual observed
numbers of students viewing the soap
operas are given in the following table,
test whether these numbers indicate a
change at the 1 percent level of
significance.
EXAMPLE (Continued)
• Solution: Given  = 0.01, n = 4,
df = 4
– 1 = 3, 20.01, 3= 11.345. The observed
and expected frequencies are given
below
EXAMPLE (Continued)
• Solution (continued): The 2 test
statistic is computed below.
EXAMPLE (Continued)
• Solution (continued):
EXAMPLE (Continued)
• Solution (continued):
Diagram showing
the rejection
region.
The Chi-Square test for Goodness of Fit
2 Test of Independence
Definition
 Contingency Table (or two-way frequency table)
A contingency table is a table in which
frequencies correspond to two variables.
(One variable is used to categorize rows, and
a second variable is used to categorize
columns.)
Contingency tables have at least
two rows and at least two
columns.
Definition
 Test of Independence
This method tests the null
hypothesis that the row
variable and column variable
in a contingency table are not
related. (The null hypothesis
is the statement that the row
and column variables are
independent.)
Assumptions
1. The sample data are randomly selected.
2.
The null hypothesis H0 is the statement that
the row and column variables are
independent; the alternative hypothesis H1 is
the statement that the row and column
variables are dependent.
3.
For every cell in the contingency table, the
expected frequency E is at least 5. (There is
no requirement that every observed
frequency must be at least 5.)
Test of Independence
Test Statistic
 =
2
(O – E)2
E
Critical Values
1.
Found in Table using
degrees of freedom = (r – 1)(c – 1)
r is the number of rows and c is the number of
columns
2.
Tests of Independence are always righttailed.
Tests of Independence
H0: The row variable is independent of the
column variable
H1: The row variable is dependent (related to)
the column variable
This procedure cannot be used to establish a direct
cause-and-effect link between variables in question.
Dependence means only there is a relationship
between the two variables.
Expected Frequency for Contingency Tables
E=
table total
n
•
row total
table total
•
•
column total
table total
p
(probability of a cell)
E=
(row total) (column total)
(table total)
E=
(row total) (column total)
(table total)
Total number of all observed frequencies
in the table
Observed and Expected Frequencies
Survived
Died
Total
Men
Women
332
318
29
27
706
1360
104
35
18
1517
1692
422
64
45
2223
Boys Girls
Total
We will use the mortality table from the Titanic to find expected
frequencies. For the upper left hand cell, we find:
E=
(706)(1692)
2223
= 537.360
Observed and Expected Frequencies
Survived
Died
Women
Men
318
332
537.360
Boys Girls
29
27
Total
706
1360
104
35
18
1517
1692
422
64
45
2223
Total
Find the expected frequency for the lower left hand cell, assuming
independence between the row variable and the column variable.
E=
(1517)(1692)
2223
= 1154.640
Observed and Expected Frequencies
Survived
Died
Women Boys Girls
Men
29
318
27
332
537.360 134.022 20.326 14.291
Total
706
35
104
18
287.978 43.674 30.709
1517
64
2223
1360
1154.64
Total
1692
422
45
To interpret this result for the lower left hand cell, we can say that although 1360
men actually died, we would have expected 1154.64 men to die if survivablility is
independent of whether the person is a man, woman, boy, or girl.
Example: Using a 0.05 significance level, test the claim
that when the Titanic sank, whether someone survived or
died is independent of whether that person is a man,
woman, boy, or girl.
H0: Whether a person survived is independent of whether
the person is a man, woman, boy, or girl.
H1: Surviving the Titanic and being a man, woman, boy,
or girl are dependent.
Example: Using a 0.05 significance level, test the claim
that when the Titanic sank, whether someone survived or
died is independent of whether that person is a man,
woman, boy, or girl.
2= (332–537.36)2 + (318–132.022)2 + (29–20.326)2 + (27–14.291)2
14.291
537.36
134.022
20.326
+ (1360–1154.64)2 + (104–287.978)2 + (35–43.674)2 + (18–30.709)2
30.709
1154.64
43.674
287.978
2=78.481 + 252.555 + 3.702+11.302+36.525+117.536+1.723+5.260
= 507.084
Example: Using a 0.05 significance level, test the claim
that when the Titanic sank, whether someone survived or
died is independent of whether that person is a man,
woman, boy, or girl.
The number of degrees of freedom are (r–1)(c–1)=
(2–1)(4–1)=3.
2.05,3 = 7.815. We reject the null hypothesis.
Survival and gender are dependent.
Test Statistic
2 = 507.084
with  = 0.05 and (r
Critical Value
– 1) (c– 1) = (2 – 1) (4 – 1) = 3 degrees of freedom
2 = 7.815 (from Table )
Example
A national survey was conducted to obtain information on the alcohol
consumption patterns of U.S. adults by marital status. A random sample of
1772 residents, 18 years old and older, yielded the data displayed in Table
13.13. Suppose we want to use the data in Table 12.13 to decide whether
marital status and alcohol consumption are associated.
a. Formulate the problem statistically by posing it as a
hypothesis test.
b. Explain the basic idea for carrying out the hypothesis test.
c. Develop a formula for computing the expected
frequencies.
d. Construct a table that provides both the observed
frequencies in Table 12.13 and the expected frequencies.
e. Discuss the details for making a decision concerning the
hypothesis test.
Table 12.13
Solution Example
a. For a chi-square independence test, the null hypothesis is that the two
variables are not associated; the alternative hypothesis is that the two
variables are associated. Thus, we want to perform the hypothesis test
H0 : Marital status and alcohol consumption are not associated.
Ha : Marital status and alcohol consumption are associated.
b. The idea behind the chi-square independence test is to compare the
observed frequencies in Table 12.13 with the frequencies we would expect if
the null hypothesis of nonassociation is true. The test statistic for making the
comparison has the same form as the one used for the goodness-of- fit test:
χ2 = (O − E)2/E, where O represents observed frequency and E
represents expected frequency.
Solution Example
c. To develop a formula for computing the expected frequencies,
consider, for instance, the cell of Table 12.13 corresponding to “Married
and Abstain,” the cell in the second row and first column. We note that
the population proportion of all adults who abstain can be estimated by
the sample proportion of the 1772 adults sampled who abstain, that is,
by
If no association exists between marital status and alcohol
consumption (i.e., if H0 is true), then the proportion of married
adults who abstain is the same as the proportion of all adults
who abstain.
Solution Example
c. Therefore, of the 1173 married adults sampled, we would
590
expect about 
1173  390.6
1772
to abstain from alcohol.
Let’s rewrite the left side of this expected-frequency computation in a
slightly different way. By using algebra and referring to Table 12.13, we
obtain
590
Expected Frequency 
1173
1772
590 1173

1772
Row total Column total


Sample size
Solution Example
c. If we let R denote “Row total” and C denote “Column total,” we can
write this equation as
R C
E
,
n
where, as usual, E denotes expected frequency and n denotes sample
size.
d. Using Equation (12.1), we can calculate the expected frequencies
for all the cells in Table 12.13. For the cell in the upper right corner of the
table, we get
R  C 354  225
E

 44.9
n
1772
Solution Example
d. In Table 12.14 (on the next slide), we have modified
Table 12.13 by including each expected frequency beneath
the corresponding observed frequency. Table 12.14 shows,
for instance, that of the adults sampled, 74 were observed to
be single and consumed more than 60 drinks per month,
whereas if marital status and alcohol consumption are not
associated, the expected frequency is 44.9.
Table 12.14
Solution Example
e. If the null hypothesis of nonassociation is true, the observed and
expected frequencies should be approximately equal, which would result
in a relatively small value of the test statistic,
χ2= (O − E)2/E.
Consequently, if χ2 is too large, we reject the null hypothesis and
conclude that an association exists between marital status and alcohol
consumption. From Table 12.14, we find that
χ2= (O − E)2/E = 94.269
Can this value be reasonably attributed to sampling error, or is it large
enough to indicate that marital status and alcohol consumption are
associated? Before we can answer that question, we must know
the distribution of the
χ2-statistic.
Key Fact
Procedure
Procedure (cont.)
EXAMPLE
• A survey was done by a car
manufacturer concerning a
particular make and model. A group
of 500 potential customers were
asked whether they purchased
their current car because of its
appearance, its performance rating,
or its fixed price (no negotiating).
The results, broken down by gender
responses, are given on the next
slide.
EXAMPLE (Continued)
Question: Do females feel differently
than males about the three different
criteria used in choosing a car, or do
they feel basically the same?
Solution
• χ2 Test for independence.
• Thus the null hypothesis will be
that the criterion used is
independent of gender, while the
alternative hypothesis will be that
the criterion used is dependent on
gender.
Solution (continued)
• The degrees of freedom is given by
(number of rows – 1)(number of
columns – 1).
• df = (2 – 1)(3 – 1) = 2.
Solution (continued)
• Calculate the row and column totals.
These row and column are called
marginal totals.
Solution (continued)
• Computation of the expected values
• The expected value for a cell is the row
total times the column total divided by
the table total.
Solution (continued)
χ 02.01, 2
Let us use  = 0.01. So df = (2 –1)(3 –1) = 2 and 20.01,
2 = 9.210.
Solution (continued)
• The 2 test statistic is computed in the
same manner as was done for the
goodness-of-fit test.
Solution (continued)
Solution (continued)
• Diagram showing the rejection region.
Problems
• Review Set 8A – 12, 13
• Review Set 8B – 10,11