1: Chi-Square Tests for Univariate Data
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12.1 Chi-Square Tests for Univariate Data
575
The hypotheses to be tested have the form
H0: p1 ϭ hypothesized proportion for Category 1
p2 ϭ hypothesized proportion for Category 2
(
pk ϭ hypothesized proportion for Category k
Ha: H0 is not true, so at least one of the true category proportions differs
from the corresponding hypothesized value.
For the example involving responses to the tax survey, let
p1 ϭ the proportion of all taxpayers who will definitely pay by credit card
p2 ϭ the proportion of all taxpayers who will probably pay by credit card
p3 ϭ the proportion of all taxpayers who will probably not pay by credit card
and
p4 ϭ the proportion of all taxpayers who will definitely not pay by credit card
The null hypothesis of interest is then
H0: p1 ϭ .25, p2 ϭ .25, p3 ϭ .25, p4 ϭ .25
A null hypothesis of the type just described can be tested by first selecting a random
sample of size n and then classifying each sample response into one of the k possible
categories. To decide whether the sample data are compatible with the null hypothesis,
we compare the observed cell counts (frequencies) to the cell counts that would have
been expected when the null hypothesis is true. The expected cell counts are
Expected cell count for Category 1 ϭ np1
Expected cell count for Category 2 ϭ np2
and so on. The expected cell counts when H0 is true result from substituting the corresponding hypothesized proportion for each pi.
Anthony Ise/PhotoDisc/Getty Images//
Cengage Learning/Getty Images
E X A M P L E 1 2 . 1 Births and the Lunar Cycle
A common urban legend is that more babies than expected are born during certain
phases of the lunar cycle, especially near the full moon. The paper “The Effect of the
Lunar Cycle on Frequency of Births and Birth Complications” (American Journal
of Obstetrics and Gynecology [2005]: 1462–1464) classified births according to the
lunar cycle. Data for a sample of randomly selected births occurring during 24 lunar
Lunar Phase
New moon
Waxing crescent
First quarter
Waxing gibbous
Full moon
Waning gibbous
Last quarter
Waning crescent
Number
of Days
Number
of Births
24
152
24
149
24
150
24
152
7,680
48,442
7,579
47,814
7,711
47,595
7,733
48,230
Data set available online
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576
Chapter 12 The Analysis of Categorical Data and Goodness-of-Fit Tests
cycles consistent with summary quantities appearing in the paper are given in the accompanying table.
Let’s define lunar phase category proportions as follows:
p1 ϭ proportion of births that occur during the new moon
p2 ϭ proportion of births that occur during the waxing crescent moon
p3 ϭ proportion of births that occur during the first quarter moon
p4 ϭ proportion of births that occur during the waxing gibbous moon
p5 ϭ proportion of births that occur during the full moon
p6 ϭ proportion of births that occur during the waning gibbous moon
p7 ϭ proportion of births that occur during the last quarter moon
p8 ϭ proportion of births that occur during the waning crescent moon
If there is no relationship between number of births and the lunar cycle, then the
number of births in each lunar cycle category should be proportional to the number
of days included in that category. Since there are a total of 699 days in the 24 lunar
cycles considered and 24 of those days are in the new moon category, if there is no
relationship between number of births and lunar cycle,
p1 5
24
5 .0343
699
Similarly, in the absence of any relationship,
p2 5
152
5 .2175
699
p3 5
24
5 .0343
699
p4 5
149
5 .2132
699
p5 5
24
5 .0343
699
p7 5
24
5 .0343
699
150
5 .2146
699
152
p8 5
5 .2175
699
p6 5
The hypotheses of interest are then
H0: p1 ϭ .0343, p2 ϭ .2175, p3 ϭ .0343, p4 ϭ .2132, p5 ϭ .0343, p6 ϭ .2146,
p7 ϭ .0343, p8 ϭ .2175
Ha: H0 is not true.
There were a total of 222,784 births in the sample, so if H0 is true, the expected
counts for the first two categories are
a
expected count
hypothesized proportion
b 5 na
b
for new moon
for new moon
5 222,784 1.03432 5 7641.49
a
expected count
hypothesized proportion
b 5 na
b
for waxing crescent
for waxing crescent
5 222,784 1.21752 5 48,455.52
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12.1 Chi-Square Tests for Univariate Data
577
Expected counts for the other six categories are computed in a similar fashion, and
observed and expected cell counts are given in the following table.
Lunar Phase
New moon
Waxing crescent
First quarter
Waxing gibbous
Full moon
Waning gibbous
Last quarter
Waning crescent
Observed Number
of Births
Expected Number
of Births
7,680
48,442
7,579
47,814
7,711
47,595
7,733
48,230
7,641.49
48,455.52
7,641.49
47,497.55
7,641.49
47,809.45
7,641.49
48,455.52
Because the observed counts are based on a sample of births, it would be somewhat surprising to see exactly 3.43% of the sample falling in the first category, exactly
21.75% in the second, and so on, even when H0 is true. If the differences between
the observed and expected cell counts can reasonably be attributed to sampling variation, the data are considered compatible with H0. On the other hand, if the discrepancy between the observed and the expected cell counts is too large to be attributed
solely to chance differences from one sample to another, H0 should be rejected in
favor of Ha. To make a decision, we need an assessment of how different the observed
and expected counts are.
The goodness-of-fit statistic, denoted by X 2, is a quantitative measure of the extent to which the observed counts differ from those expected when H0 is true. (The
Greek letter x is often used in place of X. The symbol X 2 is referred to as the chisquare [x2] statistic. In using X 2 rather than x2, we are adhering to the convention of
denoting sample quantities by Roman letters.)
The goodness-of-fit statistic, X 2, results from first computing the quantity
1observed cell count 2 expected cell count2 2
expected cell count
for each cell, where, for a sample of size n,
a
expected cell
hypothesized value of corresponding
b 5 na
b
count
population proportion
The X 2 statistic is the sum of these quantities for all k cells:
1observed cell count 2 expected cell count2 2
X2 5 a
expected cell count
all cells
The value of the X 2 statistic reflects the magnitude of the discrepancies between
observed and expected cell counts. When the differences are sizable, the value of X 2
tends to be large. Therefore, large values of X 2 suggest rejection of H0. A small value
of X 2 (it can never be negative) occurs when the observed cell counts are quite similar
to those expected when H0 is true and so would be consistent with H0.
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578
Chapter 12
The Analysis of Categorical Data and Goodness-of-Fit Tests
As with previous test procedures, a conclusion is reached by comparing a P-value
to the significance level for the test. The P-value is computed as the probability of
observing a value of X 2 at least as large as the observed value when H0 is true. This
requires information about the sampling distribution of X 2 when H0 is true.
When the null hypothesis is correct and the sample size is sufficiently large, the
behavior of X 2 is described approximately by a chi-square distribution. A chi-square
curve has no area associated with negative values and is asymmetric, with a longer tail
on the right. There are actually many chi-square distributions, each one identified
with a different number of degrees of freedom. Curves corresponding to several chisquare distributions are shown in Figure 12.1.
df = 8
df = 12
df = 20
FIGURE 12.1
Chi-square curves.
For a test procedure based on the X 2 statistic, the associated P-value is the area under
the appropriate chi-square curve and to the right of the computed X 2 value. Appendix
Table 8 gives upper-tail areas for chi-square distributions with up to 20 df. Our chisquare table has a different appearance from the t table used in previous chapters. In the
t table, there is a single “value” column on the far left and then a column of P-values (tail
areas) for each different number of degrees of freedom. A single column of t values works
for the t table because all t curves are centered at 0, and the t curves approach the z curve
as the number of degrees of freedom increases. However, because the chi-square curves
move farther and farther to the right and spread out more as the number of degrees of
freedom increases, a single “value” column is impractical in this situation.
Chi-square curve for 4 df
Shaded area = .085
FIGURE 12.2
A chi-square upper-tail area.
8.18
To find the area to the right of a particular X 2 value, locate the appropriate df
column in Appendix Table 8. Determine which listed value is closest to the X 2 value
of interest, and read the right-tail area corresponding to this value from the left-hand
column of the table. For example, for a chi-square distribution with df ϭ 4, the area
to the right of X 2 ϭ 8.18 is .085, as shown in Figure 12.2. For this same chi-square
distribution (df ϭ 4), the area to the right of 9.70 is approximately .045 (the area to
the right of 9.74, the closest entry in the table for df ϭ 4).
It is also possible to use computer software or a graphing calculator to compute
areas under a chi-square curve. This provides more accurate values for the area.
Goodness-of-Fit Tests
When H0 is true, the X 2 goodness-of-fit statistic has approximately a chi-square
distribution with df ϭ (k Ϫ 1), as long as none of the expected cell counts are too
small. When expected counts are small, and especially when an expected count is less
1observed cell count 2 expected cell count2 2
can be inflated
than 1, the value of
expected cell count
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12.1 Chi-Square Tests for Univariate Data
579
because it involves dividing by a small number. It is generally agreed that use of the
chi-square distribution is appropriate when the sample size is large enough for every expected cell count to be at least 5. If any of the expected cell frequencies are less than 5,
categories can be combined in a sensible way to create acceptable expected cell counts.
Just remember to compute the number of degrees of freedom based on the reduced
number of categories.
Goodness-of-Fit Test Procedure
Hypotheses:
Test statistic:
H0: p1 ϭ hypothesized proportion for Category 1
(
pk ϭ hypothesized proportion for Category k
Ha: H0 is not true
1observed cell count 2 expected cell count2 2
expected cell count
all cells
X2 5 a
P-values: When H0 is true and all expected counts are at least 5, X 2 has
approximately a chi-square distribution with df ϭ k Ϫ 1. Therefore, the
P-value associated with the computed test statistic value is the area to the
right of X 2 under the df 5 k 2 1 chi-square curve. Upper-tail areas for
chi-square distributions are found in Appendix Table 8.
Assumptions:
1. Observed cell counts are based on a random sample.
2. The sample size is large. The sample size is large
enough for the chi-square test to be appropriate as
long as every expected cell count is at least 5.
E X A M P L E 1 2 . 2 Births and the Lunar Cycle Revisited
We use the births data of Example 12.1 to test the hypothesis that number of births
is unrelated to lunar cycle. Let’s use a .05 level of significance and the nine-step
hypothesis-testing procedure illustrated in previous chapters.
1. Let p1, p2, p3, p4, p5, p6, p7, and p8 denote the proportions of all births falling in
the eight lunar cycle categories as defined in Example 12.1.
2. H0: p1 5 .0343, p2 ϭ .2175, p3 ϭ .0343, p4 ϭ .2132, p5 ϭ .0343, p6 ϭ .2146,
p7 ϭ .0343, p8 ϭ .2175
3. Ha: H0 is not true.
4. Significance level: a ϭ .05.
1observed cell count 2 expected cell count2 2
expected cell count
all cells
5. Test statistic: X 2 5 a
6. Assumptions: The expected cell counts (from Example 12.1) are all greater than
5. The births represent a random sample of births occurring during the lunar
cycles considered.
7. Calculation:
X2 5
17680 2 7641.492 2
148230 2 48455.522 2
148442 2 48455.522 2
1
1%1
7641.49
48455.52
48455.52
5 .194 1 .004 1 .511 1 2.108 1 .632 1 .962 1 1.096 1 1.050
5 6.557
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580
Chapter 12
The Analysis of Categorical Data and Goodness-of-Fit Tests
8. P-value: The P-value is based on a chi-square distribution with df ϭ 8 Ϫ 1 ϭ
7. The computed value of X 2 is smaller than 12.01 (the smallest entry in the
df ϭ 7 column of Appendix Table 8), so P-value Ͼ .10.
9. Conclusion: Because P-value Ͼ a, H0 cannot be rejected. There is not sufficient
evidence to conclude that number of births and lunar cycle are related. This is
consistent with the conclusion in the paper: “We found no statistical evidence that
deliveries occurred in a predictable pattern across the phases of the lunar cycle.”
Statistical software can be used to perform a chi-square goodness-of-fit test.
Minitab output for the data and hypothesized proportions of this example is shown
here.
Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Number of Births
Using category names in Lunar Phase
Category
First Quarter
Full Moon
Last Quarter
New Moon
Waning Crescent
Waning Gibbous
Waxing Crescent
Waxing Gibbous
N
DF
222784
7
Observed
7579
7711
7733
7680
48230
47595
48442
47814
Chi-Sq
6.55683
Test
Proportion
0.0343
0.0343
0.0343
0.0343
0.2175
0.2146
0.2175
0.2132
P-Value
0.476
Expected
7641.5
7641.5
7641.5
7641.5
48455.5
47809.4
48455.5
47497.5
Contribution
to Chi-Sq
0.51105
0.63227
1.09584
0.19406
1.04961
0.96189
0.00377
2.10835
Note that Minitab has reordered the categories from smallest to largest based
on the observed count. Minitab also carried a bit more decimal accuracy in the computation of the chi-square statistic, reporting X 2 ϭ 6.55683 and an associated
P-value of .476. The computed P-value ϭ .476 is consistent with the statement
P-value Ͼ .10 from Step 8 of the hypothesis test.
E X A M P L E 1 2 . 3 Hybrid Car Purchases
USA Today (“Hybrid Car Sales Rose 81% Last Year,” April 25, 2005) reported the
top five states for sales of hybrid cars in 2004 as California, Virginia, Washington,
Florida, and Maryland. Suppose that each car in a sample of 2004 hybrid car sales is
classified by state where the sale took place. Sales from states other than the top five
were excluded from the sample, resulting in the accompanying table.
State
California
Virginia
Washington
Florida
Maryland
Total
Data set available online
Observed Frequency
250
56
34
33
33
406
(The given observed counts are artificial, but they are consistent with hybrid sales
figures given in the article.)
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12.1 Chi-Square Tests for Univariate Data
581
We will use the X 2 goodness-of-fit test and a significance level of a ϭ .01 to
test the hypothesis that hybrid sales for these five states are proportional to the
2004 population for these states. 2004 population estimates from the Census Bureau web site are given in the following table. The population proportion for each
state was computed by dividing each state population by the total population for
all five states.
State
2004 Population
Population Proportion
35,842,038
7,481,332
6,207,046
17,385,430
5,561,332
72,477,178
0.495
0.103
0.085
0.240
0.077
California
Virginia
Washington
Florida
Maryland
Total
If these same population proportions hold for hybrid car sales, the expected
counts are
Expected count for California
Expected count for Virginia
Expected count for Washington
Expected count for Florida
Expected count for Maryland
5
5
5
5
5
406 1.4952
406 1.1032
406 1.0852
406 1.2402
406 1.0772
5
5
5
5
5
200.970
41.818
34.510
97.440
31.362
These expected counts have been entered in Table 12.1.
T AB LE 12 .1 Observed and Expected Counts for Example 12.3
State
California
Virginia
Washington
Florida
Maryland
Observed Counts
Expected Counts
250
56
34
33
33
200.970
41.818
34.510
97.440
31.262
1. Let p1, p2, ... , p5 denote the actual proportion of hybrid car sales for the five states
in the following order: California, Virginia, Washington, Florida, and
Maryland.
2. H0: p1 5 .495, p2 ϭ .103, p3 ϭ .085, p4 ϭ .240, p5 ϭ .077
3. Ha: H0 is not true.
4. Significance level: a ϭ .01
1observed cell count 2 expected cell count2 2
expected cell count
all cells
5. Test statistic: X 2 5 a
6. Assumptions: The sample was a random sample of hybrid car sales. All expected
counts are greater than 5, so the sample size is large enough to use the chi-square
test.
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582
Chapter 12 The Analysis of Categorical Data and Goodness-of-Fit Tests
7. Calculation: From Minitab
Chi-Square Goodness-of-Fit Test for Observed
Counts in Variable: Hybrid Sales
Using category names in State
Category
California
Florida
Maryland
Virginia
Washington
N
DF
406
4
Observed
250
33
33
56
34
Chi-Sq
59.4916
Test
Proportion
0.495
0.240
0.077
0.103
0.085
P-Value
0.000
Expected
200.970
97.440
31.262
41.818
34.510
Contribution
to Chi-Sq
11.9617
42.6161
0.0966
4.8096
0.0075
8. P-value: All expected counts exceed 5, so the P-value can be based on a chisquare distribution with df ϭ 5 Ϫ 1 ϭ 4. From Minitab, the P-value is 0.000.
9. Conclusion: Since P-value Յ a, H0 is rejected. There is convincing evidence that
hybrid sales are not proportional to population size for at least one of the five
states.
Based on the hybrid sales data, we have determined that there is convincing evidence
that at least one of these five states has hybrid sales that are not proportional to population size. Looking back at the Minitab output, notice that there is a column
labeled “Contribution to Chi-Sq.” This column shows the individual values of
1observed cell count 2 expected cell count2 2
, which are summed to produce the
expected cell count
value of the chi-square statistic. Notice that the two states with the largest contribution to the chi-square statistic are Florida and California. For Florida, observed hybrid sales were smaller than expected (observed 5 33, expected 5 97.44), whereas for
California observed sales were higher than expected (observed 5 250, expected 5
200.970).
EX E RC I S E S 1 2 . 1 - 1 2 . 1 3
12.1 From the given information in each case below,
state what you know about the P-value for a chi-square
test and give the conclusion for a significance level of
a ϭ .01.
d. X 2 ϭ 21.3, df ϭ 4
a. X 2 ϭ 7.5, df ϭ 2
2
e. X 2 ϭ 5.0, df ϭ 3
b. X ϭ 13.0, df ϭ 6
2
c. X ϭ 18.0, df ϭ 9
12.2 A particular paperback book is published in a
choice of four different covers. A certain bookstore keeps
copies of each cover on its racks. To test the hypothesis
that sales are equally divided among the four choices, a
random sample of 100 purchases is identified.
Bold exercises answered in back
Data set available online
a. If the resulting X 2 value were 6.4, what conclusion
would you reach when using a test with significance
level .05?
b. What conclusion would be appropriate at significance level .01 if X 2 5 15.3?
c. If there were six different covers rather than just
four, what would you conclude if X 2 5 13.7 and a
test with a ϭ .05 was used?
12.3 Packages of mixed nuts made by a certain company contain four types of nuts. The percentages of nuts
of Types 1, 2, 3, and 4 are supposed to be 40%, 30%,
20%, and 10%, respectively. A random sample of nuts is
selected, and each one is categorized by type.
Video Solution available
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12.1 Chi-Square Tests for Univariate Data
a. If the sample size is 200 and the resulting test statistic value is X 2 5 19.0, what conclusion would be
appropriate for a significance level of .001?
b. If the random sample had consisted of only 40 nuts,
would you use the chi-square test here? Explain your
reasoning.
12.4
The article “In Bronx, Hitting Home Runs Is A
Breeze” (USA Today, June 2, 2009) included a classification of 87 home runs hit at the new Yankee Stadium
according to direction that the ball was hit, resulting in
the accompanying data.
Direction
Number of
Home Runs
Left Left
Center Right Right
Field Center
Center Field
18
10
7
18
The authors of the paper “Racial Stereotypes in
Children’s Television Commercials” (Journal of Advertising Research [2008]: 80–93) counted the number of
times that characters of different ethnicities appeared in
commercials aired on Philadelphia television stations, resulting in the data in the accompanying table.
Ethnicity
Observed
Frequency
AfricanAmerican
57
Asian Caucasian Hispanic
11
330
6
Number of Correct Identifications
0
1
2
Observed Count
21
10
13
a. Can a person identify her roommate by smell? If not,
the data from the experiment should be consistent
with what we would have expected to see if subjects
were just guessing on each trial. That is, we would
expect that the probability of selecting the correct shirt
would be 1/3 on each of the two trials. It would then
be reasonable to regard the number of correct identifications as a binomial variable with n ϭ 2 and p ϭ 1/3.
Use this binomial distribution to compute the proportions of the time we would expect to see 0, 1, and 2
correct identifications if subjects are just guessing.
b. Use the three proportions computed in Part (a) to
carry out a test to determine if the numbers of correct identifications by the students in this study are
significantly different than what would have been
expected by guessing. Use a ϭ .05. (Note: One of
the expected counts is just a bit less than 5. For purposes of this exercise, assume that it is OK to proceed with a goodness-of-fit test.)
The paper “Cigarette Tar Yields in Relation to
Mortality from Lung Cancer in the Cancer Prevention
Study II Prospective Cohort” (British Medical Journal
[2004]: 72–79) included the accompanying data on the
12.7
Based on the 2000 Census, the proportion of the U.S.
population falling into each of these four ethnic groups are
.177 for African-American, .032 for Asian, .734 for Caucasian, and .057 for Hispanic. Do the data provide sufficient
evidence to conclude that the proportions appearing in
commercials are not the same as the census proportions?
Test the relevant hypotheses using a significance level of .01.
12.6 The paper “Sociochemosensory and Emotional
Functions” (Psychological Science [2009]: 1118–1124)
Bold exercises answered in back
describes an interesting experiment to determine if college
students can identify their roommates by smell. Forty-four
female college students participated as subjects in the experiment. Each subject was presented with a set of three
t-shirts that were identical in appearance. Each of the three
t-shirts had been slept in for at least 7 hours by a person
who had not used any scented products (like scented deodorant, soap, or shampoo) for at least 48 hours prior to
sleeping in the shirt. One of the three shirts had been worn
by the subject’s roommate. The subject was asked to identify the shirt worn by her roommate. This process was then
repeated with another three shirts, and the number of times
out of the two trials that the subject correctly identified the
shirt worn by her roommate was recorded. The resulting
data is given in the accompanying table.
34
a. Assuming that it is reasonable to regard this sample
of 87 home runs as representative of home runs hit
at Yankee Stadium, carry out a hypothesis test to
determine if there is convincing evidence that the
proportion of home runs hit is not the same for all
five directions.
b. Write a few sentences describing how the observed
counts for the five directions differ from what would
have been expected if the proportion of home runs is
the same for all five directions.
12.5
583
Data set available online
tar level of cigarettes smoked for a sample of male smokers who subsequently died of lung cancer.
Tar Level
Frequency
0–7 mg
8–14 mg
15–21 mg
Ն22 mg
103
378
563
150
Video Solution available
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584
Chapter 12
The Analysis of Categorical Data and Goodness-of-Fit Tests
Assume it is reasonable to regard the sample as representative of male smokers who die of lung cancer. Is there
convincing evidence that the proportion of male smoker
lung cancer deaths is not the same for the four given tar
level categories?
12.8
The paper referenced in the previous exercise
also gave the accompanying data on the age at which
smoking started for a sample of 1031 men who smoked
low-tar cigarettes.
Age
Frequency
Ͻ16
16–17
18–20
Ն21
237
258
320
216
not equally likely to occur in each of the 3-hour time
periods used to construct the table? Test the relevant
hypotheses using a significance level of .05.
b. Suppose a safety office proposes that bicycle fatalities
are twice as likely to occur between noon and midnight as during midnight to noon and suggests the
following hypothesis: H0: p1 5 1/3, p2 ϭ 2/3, where
p1 is the proportion of accidents occurring between
midnight and noon and p2 is the proportion occurring between noon and midnight. Do the given data
provide evidence against this hypothesis, or are the
data consistent with it? Justify your answer with an
appropriate test. (Hint: Use the data to construct a
one-way table with just two time categories.)
12.10
The report referenced in the previous exercise
(“Fatality Facts 2004: Bicycles”) also classified 719 fatal
a. Use a chi-square goodness-of-fit test to test the null
hypothesis H0: p1 ϭ .25, p2 ϭ .2, p3 ϭ .3, p4 ϭ .25,
where p1 ϭ proportion of male low-tar cigarette
smokers who started smoking before age 16, and p2,
p3, and p4 are defined in a similar way for the other
three age groups.
b. The null hypothesis from Part (a) specifies that half
of male smokers of low-tar cigarettes began smoking
between the ages of 16 and 20. Explain why p2 ϭ .2
and p3 ϭ .3 is consistent with the ages between 16
and 20 being equally likely to be when smoking
started.
The report “Fatality Facts 2004: Bicycles” (Insurance Institute, 2004) included the following table
12.9
classifying 715 fatal bicycle accidents according to time
of day the accident occurred.
Time of Day
Midnight to 3 a.m.
3 a.m. to 6 a.m.
6 a.m. to 9 a.m.
9 a.m. to Noon
Noon to 3 p.m.
3 p.m. to 6 p.m.
6 p.m. to 9 p.m.
9 p.m. to Midnight
Number of Accidents
38
29
66
77
99
127
166
113
a. Assume it is reasonable to regard the 715 bicycle accidents summarized in the table as a random sample
of fatal bicycle accidents in 2004. Do these data support the hypothesis that fatal bicycle accidents are
Bold exercises answered in back
Data set available online
bicycle accidents according to the month in which the
accident occurred, resulting in the accompanying table.
Month
Number of Accidents
January
February
March
April
May
June
July
August
September
October
November
December
38
32
43
59
78
74
98
85
64
66
42
40
a. Use the given data to test the null hypothesis
H0: p1 5 1/12, p2 ϭ 1/12, ... , p12 ϭ 1/12, where p1
is the proportion of fatal bicycle accidents that occur
in January, p2 is the proportion for February, and so
on. Use a significance level of .01.
b. The null hypothesis in Part (a) specifies that fatal
accidents were equally likely to occur in any of the
12 months. But not all months have the same number of days. What null and alternative hypotheses
would you test to determine if some months are
riskier than others if you wanted to take differing
month lengths into account? (Hint: 2004 was a leap
year, with 366 days.)
c. Test the hypotheses proposed in Part (b) using a .05
significance level.
Video Solution available
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12.2 Tests for Homogeneity and Independence in a Two-way Table
12.11 An article about the California lottery that appeared in the San Luis Obispo Tribune (December 15,
1999) gave the following information on the age distribution of adults in California: 35% are between 18 and
34 years old, 51% are between 35 and 64 years old, and
14% are 65 years old or older. The article also gave information on the age distribution of those who purchase
lottery tickets. The following table is consistent with the
values given in the article:
Age of Purchaser
18–34
35–64
65 and over
Frequency
36
130
34
12.12 A certain genetic characteristic of a particular
plant can appear in one of three forms (phenotypes). A
researcher has developed a theory, according to which
12.2
the hypothesized proportions are p1 5 .25, p2 5 .50, and
p3 5 .25. A random sample of 200 plants yields X 2 5
4.63.
a. Carry out a test of the null hypothesis that the theory is correct, using level of significance a ϭ .05.
b. Suppose that a random sample of 300 plants had
resulted in the same value of X 2. How would your
analysis and conclusion differ from those in Part (a)?
The article “Linkage Studies of the Tomato”
(Transactions of the Royal Canadian Institute [1931]:
1–19) reported the accompanying data on phenotypes
12.13
resulting from crossing tall cut-leaf tomatoes with dwarf
potato-leaf tomatoes. There are four possible phenotypes: (1) tall cut-leaf, (2) tall potato-leaf, (3) dwarf cutleaf, and (4) dwarf potato-leaf.
Suppose that the data resulted from a random sample of
200 lottery ticket purchasers. Based on these sample
data, is it reasonable to conclude that one or more of
these three age groups buys a disproportionate share of
lottery tickets? Use a chi-square goodness-of-fit test with
a ϭ .05.
Bold exercises answered in back
585
Phenotype
Frequency
1
2
3
4
926
288
293
104
Mendel’s laws of inheritance imply that p1 ϭ 9/16, p2 ϭ
3/16, p3 ϭ 3/16, and p4 ϭ 1/16. Are the data from this
experiment consistent with Mendel’s laws? Use a .01
significance level.
Data set available online
Video Solution available
Tests for Homogeneity and Independence
in a Two-way Table
Data resulting from observations made on two different categorical variables can also
be summarized using a tabular format. As an example, suppose that residents of a
particular city can watch national news on affiliate stations of ABC, CBS, NBC, or
PBS. A researcher wishes to know whether there is any relationship between political
philosophy (liberal, moderate, or conservative) and preferred news program among
those residents who regularly watch the national news. Let x denote the variable political philosophy and y the variable preferred network. A random sample of 300 regular
watchers is selected, and each individual is asked for his or her x and y values. The
data set is bivariate and might initially be displayed as follows:
Observation
1
2
3
(
299
300
x Value
Liberal
Conservative
Conservative
(
Moderate
Liberal
y Value
CBS
ABC
PBS
(
NBC
PBS
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.