CHAPTER 13 REVIEW
1. The following are the approximate percentages for the different blood types among white Americans:
A: 40%; B: 11%; AB: 4%; O: 45%. A random sample of 1000 black Americans yielded the following
blood type data: A: 270; B: 200; AB: 40; O: 490. Does this sample indicate that the distribution of
blood types among black Americans differs from that of white Americans?
ANS:
H0: ρA = 0.40, ρB = 0.11, ρAB = 0.04, ρO = 0.45
Ha: the actual population proportions are different from those in H0.
Blood type
A
B
AB
O
Observed values
270
200
40
490
Expected Values
(0.4)(1000) = 400
(0.11)(1000) = 110
(0.04)(1000) = 40
(0.45)(1000) = 450
(O  E )2 (270  400) 2
(490  450) 2

 ... 
 119.44
E
400
450
df = 4 – 1 = 3
From the χ2 table, we see that 119.44 is larger than any value for 3 df. Hence, P value < 0.0005. [The
TI-83 gives a P value of 1.02 x 10-25 (χ2cdf(119.44, 1E99, 3).]
Because P value < 0.05, we reject the null and conclude that not all the proportions are as hypothesized
in the null. We have very strong evidence that the proportions of the various blood types among black
Americans differ from the proportions among white Americans.
2  
2. The statistics teacher, Ms. Uong, used her calculator to simulate rolling a die 96 times and storing the
results in a list L1. She did this by entering MATH PRB randInt(1, 6, 96)  L1. Next she stored the
list (STAT SortA (L1)). She then counted the number of each face value. The results were as follows:
Face value
1
2
3
4
5
6
Observed
19
15
10
14
17
21
Does it appear that the teacher’s calculator is simulating a fair die?
ANS:
H0: p0 = p2 = p3 = p4 = p5 = p6 = 1/6
Ha: Not all of the proportions are equal to 1/6.
Face value
1
2
3
4
5
6
Observed
19
15
10
14
17
21
Expected
(1/6)(96) = 16
(1/6)(96) = 16
(1/6)(96) = 16
(1/6)(96) = 16
(1/6)(96) = 16
(1/6)(96) = 16
χ2 = 4.75, df = 6 – 1 = 5  P-value > 0.25 (table) or P value = 0.45 (calculator [χ2cdf(4.75, 1E99, 5)])
Because the P value is > 0.05, we have no evidence to reject the null. We cannot conclude that the
calculator is failing to simulate a fair die.
3. A study of 150 cities wanted to determine if crime rate is related to outdoor temperature. The results
of the study are summarized in the following table:
Temperature
Below
Normal
Above
Crime Rate
Below
Normal
12
8
35
41
4
7
Above
5
24
14
Do these data provide evidence, at the 0.02 level of significance, that the crime rate is related to the
temperature at the time of the crime?
ANS:
H0: The crime rate is independent of temperature.
Ha: The crime rate is not independent of temperature.
Below
Normal
Temperature
Above
Total
Crime Rate
Below
Normal
12
8
(8.5)
(9.33)
35
41
(34)
(37.33)
4
7
(8.5)
(9.33)
51
56
Above
5
(7.17)
24
(28.67)
14
(7.17)
43
Total
25
100
25
150
(12  8.5) 2 (8  9.33) 2
(14  7.17) 2
 

 ... 
 12.92
8.5
9.33
7.17
df = (3 – 1)(3 – 1) = 4  0.01 < P value < 0.02 (table). [P value = 0.012 (calculator).]
Because P value < 0.02, we reject H0. We have strong evidence that the number of crimes committed is
related to the temperature at the time of the crime.
2
4. A study yields a chi-square statistic value of 20 (χ2 = 20). What is the P value of the test if
(a) the study was a goodness-of-fit test with n = 12?
(b) the study was a test of independence between two categorical variables, the row variable with 3
values and the column variable with 4 values?
ANS:
(a) n = 12  df = 12 – 1 = 11  0.025 < P value < 0.05
[Using the calculator: χ2cdf(20, 1E99, 11) = 0.045.]
(b) r = 3, c = 4  df = (3 – 1)(4 – 1) = 6  0.0025 < P value < 0.005
[Using the calculator: χ2cdf(20, 1E99, 6) = 0.0028.]
5 – 7 The following data were collected while conducting a chi-square test for independence:
Preference
Male
Female
Brand A
16
18*
Brand B
22
30
Brand C
15
28
5. What null hypothesis is being tested?
ANS:
H0: Gender and preference are independent in the population.
6. What is the expected value of the cell marked with the *?
ANS:
Preference
Brand A
Brand B
Brand C
Male
16
22
15
Female
18*
30
28
(20.03)
Total
34
52
43
The expected value =
Total
53
76
129
76(34)
 20.03
129
7. How many degrees of freedom are involved in the test?
ANS:
df = (2 – 1)(3 – 1) = 2
8. A χ2 goodness-of-fit is performed on a random sample of 360 individuals to see if the number of
birthdays each month is proportional to the number of days in the month. χ2 is determined to be 23.5.
The P value for this test is
(a) 0.001 < P < 0.005
(b) 0.02 < P < 0.025
(c) 0.025 < P < 0.05
(d) 0.01 < P < 0.02
(e) 0.05 < P < 0.10
ANS: D
Because n = 12 months, df = 12 – 1 = 11. Reading from the χ2 table, we have 0.01 < P < 0.02. [Using
the calculator, χ2cdf(23.5, 1E99, 11) = 0.0150
9. For the following two-way table, compute the value of χ2.
A
B
(a) 2.63
(b) 1.22
(c) 1.89
(d) 2.04
(e) 1.45
ANS: E
C
15
10
D
25
30
C
15
(12.5)
10
(12.5)
25
A
B
Total
D
25
(27.5)
30
(27.5)
55
Total
40
40
80
(15  12.5) 2
(30  27.5) 2
 
 ... 
 1.45
12.5
27.5
2
10. An AP Statistics students noted that the probability distribution for a binomial random variable with
n = 4 and p = 0.3 is approximately given by:
n
0
1
2
3
4
P
0.24
0.41
0.27
0.08
0.01
The student decides to test the randBin function on her calculator by putting 500 values into a list
from this function (randBin(4, 0.3, 500)  L1) and counting the number of each outcome. She obtained
n
0
1
2
3
4
Observed
110
190
160
36
4
Do these data indicate that the “randBin” function on the calculator is failing to correctly generate the
correct quantities of 0, 1, 2, and 3 from this distribution?
ANS:
Ho: p0 = 0.24, p1 = 0.41, p2 = 0.27, p3 = 0.07, p4 = 0.01
Ha: Not all of the proportions stated in Ho are correct.
n
0
1
2
3
4
Expected
0.24(500) = 120
0.41(500) = 205
0.27(500) = 135
0.08(500) = 40
0.01(500) = 5
(O  E ) 2
 7.16, df  5  1  4  0.10  P  0.15
E
[On the calculator, χ2cdf(7.16, 1E99, 4) = 0.128
The P-value is greater than 0.05, which is too high to reject the null. We do not have strong evidence
that the calculator is not performing as it should.
2  
11. A chi-square test for the homogeneity of proportions is conducted on three populations and one
categorical variable that has four values. Computation of the chi-square statistic yields χ2 = 17.2. Is this
finding significant and the 0.01 level of significance?
ANS:
df = 4 – 1 = 3
χ2cdf(17.2, 1E99, 3) = 0.009
The finding is significant at the 0.01 level of significance because 0.009 < 0.01.
12. Which of the following best described the difference between a test for independence and a test for
homogeneity of proportions?
(a) There is no difference because they both produce the same value of the chi-square test statistic.
(b) A test for independence has one population and two categorical variables, whereas a test for
homogeneity of proportions has more than one population and only one categorical variable.
(c) A test for homogeneity of proportions has one population and two categorical variables; whereas, a
test for independence has more than one population and only one categorical variable.
(d) A test for independence uses count data when calculating chi-square and a test for homogeneity uses
percentages or proportions when calculating chi-square.
ANS: B
13. Restaurants in two parts of a major city were compared on customer satisfaction to see if location
influences customer satisfaction. A random sample of 38 patrons from the Big Steak Restaurant in the
eastern part of town and another random sample of 36 patrons from the Big Steak Restaurant on the
western side of town were interviewed for the study. The restaurants are under the same management,
and the researcher established that they are virtually identical in terms of décor and service. The results
are presented in the following table.
Patron’s Ratings of Restaurants
Excellent
Good
Fair
Poor
Eastern
10
12
11
5
Western
6
15
7
8
Do these data provide good evidence that location influences customer satisfaction?
ANS:
Ho: The proportions of patrons from each side of town that rate the restaurant Excellent, Good, Fair,
and Poor are same for eastern and western.
Ha: Not all the proportions are the same.
Eastern
Western
Total
Excellent
10
(8.22)
6
(7.78)
16
Good
12
(13.86)
15
(13.14)
27
Fair
11
(9.24)
7
(8.76)
18
Poor
5
(6.68)
8
(6.32)
13
Total
38
(O  E ) 2
 2.86, df  (2  1)( 4  1)  3  P  0.25 (from table)
E
[Use calculator, χ2cdf(2.86, 1E99, 3) = 0.41]
The P-value is large. We cannot reject the null. These data do not provide us with evidence that
location influences customer satisfaction.
2  
36
74
14. The number of defects from a manufacturing process by day of the week are as follows:
Number
Monday
36
Tuesday
23
Wednesday
26
Thursday
25
Friday
40
The manufacturing is concerned that the number of defects is greater on Monday and Friday. Test, at
the 0.05 level of significance, the claim that the proportion of defects is the same each day of the week.
ANS:
Ho: p1 = p2 = p3 = p4 = p5 (the proportion of defects is the same each day of the week.)
Ha: At least one proportion is not equal to the others (the proportion of defects is not the same each day
of the week.)
The number of expected defects is the same for each day of the week. Because there were a total of 150
defects during the week, we would expect, if the null is true, to have 30 defects each day.
Observed
Expected
Monday
36
30
Tuesday
23
30
Wednesday
26
30
Thursday
25
30
Friday
40
30
(O  E ) 2
 7.533, df  5  1  4  0.10  P  0.15 (from table.)
E
[Use calculator, χ2cdf(7.533, 1E99, 4) = 0.11]
The P value is too large to reject the null. We do not have strong evidence that there are more defects
produced on Monday and Friday than on other days of the week.
2  
15. A study was done on opinions concerning the legalization on marijuana at Mile High College. One
hundred fifty-seven respondents were randomly selected from a large pool of faculty, students, and
parents at the college. Respondents were given a choice of favoring the legalization of marijuana,
opposing the legalization of marijuana, or favoring making marijuana a legal but controlled substance.
The results of the survey were as follows.
Students
Faculty
Parents
Favor
Organization
Oppose
Legalization
17
33
5
9
40
8
Favor
Legalization
with Control
6
27
12
Do these data support, at the 0.05 level, the contention that they type of respondent (student, faculty, or
parent) is related to the opinion toward legalization? Is this a test of independence or a test of
homogeneity of proportions?
ANS:
Because we have a single proportion from which we drew our sample and we are asking if two variables
are related within that population, this is a chi-square test for independence.
Ho: Type of respondent and opinion toward the legalization of marijuana are independent.
Ha: Type of respondent and opinion toward the legalization of marijuana are not independent.
Students
Favor
Organization
Oppose
Legalization
17
9
Favor
Legalization
with Control
6
Total
32
Faculty
Parents
Total
(11.21)
33
(35.03)
5
(18.76)
55
(11.62)
40
(36.31)
8
(36.31)
57
(9.17)
27
(28.66)
12
(7.16)
45
100
25
157
(O  E ) 2
 
 10.27, df  (3  1)(3  1)  4  0.025  P  0.05 (from table).
E
[Use calculator, χ2cdf(10.27, 1E99, 4) = 0.036]
Because P-value < 0.05, we reject Ho. We have evidence that the type of respondent is related to
opinion concerning the legalization of marijuana.
2
16. In a recent year, at the 6 p.m. time slot, television channels 2, 3, 4, and 5 captured the entire
audience with 30%, 25%, 20%, and 25%, respectively. During the first week of the next season, 500
viewers are interviewed.
a) If viewer preferences have not changed, what number of persons is expected to watch each channel?
b) Suppose that the actual observed numbers are as follows:
2
3
4
5
139
138
112
111
Do these numbers indicate a change? Are the differences significant?
ANS:
a) Expected:
2
3
4
5
(0.30)(500) = (0.25)(500) = (0.20)(500) = (0.25)(500) =
150
125
100
125
b) Ho: The television audience is distributed over channels 2, 3, 4, and 5 with percentages 30%, 25%,
20%, and 25%, respectively.
Ha: The audience distribution is not 30%, 25%, 20%, and 25%, respectively.
(139  150) 2
(111  125) 2
2 
 ... 
 5.167
150
125
 02.05,3  7.81
Since 5.167 < 7.81, we do not have sufficient evidence to reject the null hypothesis.
17. A grocery store manager wishes to determine whether a certain product will sell equally well in any
of five locations in the store. Five displays are set up, one in each location, and the resulting numbers of
the product sold are noted.
1
43
2
29
3
52
4
34
5
48
Is there enough evidence that location makes a difference? Test at both the 5% and 10% significance
levels.
ANS
Ho: Sales of the product are uniformly distributed over the five locations.
Ha: Sales are not uniformly distributed over the five locations.
A total of 43 + 29 + 52 + 34 + 48 = 206 units were sold. If location does not matter, we would expect
206/5 = 41.2 units sold per location (uniform distribution.)
Expected:
1
2
3
4
5
41.2
41.2
41.2
41.2
41.2
Thus,
(43  41.2) 2
(48  41.2) 2
2 
 ... 
 8.903
41.2
41.2
 02.05, 4  9.49
 02.10, 4  7.78
Since 8.903 < 9.49, we do not have sufficient evidence to reject the null hypotheses at the 5% level.
However, at the 10% level, 8.903 > 7.78, we do have sufficient evidence to reject the null hypothesis.
18. In a nationwide telephone poll of 1000 adults representing Democrats, Republicans, and
Independent, respondents were asked if their confidence in the U.S. banking system had been shaken by
the savings and loan crisis. The answers, cross-classified by party affiliation, are given in the following
two-way table.
Democrats
Republicans
Independents
Yes
175
150
75
No
220
165
105
No Opinion
55
35
20
Test the null hypothesis that shaken confidence in the banking system is independent of party affiliation.
Use a 10% significance level.
ANS:
Ho: Party affiliation and shaken confidence in the banking system are independent.
Ha: Party affiliation and shaken confidence in the banking system are not independent.
Democrats
Republicans
Independents
Total
Yes
175
(180)
150
(140)
75
(80)
400
No
220
(220.5)
165
(171.5)
105
(98)
490
No Opinion
55
(49.5)
35
(38.5)
20
(22)
110
Total
450
350
200
1000
(175  180) 2
(20  22) 2
 
 ... 
 3.024
180
22
df  (3  1)(3  1)  4
2
 02.10, 4  7.78
Since 3.024 < 7.78, there is not sufficient evidence to reject the null hypothesis of independence. Thus,
at the 10% significance level, we cannot claim a relationship between party affiliation and shaken
confidence in the banking system.
19. To determine whether men with a combination of childhood abuse and a certain abnormal gene are
more likely to commit violent crimes, a study is run on a simple random sample of 575 males in the 25
to 35 age group. The data are summarized in the following table:
Criminal behavior
Normal behavior
Not abused,
normal gene
48
201
Abused, normal
gene
21
79
Not abused,
abnormal gene
32
118
Abused, abnormal
gene
26
50
Is there evidence of a relationship between the four categories (based on childhood abuse and abnormal
genetics) and behavior (criminal versus normal)? Explain.
ANS:
Criminal
behavior
Normal
behavior
Total
Not abused,
normal gene
48
(55)
201
(194.0)
249
Abused,
normal gene
21
(22.1)
79
(77.9)
100
Not abused,
abnormal gene
32
(33.1)
118
(116.9)
150
Abused,
abnormal gene
26
(16.8)
50
(59.2)
76
Total
127
448
575
Ho: The four categories (based on childhood abuse and abnormal genetics) and behavior (criminal
versus normal) are independent.
Ha: The four categories (based on childhood abuse and abnormal genetics) and behavior (criminal
versus normal) are not independent.
(48  55) 2
(50  59.2) 2
 
 ... 
 7.752
55
59.2
df  (2  1)( 4  1)  3
2
 02.05,3  7.81
Since 7.752 < 7.81, the data do not provide some evidence to reject Ho and conclude that there is not
evidence of a relationship between the four categories (based on childhood and abnormal genetics) and
behavior (criminal versus normal).
20. In a large city, a group of AP Statistics students work together on a project to determine which
group of school employees has the greatest proportion who are satisfied with their jobs. In independent
simple random sample of 100 teachers, 60 administrators, 45 custodians, and 55 secretaries, the numbers
satisfied with their jobs were found to be 82, 38, 34, and 36, respectively. Is there evidence that the
proportion of employees satisfied with their jobs is different in different school system job categories?
ANS:
Ho: The proportion of employees satisfied with their jobs is the same across the various school system
job categories.
Ha: At least two of the job categories differ in the proportion of employees satisfied with their jobs.
The observed and expected counts are as follows:
Teachers
Administrators
Satisfied
82
(73.1)
28
Not satisfied
18
(26.9)
22
Total
100
60
Custodians
Secretaries
Total
(43.8)
34
(32.9)
36
(40.2)
190
(16.2)
11
(12.1)
19
(14.8)
70
45
55
260
(82  73.1) 2
(19  14.8) 2
 
 ... 
 8.640
73.1
14.8
df  (4  1)( 2  1)  3
2
 02.05,3  7.81
Since 8.640 > 7.81, there is sufficient evidence to reject Ho and we can conclude that there is evidence
that the proportion of employees satisfied with their jobs is not the same across all the school system job
categories.
21. The following table provides the responses of a group of 100 children shown three different toys
and asked which one they liked the best. Based on the data, is there evidence that one of the three shows
a difference in the preference between the boys and girls at the 5% significance level?
Boys
Girls
Toy A
25
9
Toy B
24
25
Toy C
11
6
ANS:
Ho. There is no difference in the preference for toys between boys and girls.
Ha: There is a difference in the preference for toys between boys and girls.
Boys
Girls
Total
Toy A
25
(20.4)
9
(13.6)
34
Toy B
24
(29.4)
25
(19.6)
49
Toy C
11
(10.2)
6
(6.8)
17
Total
60
40
100
(25  20.4) 2
(6  6.8) 2
 ... 
 5.23
20.4
6.8
df  (2  1)(3  1)  2
2 
 02.05, s  7.81
Since 5.23 < 7.81, we do not have evidence against the null hypothesis; therefore, boys and girls do not
differ in their choice of toys.
22. In a study of exercise habits in men working in the health care profession in Chicago, researchers
classified the 356 sampled employees according to the level of education they completed and their
exercise habits. The researchers want to ascertain if there is an association between the level of
education completed and exercise habits at the 5% significance level. The data from the study are
provided in the table below.
Regularly
Occasionally
Never
College
51
92
68
Some College
22
21
9
HS
43
28
22
ANS:
Ho: There is no association between educational level completed and exercise habits for men working
in the health care profession in Chicago.
Ha: There is an association between educational level completed and exercise habits for men working in
the health care profession in Chicago.
Regularly
Occasionally
Never
Total
College
51
(68.75)
92
(83.57)
68
(58.68)
211
Some College
22
(16.94)
21
(20.60)
9
(14.46)
52
HS
43
(30.30)
28
(36.83)
22
(25.86)
93
Total
116
141
99
356
(51  68.75) 2
(22  25.86) 2
 ... 
 18.5097
68.75
25.86
df  (3  1)(3  1)  4
2 
 02.05, 4  9.49
Since 18.5097 > 9.49, there is evidence to reject the null hypothesis; in other words, there is evidence of
an association between level of education and exercise habits among men working in the health care
profession in Chicago.
The next set of questions refers to the following situation:
A survey was conducted to investigate the severity of rodent problems in egg and poultry operations. A
random sample of operators was selected, and the operators were classified according to the type of
operation and the extent of the rodent population. A total of 78 egg operators and 53 turkey operators
were classified and the summary information is:
23. The value of the test statistic is:
(a) about 5.99
(b) about 9.71
(c) about 6.81
(d) about 5.64
(e) about 8.60
Solution: d
24. The expected count in the (egg, mild infestation) cell is:
(a) about 26.00
(b) about 33.33
(c) about 53.00
(d) about 31.55
(e) about 78.00
Solution: d
25. The approximate p-value is found to be:
(a) about .060
(b) about .014
(c) about .032
(d) about .008
(e) about .05
Solution: a
The next set of questions refers to the following situation
In the paper “Color Association of Male and Female Fourth-Grade School Children” (J. Psych., 1988,
383-8), children were asked to indicate what emotion they associated with the color red. The response
and the sex of the child are noted and summarized below. The first number in each cell is the count, the
second number is the row percent.
Frequency|
Row Pct | anger | happy | love | pain | Total
-------------+---------+---------+-------+-------+
f
| 27
| 19 | 39 | 17 | 102
| 26.47 | 18.63 | 38.24 | 16.67 |
-------------+---------+---------+-------+-------+
m
| 34
| 12 | 38 | 28 | 112
| 30.36 | 10.71 | 33.93 | 25.00 |
-------------+---------+--------+--------+-------+
Total
61
31
77
45
214
26. Under a suitable null hypothesis, the expected frequency for the cell corresponding to Anger and
Males is:
(a) 15.9
(b) 55.7
(c) 30.4
(d) 31.9
(e) 29.1
Solution: d
27. The null hypothesis will be rejected at α =0.05 if the test statistic exceeds:
(a) 3.84
(b) 5.99
(c) 7.81
(d) 9.49
(e) 14.07
Solution: c
28. The approximate p-value is:
(a) Between .100 and .900
(b) Between .050 and .100
(c) Between .025 and .050
(d) Between .010 and .025
(e) Between .005 and .010
Solution: a
29. A random sample of 100 members of a union are asked to respond to two questions: Question 1.
Are you happy with your financial situation? Question 2. Do you approve of the Federal government’s
economic policies?
The responses are:
Question 1.
Yes No
|
Total
Question
Yes 22
48
|
70
2
No
12
18
|
30
Total 34
66
|
100
To test the null hypothesis that response to Question 1 is independent of response to Question 2 at 5%
level, the expected frequency for the cell (Yes, Yes) and the critical value of the associated test statistic
are:
(a) 23.8 and 1.96 respectively
(b) 10.2 and 3.84 respectively
(c) 23.8 and 3.84 respectively
(d) 23.8 and 7.81 respectively
(e) 10.2 and 7.81 respectively
Solution: c
30. A survey was conducted to investigate whether alcohol consumption and smoking are related. The
following information was compiled for 600 individuals:
Smoker
Non-smoker
Drinker
193
165
Non-drinker
89
153
Which of the following statements is true?
(a) The appropriate alternative hypothesis is A: Smoking and Alcohol Consumption are independent.
(b) The appropriate null hypothesis is H: Smoking and Alcohol Consumption are not independent.
(c) The calculated value of the test statistic is 3.84.
(d) The calculated value of the test statistic is 7.86.
(e) At level .01 we conclude that smoking and alcohol consumption are related.
Solution: e
31. Doctors’ practices have been categorized as to being Urban, Rural, or Intermediate. The number of
doctors who prescribed tetracycline to at least one patient under the age of 8 were recorded for each of
these practice areas. The results are:
Urban Intermediate Rural
Tetracycline
95
74
31
No tetracycline
126
84
30
If the county type of practice and the use of tetracycline are independent, then the expected number of
rural doctors who prescribe tetracycline is:
(a) 31.0
(b) 27.7
(c) 1.37
(d) 51
(e) 62
Solution: b
32. For the problem outlined above, the critical value (table E) of the test statistic when the level of
significance is α =0.05, is:
(a) 0.1026
(b) 7.3778
(c) 5.9915
(d) 12.5916
(e) 7.8147
Solution: c
The next set of questions refers to the following situation:
A study was conducted to determine if the fatality rate depends on the size of the automobile. The
analysis of accidents is as follows (with some values hidden):
DEATH
SIZE
FREQUENCY |
m | s | L | TOTAL
-------------------+-----------+-------+-------+
no
|
63 | 128 | 46 | 237
-------------------+-----------+--------+--------+
yes
|
26 | 95 | 16 | 137
-------------------+-----------+-------+--------+
TOTAL
89
223
62
374
33. Under a suitable null hypothesis, the expected frequency for the cell corresponding to fatal type of
accident and small size automobile is:
(a) 81.68
(b) 67.00
(c) 61.43
(d) 63.41
(e) 59.72
Solution: a
34. The null hypothesis will be rejected at α =0.05 if the test statistic exceeds:
(a) 12.59
(b) 7.81
(c) 5.99
(d) 3.84
(e) 9.49
Solution: c
35. The approximate p-value is:
(a) less than 0.005
(b) between 0.005 and 0.010
(c) between 0.010 and 0.025
(d) between 0.025 and 0.050
(e) between 0.050 and 0.100
Solution: c