CHAPTER 15: CHI-SQUARED TESTS
TRUE/FALSE
1. A chi-squared distribution is symmetric.
2. A chi-squared test is used to describe a population of nominal data.
3. The total of the observed frequencies in a multinomial experiment must equal nk where n is the
number of trials and k is the number of categories.
4. In conducting a chi-squared goodness-of-fit test, an essential condition is that all expected frequencies
are at least five.
5. In a goodness-of-fit test, all of the proportions specified in the null hypothesis must be equal to each
other.
6. The alternative hypothesis in a goodness-of-fit test is that none of the pi values are equal to their values
specified in H0.
7. If the expected frequency of a cell is less than 5, you should increase the significance level.
8. If the expected frequency of a cell is less than 5, you should combine cells of the table.
9. For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance
of .025, the chi-squared table value is 20.4831. Suppose the value of your test statistic is 16.857. This
will lead you to reject the null hypothesis.
10. A left-tailed area in the chi-squared distribution equals .10. For 5 degrees of freedom the table value
equals 9.23635.
11. A left-tailed area in the chi-squared distribution equals .90. For 10 degrees of freedom the table value
equals 15.9871.
12. For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance
of .05, the test statistics is 25.168. The chi-squared value from the table is 21.0261. These results will
lead us to reject the null hypothesis.
13. A chi-squared goodness-of-fit test is always a two-tailed test.
14. If the observed frequencies are all smaller than the expected frequencies in a goodness-of-fit test, the
chi-squared test statistic will be negative.
15. The only way the chi-squared test statistic can be zero is if the observed frequencies are all exactly the
same as the expected frequencies.
16. All of the expected frequencies in a chi-squared goodness-of-fit test must be equal to each other.
17. A small chi-squared test statistic in a goodness-of-fit test supports the null hypothesis.
18. You cannot use a chi-squared goodness-of-fit test when there are only two possible outcomes for each
trial in your experiment.
19. If there are only two categories, the chi-squared goodness-of-fit test is the same as the z-test for p, the
population proportion (as long as the sample/cell sizes meet the conditions).
20. A multinomial experiment with two categories is identical to a binomial experiment.
MULTIPLE CHOICE
1. To determine the critical values in the chi-squared distribution table, you need to know the:
a.
sample size.
c.
probability of Type II error.
b.
degrees of freedom
d.
All of these choices are true.
2. Suppose the value of your chi-squared test statistic in a goodness-of-fit test is equal to 0. What do you
conclude?
a.
Reject H0. Conclude that at least one proportion is not equal to its specified value.
b.
Fail to reject H0. Not enough evidence to say the proportions are different from what is
listed in H0.
c.
Not enough information; need the degrees of freedom for the test.
d.
None of these choices.
3. To determine whether data were drawn from a multinomial distribution with certain proportions, you
use a:
a.
chi-squared goodness-of-fit test.
c.
chi-squared test for normality.
b.
chi-squared test of a contingency table.
d.
None of these choices.
4. A chi-squared goodness-of-fit test is always conducted as a(n):
a.
lower-tail test.
c.
two-tail test.
b.
upper-tail test.
d.
All of these choices are true.
5. If each element in a population is classified into one and only one of several categories, the population
is:
a.
normal.
c.
chi-squared.
b.
multinomial.
d.
None of these choices.
6. Of the values for a chi-squared test statistic listed below, which one is most likely to lead to rejecting
the null hypothesis in a goodness-of-fit test?
a.
0
c.
1.96
b.
.05
d.
45
7. Which of the following represents H1 in a chi-squared goodness-of-fit test to see if all 5 colors of a
certain candy appear in the same proportion in the population?
a.
H1: p1 = p2 = p3 = p4 = p5 = .20.
b.
H1: At least one proportion is not equal to .20.
c.
H1: None of these proportions are equal.
d.
None of these choices.
8. How do you calculate the expected frequency for one cell in a goodness-of-fit test?
a.
The expected frequency is equal to the proportion specified in H0 for that cell.
b.
Use the total number of observations divided by the number of categories.
c.
Multiply the specified proportion for that cell (found in H0) by the total sample size.
d.
None of these choices.
9. How does a multinomial distribution differ from a binomial distribution?
a.
A binomial has only two possible categories and a multinomial can have more.
b.
A binomial has a fixed number of n trials. A multinomial has a fixed number of nk trials,
where k is the number of categories.
c.
The probabilities in a binomial distribution are always p and 1 − p. The trials in a
multinomial distribution are always p/k and (1 − p/k).
d.
All of these choices are true.
10. If the expected frequency ei for any cell i is less than 5, we should:
a.
choose another sample with five or more observations.
b.
use the normal distribution instead of the chi-squared distribution.
c.
combine cells such that each observed frequency fi is 5 or more.
d.
increase the degrees of freedom by 5.
11. The sampling distribution of the test statistic for a goodness-of-fit test with k categories is a:
a.
chi-squared distribution with k − 1 degrees of freedom.
b.
normal distribution.
c.
Student t-distribution with k − 1 degrees of freedom.
d.
None of these choices.
12. Which of the following conditions indicate that H0 should be rejected in a goodness-of-fit test?
a.
The observed frequencies are equal to their expected frequencies.
b.
The test statistic is large.
c.
The degrees of freedom is large.
d.
All of these choices are true.
13. Consider a multinomial experiment with 200 trials, where the outcome of each trial is classified into
one of 5 categories. The number of degrees of freedom associated with the chi-squared goodness-of-fit
test equals:
a.
195
c.
5
b.
40
d.
4
14. A left tail area in the chi-squared distribution equals .99. For df = 8, the table value equals:
a.
20.090
c.
2.7326
b.
3.4895
d.
15.5073
15. Which of the following statements regarding the chi-squared distribution is true?
a.
The chi-squared distribution is skewed to the right.
b.
All values of the chi-squared distribution are greater than or equal to zero.
c.
The critical region for a goodness-of-fit test with k categories is
2 >
2, k-1, where
2 is
the value of the test statistic.
d.
All of these choices are true.
16. Which of the following is not a characteristic of a multinomial experiment?
a.
The experiment consists of a fixed number of trials.
b.
The outcome of each trial is classified into one of two possible categories.
c.
The probability pi that the outcome will fall into cell i remains constant for each trial.
d.
Each trial of the experiment is independent of the other trials.
17. If we use the
is equal to:
a.
3
c.
5
b.
4
d.
None of these choices.
18. The rule of five requires that the:
a.
observed frequency for each cell must be at least 5.
b.
degrees of freedom for the test must be at least 5.
c.
expected frequency for each cell must be at least 5.
d.
difference between the observed and expected frequency for each cell must be at least 5.
COMPLETION
1. A(n) ____________________ experiment is like a binomial experiment except it contains two or more
categories.
2. When k = 2 the ____________________ experiment is identical to the ____________________
experiment.
3. The chi-squared goodness-of-fit test compares the ____________________ frequencies in the table to
the ____________________ frequencies based on the null hypothesis.
4. If the expected frequencies and the observed frequencies are quite different, you are likely to
____________________ the null hypothesis.
5. A chi-squared distribution has a shape that is ____________________.
6. The values of a chi-squared distribution are always ____________________ zero.
7. A test statistic that lies in the far right tail of the chi-squared distribution indicates you will
____________________ H0.
8. The rule of ____________________ states that in order to conduct the chi-squared goodness-of-fit
test, the expected value for each cell must be ____________________ or more.
9. The rule of five states that in order to conduct the chi-squared goodness-of-fit test, the
____________________ value for each cell must be five or more.
10. The alternative hypothesis of a goodness-of-fit test states that ____________________ of the
proportions is not equal to its value specified in H0.
SHORT ANSWER
1. Five types of apples are displayed side by side in several supermarkets in the city of Miami. It was
noted that in one day, 180 customers purchased apples. Of these, 30 picked type A, 40 picked type B,
25 picked type C, 35 picked type D, and 50 picked type E. In Miami, can you conclude at the 5%
significance level that there is a preferred type of apples?
2. In 2011, the student body of a state university in Alabama consists of 30% freshmen, 25%
sophomores, 27% juniors, and 18% seniors. A sample of 400 students taken from the 2012 student
body showed that there are 138 freshmen, 88 sophomores, 94 juniors, and 80 seniors. Test with 5%
significance level to determine whether the student body proportions have changed.
3. In 2011, Brand A MP3 Players had 45% of the market, Brand B had 35%, and Brand C had 20%. This
year the makers of brand C launched a heavy advertising campaign. A random sample of electronic
stores shows that of 10,000 MP3 Players sold, 4,350 were Brand A, 3,450 were Brand B, and 2,200
were Brand C. Has the market changed? Test at
= .01.
4. Consider a multinomial experiment involving 100 trials and 3 categories (cells). The observed
frequencies resulting from the experiment are shown in the accompanying table.
Category
1
2
3
Frequency
38
35
27
Use the 5% significance level to test the hypotheses H0: p1 = .45, p2 = .30, p3 = .25 vs. H1: At least one
proportion differs from their specified values.
5. Consider a multinomial experiment involving 160 trials 4 categories (cells). The observed frequencies
resulting from the experiment are shown in the accompanying table.
Category
1
2
3
4
Frequency
53
35
30
42
Use the 10% significance level to test the hypotheses: H0: p1 = p2 = p3 = p4 = .25 vs. H1: At least one
proportion differs from their specified values.
6. A calculus professor posted the following grade distribution guidelines for her elementary calculus
class: 8% A, 35% B, 40% C, 12% D, and 5% F. A sample of 100 elementary statistics grades at the
end of last semester showed 12 As, 30 Bs, 35 Cs, 15 Ds, and 8 Fs. Test at the 5% significance level to
determine whether the actual grades deviate significantly from the posted grade distribution guidelines.
7. In 2011, computers of Brand A controlled 25% of the market, Brand B 20%, Brand C 10%, and brand
D 45%. In 2015, sample data was collected from many randomly selected stores throughout the
country. Of the 1,200 computers sold, 280 were Brand A, 270 were Brand B, 90 were Brand C, and
560 were Brand D. Has the market changed since 2011? Test at the 1% significance level.
8. A cable company prepared four versions of a set of instructions for hooking up a TV. The company
asked a sample of 1,600 people which one of the four forms was easiest to understand. In the sample,
425 people preferred Form A, 385 preferred Form B, 375 preferred Form C, and 415 preferred Form D.
At the 5% level of significance, can one conclude that in the population there is a preferred form?
9. Explain what is meant by the rule of five and what you should do if this rule is not met.
10. Consumer panel preferences for three proposed Cafes are as follows:
Cafe A
Cafe B
Cafe C
48
62
40
Use 0.05 level of significance and test to see if there is a preference among the three Cafes, according
to the data.
11. A Deli proposes to serve 4 main Sandwiches. For planning purposes, the manager expects that the
proportions of each that will be selected by her customers will be:
Selection
Proportion
Turkey
.50
Roast Beef
.20
Pastrami
.10
Tuna
.20
Of a random sample of 100 customers, 44 selected chicken, 24 selected roast beef, 13 selected
Pastrami, and 10 selected tuna. Should the manager revise her estimates? Use
= .01.
12. To determine whether a single coin is fair, the coin was tossed 200 times. The observed frequencies
with which each of the two sides of the coin turned up are recorded as 112 heads and 88 tails. Is the
coin fair? Use a=.05
Student Absenteeism
Consider a multinomial experiment involving n = 200 students of a large high school. The attendance
department recorded the number of students who were absent during the weekdays. The null
hypothesis to be tested is: H0: p1 = .10, p2 = .25, p3 = .30, p4 = .20, p5 = .15.
13. {Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the
following frequencies:
Day of the Week
Mon.
Tues.
Wed.
Thurs.
Fri.
Number Absent
16
44
56
48
36
14. {Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the
following frequencies: (n = 100)
Day of the Week
Mon.
Tues.
Wed.
Thurs.
Fri.
Number Absent
8
22
28
24
18
15. {Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the
following frequencies: (n = 50)
Day of the Week
Mon.
Tues.
Wed.
Thurs.
Fri.
Number Absent
4
11
14
12
9
16. {Student Absenteeism Narrative} Review the previous results. What is the effect of decreasing the
sample size?
Sales Volumes
A telemarketer makes five calls per day. A sample of 200 days gives the frequencies of sales volumes
listed below:
Number of Sales
Observed Frequency (days)
0
10
1
38
2
69
3
63
4
18
5
2
Assume the population is binomial distribution with a probability of purchase p equal to .50.
17. {Sales Volumes Narrative} Compute the expected frequencies for x = 0, 1, 2, 3, 4, and 5 by using the
binomial probability function or the binomial tables. Combine categories if necessary to satisfy the
rule of five.
18. {Sales Volumes Narrative} Should the assumption of a binomial distribution be rejected at the 5%
significance level?
19. In a test of a contingency table, rejecting the null hypothesis concludes the variables are not
independent.
20. To produce expected values for a test of a contingency table, you multiply estimated joint probabilities
for each cell by the total sample size, n.
21. To calculate the expected values in a test of a contingency table, you assume that the null hypothesis is
true.
22. In a goodness-of-fit test, H0 lists specific values for proportions and the test of a contingency table
does not.
23. If two events A and B are independent, the P(A and B) = P(A) + P(B).
24. The expected frequency for the cell in row i and column j is the row i total plus the row j total, all
divided by n.
25. In the test of a contingency table, the expected cell frequencies must satisfy the rule of 5.
26. In the test of a contingency table, the observed cell frequencies must satisfy the rule of 5.
27. The test statistic for the chi-squared test of a contingency table is the same as the test statistic for the
goodness-of-fit test.
28. A chi-squared test of a contingency table is applied to a contingency table with 3 rows and 4 columns
for two qualitative variables. The degrees of freedom for this test must be 12.
29. A chi-squared test of a contingency table is applied to a contingency table with 4 rows and 4 columns
for two qualitative variables. The degrees of freedom for this test must be 9.
30. A chi-squared test of a contingency table with 10 degrees of freedom results in a test statistic of
17.894. Using the chi-squared table, the most accurate statement that can be made about the p-value
for this test is that .05 < p-value < .10.
31. In a chi-squared test of a contingency table, the value of the test statistic was
2 = 15.652, and the
critical value at
= .025 was 11.1433. Thus, we must reject the null hypothesis at
= .025.
32. A chi-squared test of a contingency table with 6 degrees of freedom results in a test statistic of 13.25.
Using the chi-squared table, the most accurate statement that can be made about the p-value for this
test is that p-value is greater than .025 but smaller than .05.
33. The degrees of freedom for the test statistic in a test of a contingency table is (r − 1)(c − 1) where r is
the number of rows in the table, and c is the number of columns.
34. Contingency tables are used in:
a.
testing independence of two samples.
b.
testing dependence in matched pairs.
c.
testing independence of two qualitative variables in a population.
d.
describing a single population.
35. The chi-squared test of a contingency table is based upon:
a.
one quantitative variable.
b.
two quantitative variables.
c.
one qualitative variable.
d.
two qualitative variables.
36. To address whether two variables are related in a contingency table, the null hypothesis, H0, says that
a.
The two variables are independent.
b.
The two variables are dependent.
c.
The two variables are equal.
d.
None of these choices.
37. To address whether two variables are related in a contingency table, the alternative hypothesis, H1, is:
a.
The two variables are independent.
b.
The two variables are dependent.
c.
The two variables are equal.
d.
None of these choices.
38. How do you find the probabilities needed to obtain expected frequencies for a test of a contingency
table?
a.
If there are r rows and c columns, let each probability be 1/rc.
b.
Use the probabilities specified in the null hypothesis.
c.
Assume H0 is true and use your data to calculate them.
d.
None of these choices.
39. The number of degrees of freedom for a contingency table with r rows and c columns is:
a.
r + c
b.
rc
c.
(r − 1)(c − 1)
d.
None of these choices.
40. If you reject H0 in a test of a contingency table, you conclude that based on your data:
a.
The two nominal variables are independent.
b.
The two nominal variables are equal.
c.
The two nominal variables have the same proportions listed in H0.
d.
None of these choices.
41. A large chi-squared test statistic in a test of a contingency table means you conclude:
a.
The two nominal variables are dependent.
b.
The two nominal variables are equal.
c.
The two nominal variables have the same proportions listed in H0.
d.
None of these choices.
42. A chi-squared test statistic in a test of a contingency table that is equal to zero means:
a.
The two nominal variables are equal.
b.
The two nominal variables are independent.
c.
The two nominal variables have the same proportions listed in H0.
d.
All of these choices.
43. The number of degrees of freedom for a contingency table with 4 rows and 8 columns is
a.
32
b.
12
c.
21
d.
10
44. A chi-squared test of a contingency table with 6 degrees of freedom results in a test statistic
2 =
13.58. Using the
2 tables, the most accurate statement that can be made about the p-value for this test
is that:
a.
p-value > .10
b.
p-value > .05
c.
.05 < p-value < .10
d.
.025 < p-value < .05
45. Which statistical technique is appropriate when we wish to analyze the relationship between two
qualitative variables with two or more categories?
a.
The chi-squared test of a multinomial experiment.
b.
The chi-squared test of a contingency table.
c.
The t-test of the difference between two means.
d.
The z-test of the difference between two proportions.
46. A chi-squared test of a contingency table with 4 rows and 5 columns shows that the value of the test
statistic is 22.18. Using a chi-squared table, the most accurate statement that can be made is:
a.
p-value is greater than 0.05
b.
p-value is smaller than 0.025
c.
p-value is greater than 0.025 but smaller than 0.05
d.
p-value is greater than 0.10
47. The chi-squared test of a(n) ____________________ is used to determine whether there is enough
evidence to say two nominal variables are related.