CHAPTER 15

CHI­SQUARED TESTS

MULTIPLE CHOICE QUESTIONS

In the following  multiple­choice questions,  please  circle the  correct answer.

1. Which   statistical   technique   is   appropriate   when   we   describe   a   single   population   of  

qualitative  data  with  exactly two  categories?
a. z­test  of a population  proportion
b. The chi­squared  test of a multinomial  experiment
c. The chi­squared  test of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: d

2. If we  want  to conduct  a two­tail test of a population  proportion,  we  can employ:

a. z­test  of a population  proportion
b. the chi­squared  test  of a binomial  experiment  since  z 2 = χ 2
c. the chi­squared  test  of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: d

3. Which  statistical   technique   is   appropriate   when  we  wish   to   analyze  the   relationship  
between  two  qualitative  variables  with  two  or more  categories?
a. The chi­squared  test of a multinomial  experiment
b. The chi­squared  test of a contingency  table
c. The t­test of the difference  between  two  means
d. Both a and  b
e. Both b and  c
ANSWER: b

72
73 Chapter Sixteen

4. If we  want  to conduct  a one­tail test of a population  proportion,  we  can employ:

a. z­test  of a population  proportion
b. the chi­squared  test  of a binomial  experiment  since  z 2 = χ 2
c. the chi­squared  test  of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: a

5. In a goodness­ of­fit test, suppose  that  the  value  of the  test statistic is 13.08 and  df = 6. At 

the 5% significance level, the  null hypothesis  is
a. rejected  and  p­value  for the  test  is smaller  than  0.05
b. not  rejected,  and  p­value  for the  test is greater  than  0.05
c. rejected,  and  p­value  for the  test is greater  than  0.05
d. not  rejected,  and  p­value  for the  test is smaller  than  0.05
ANSWER: a

6. In a goodness­ of­fit test, suppose  that  a sample  showed  that  the  observed  frequency   f i  

and  expected  frequency   ei  were  equal  for each cell i. Then, the  null hypothesis  is
a. rejected  at  α =  0.05 but  is not  rejected  at  α = 0.025
b. not  rejected  at  α =  0.05 but  is rejected  at  α =  0.025
c. rejected  at any  level  α
d. not  rejected  at any   α  level
ANSWER: d

7. The  president  of a  state  university  collected  data  from  students  concerning  building  a  
new  library,  and  classified  the  responses  into  different  categories  (strongly  agree,  agree,  
undecided,  disagree,  strongly  disagree)  and  according  to whether  the  student  was  male  
or  female.  To  determine  whether  the  data  provide  sufficient  evidence  to  indicate  that  
the responses  depend  upon  gender,  the  most  appropriate  test is:
a. chi­squared  goodness­ of­fit test
b. chi­squared  test of a contingency  table  (test of independence)
c. chi­squared  test of normality
d. chi­squared  test for comparing  five proportions
ANSWER: b 

8. The number  of degrees  of freedom  for a contingency  table with  6 rows  and  6 columns  is

a. 36
b. 25
c. 12
d. 6
ANSWER: b
 Chi-Squared Tests 74

9. A  chi­squared  test  of  a  contingency  table  with  4 rows  and  5 columns  shows  that  the  
value  of the  test  statistic  is 22.18. The  most  accurate  statement  that  can  be  made  about  
the p­value  for this test is that  
a. p­value  is greater  than  0.05
b. p­value  is smaller  than  0.025
c. p­value  is greater  than  0.025 but  smaller  than  0.05
d. p­value  is greater  than  0.10
ANSWER: c

10. The number  of degrees  of freedom  for a contingency  table with  4 rows  and  8 columns  is
a. 32
b. 28
c. 24
d. 21
ANSWER: d

11. The chi­squared  distribution  is used:

a. in a goodness­ of­fit test
b. in a test of a contingency  table
c. in making  inferences  about  a single  population  variance
d. All of the  above  answers  are correct
ANSWER: d

12. To determine  whether  a single  coin  is fair, the  coin  was  tossed  100 times,  and  head  was  
observed  60 times. The value  of the  test  statistic is
a. 40
b. 4
c. 60
d. 6
ANSWER: b

13. To determine  whether  data  were  drawn  from  any  distribution,  we  use
a. a chi­squared  goodness­ of­fit test
b. a chi­squared  test  of a contingency  table
c. a chi­square  test for normality
d. None  of the  above  answers  is correct
ANSWER: a

14. A chi­squared  goodness­ of­fit test is always  conducted  as:

a. a lower­ tail test
b. an upper­ tail test
c. a two­ tail test
d. Either  a or b
ANSWER: b
75 Chapter Sixteen

15. 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
ANSWER: c

16. The number  of degrees  of freedom  in testing  for normality  is the

a. number  of intervals  used  to test the  hypothesis  minus  1
b. number  of parameters  estimated  minus  1
c. number   of   intervals   used   to   test   the   hypothesis   minus   number   of   parameters  
estimated  minus  1
d. number   of   intervals   used   to   test   the   hypothesis   minus   number   of   parameters  
estimated  minus  2
ANSWER: c

17. If the expected  frequency   ei  for any  cell i is less than  5, we

a. must  choose  another  sample  of five or more  observations
b. should  use  the normal  distribution  instead  of the  chi­squared  distribution
c. should  combine  the cells such  that  each  observed  frequency   f i  is 5 or more
d. increase  the  number  of degrees  of freedom  for the  test by 5
ANSWER: c

18. If each  element  in a population  is classified  into  one  and  only  one  of several  categories,  
the population  is a:
a. normal  population
b. multinomial  population
c. chi­squared  population  
d. binomial  population
ANSWER: b

19. To   determine   the   critical   values   in   the   chi­squared   distribution   table,   the   process  
requires  the following:
a. degrees  of freedom  
b. probability  of Type  I error
c. probability  of Type  II error
d. Both a and  b
ANSWER: d

20. Of the  values  for  a chi­squared  test  statistic  listed  below,  which  one  is likely  to  lead  to  
rejecting  the null hypothesis  in a goodness­ of­fit test?
a. 0
b. 1
c. 2
 Chi-Squared Tests 76

d. 40
ANSWER: d

21. The  number  of degrees  of  freedom  in  a  test  of a  contingency  table  with  4 rows  and  3  
columns  equals:
a. 4
b. 7
c. 6
d. 3
ANSWER: c

22. Which   statistical   technique   is   appropriate   when   we   describe   a   single   population   of  

qualitative  data  with  two  or more  categories?
a. z­test  of the difference  between  two  proportions
b. The chi­squared  test of a multinomial  experiment
c. The chi­squared  test of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: b

23. The  sampling  distribution  of the  test  statistic for a goodness­ of­fit test  with  k categories  
is:
a. Student  t distribution  with  k­1 degrees  of freedom
b. normal  distribution
c. chi­squared  distribution  with  k­1 degrees  of freedom
d. approximately  chi­squared  distribution  with   k­1 degrees  of freedom
ANSWER: d

24. The chi­squared  test of a contingency  table is based  upon:

a. two  qualitative  variables
b. two  quantitative  variables
c. three  or more  qualitative  variables
d. three  or more  quantitative  variables
ANSWER: a

25. Which  of the following  statements  is not  correct?

a. The chi­squared  distribution  is symmetrical
b. The chi­squared  distribution  is skewed  to the  right
c. All values  of the chi­squared  distribution  are positive
d. The critical region  for a goodness­ of­fit test with  k categories  is  χ > χ α ,k −1
2 2

ANSWER: a

26. Which   statistical   technique   is   appropriate   when   we   compare   two   populations   of  

qualitative  data  with  exactly two  categories?
a. z­test  of  a population  proportion
b. z­test  of the difference  between  two  proportions
c. The chi­squared  test of a contingency  table
77 Chapter Sixteen

d. Both a and  b
e. Both b and  c
ANSWER: e
27. Which  of the following  statements  is not  correct?
a. The chi­squared  test of independence  is a one­sample  test
b. Both variables  in the chi­squared  test of independence  are qualitative  variables
c. The chi­squared  goodness­ of­fit test involves  two  categorical variables
d. The chi­squared  distribution  is skewed  to the  right
ANSWER: c

28. A  left  tail  area  in  the  chi­squared  distribution  equals  0.99. For  df  =  8, the  table  value  
equals:
a. 20.0902
b. 3.4895
c. 2.7326
d. 15.5073
ANSWER: a

29. Which  of the following  statements  is true  for the  chi­squared  tests?

a. Testing  for equal  proportions  is identical to testing  for goodness­ of­fit
b. The number  of degrees  of freedom  in a test  of a contingency  table  with  r rows  and  c 
columns  is (r­1)(c­1).
c. The number  of degrees  of freedom  in a goodness­ of­fit test with  k categories  is k­1
d. All of the  above  statements  are true.
ANSWER: d

30. The   degrees   of   freedom   in   a   chi­squared   test   for   normality,   where   the   number   of  
standardized  intervals  is 5 and  there  are  2 population  parameters  to be  estimated  from  
the data,  is equal  to:
a. 5
b. 4
c. 3
d. 2
ANSWER: d

31. A chi­squared  test  for independence  with  6 degrees  of freedom  results  in a test  statistic  
χ 2 = 13.58 . Using  the χ 2 tables, the  most  accurate  statement  that  can be made  about  the  
p­value  for this test is that:
a. p­value  > 0.10
b. p­value  > 0.05
c. 0.05 < p­value  < 0.10
d. 0.025 < p­value  < 0.05
ANSWER: d
 Chi-Squared Tests 78

32. In  a goodness­ of­fit test,  the  null  hypothesis  states  that  the  data  came  from  a  normally  
distributed  population.   The  researcher  estimated  the  population  mean  and  population  
standard  deviation  from  a sample  of 500 observations.  In addition,  the  researcher  used  6  
standardized  intervals  to test  for normality.  Using  a 5% level  of significance, the  critical  
value  for this test is:
a. 11.1433
b. 9.3484
c. 7.8147
d. 9.4877
ANSWER: c

33. In   a   chi­squared   test   of   a   contingency   table,   the   value   of   the   test   statistic   was  
χ 2 = 12.678 , and  the critical value  at  α = 0.025  was  14.4494.  Thus,
a.   we  fail to reject the null hypothesis  at  α = 0.025
b. we  reject the null hypothesis  at  α = 0.025
c. we  don’t have  enough  evidence  to accept  or reject the  null  hypothesis  at  α = 0.025
d. we  should  decrease  the level of significance  in order  to reject the  null  hypothesis
ANSWER: a

34. Which  statistical technique  is appropriate  when  we  compare  two  or more  populations  of  
qualitative  data  with  two  or more  categories?
a. z­test  of the difference  between  two  proportions
b. The chi­squared  test of a multinomial  experiment
c. The chi­squared  test of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: c

35. Which  of the following  tests  does  not  use  the  chi­squared  distribution?
a. Test of a contingency  table
b. Goodness­ of­fit test
c. Difference  between  two  population  means  test
d. All of the  above  tests  use  the  chi­squared  distribution
ANSWER: c

36. Which   statistical   technique   is   appropriate   when   we   compare   two   populations   of  

qualitative  data  with  two  or more  categories?
a. z­test  of the difference  between  two  proportions
b. The chi­squared  test of a multinomial  experiment
c. The chi­squared  test of a contingency  table
d. Both a and  b
e. Both b and  c
ANSWER: c
79 Chapter Sixteen

37. Which  of the following  is not  a characteristic of a multinomial  experiment?

a. The experiment  consists  of a fixed  number,  n, of trials.
b. The outcome  of each  trial can be classified  into  one  of two  categories  called  successes  
and  failure.
c. The probability   pi  that  the  outcome  will fall into cell i remain  constant  for each  trial.
d. Each trial of the experiment  is independent  of the  other  trials
ANSWER: b

38. In  a  chi­squared  goodness­ of­fit  test,  if the  expected  frequencies   ei   and  the  observed  
frequencies   f i  were  quite  different,  we  would  conclude  that:
a. the null  hypothesis  is false, and  we  would  reject it
b. the null  hypothesis  is true, and  we  would  not  reject it
c. the alternative  hypothesis  is false, and  we  would  reject it
d. the chi­squared  distribution  is invalid,  and  we  would  use  the  t­distribution  instead
ANSWER: a

39. In chi­squared  tests, the  conventional  and  conservative  rule  – known  as the   rule of five – 
is to require  that  the:
a. observed  frequency  for each cell be at least five 
b. degrees  of freedom  for the test be at least  five
c. expected  frequency  for each cell be at least five
d. difference  between  the observed  and  expected  frequency  for each cell be at least five
ANSWER: c

40. Consider  a multinomial  experiment  with  200 trials, and  the  outcome  of each  trial can  be  
classified  into  one  of 5 categories.    The  number  of degrees  of freedom  associated  with  
the chi­squared  goodness­ of­fit test equals:
a. 195
b. 205
c. 199
d. 4
ANSWER: d
 Chi-Squared Tests 80

TRUE/FALSE QUESTIONS

41. The   null   hypothesis   states   that   the   sample   data   came   from   a   normally   distributed  
population.   The   researcher   calculates   the   sample   mean   and   the   sample   standard  
deviation  from  the  data.    The  data  arrangement  consisted  of  five  categories.    Using  a  
0.05 significance level, the  appropriate  critical value  for this chi­square  test for normality  
is 5.99147
ANSWER: T

42. A test  for independence  is applied  to a contingency  table  with  3 rows  and  4 columns  for  
two  qualitative  variables. The degrees  of freedom  for this chi­square  test must  equal  12.
ANSWER: F

43. A chi­square  test  for independence  with  6 degrees  of freedom  results  in a test  statistic of  
13.25.  Using  the  chi­square  table,  the  most  accurate  statement  that  can  be  made  about  
the p­value  for this test is that  p­value  is greater  than  0.025 but  smaller  than  0.05.
ANSWER: T

44. Whenever  the expected  frequency  of a cell is less than  5, one  remedy  for this condition  is  

to increase  the significance level.
ANSWER: F

45. In  testing  a  population  mean  or  constructing  a  confidence  interval  for  the  population  
mean,  an essential  assumption  is that  expected  frequencies  are at least five.
ANSWER: F

46. A right­tailed  area  in  the  chi­square  distribution  equals  0.05.  For  6 degrees  of freedom  
the table value  equals  12.5916.
ANSWER: T

47. Whenever  the expected  frequency  of a cell is less than  5, one  remedy  for this condition  is  

to increase  the size of the sample.
ANSWER: T

48. A left­tailed  area  in the chi­square  distribution  equals  0.10.  For 5 degrees  of freedom  the  

table value  equals  9.23635.
ANSWER: F

49. For a chi­square  distributed  random  variable  with  10 degrees  of freedom  and  a level  of  
significance of 0.025, the chi­square  value  from  the  table  is 20.4831.  The computed  value  
of the test statistics is 16.857. This will lead  us  to reject the  null  hypothesis.
ANSWER: F

50. A test  for independence  is applied  to a contingency  table  with  4 rows  and  4 columns  for  
two  qualitative  variables.  The degrees  of freedom  for this test will be 9.
ANSWER: T
81 Chapter Sixteen

51. A chi­square  test  for independence  with  10 degrees  of freedom  results  in a test  statistic  
of  17.894.   Using  the  chi­square  table,  the  most  accurate  statement  that  can  be  made  
about  the p­value  for this test is that  0.05 < p­value  < 0.10.
ANSWER: T

52. Whenever  the expected  frequency  of a cell is less than  5, one  remedy  for this condition  is  

to decrease  the size of the sample.
ANSWER: F

53. The  middle  0.95 portion  of  the  chi­square  distribution  with  9 degrees  of  freedom  has  
table values  of 3.32511 and  16.9190, respectively.
ANSWER: F

54. In   applying   the   chi­square   goodness­ of­fit   test,   the   rule   of   thumb   for   all   expected  
frequencies  is that  each  expected  frequency  equal  or exceeds  5.
ANSWER: T

55. In a chi­squared  test  of independence,  the  value  of the  test  statistic was   χ 2 = 15.652, and  
the  critical  value  at   α = 0.025  was  11.1433.  Thus,  we  must  reject the  null  hypothesis  at  
α = 0.025 .
ANSWER: T

56. The   chi­squared   test   of   independence   is   based   upon   three   or   more   quantitative  
variables.
ANSWER: F

57. In  a goodness­ of­fit test,  the  null  hypothesis  states  that  the  data  came  from  a  normally  
distributed  population.   The  researcher  estimated  the  population  mean  and  population  
standard  deviation  from  a sample  of 300 observations.  In addition,  the  researcher  used  6  
standardized   intervals   to   test   for   normality.   Using   a   2.5%   level   of   significance,   the  
critical value  for this test is 14.4494.
ANSWER: F

58. A chi­square  goodness­ of­fit test is always  conducted  as a two­ tail test.

ANSWER: F

59. A left­tailed  area  in  the  chi­square  distribution  equals  0.90.  For  10 degrees  of freedom  
the table value  equals  15.9871.
ANSWER: T

60. For a chi­square  distributed  random  variable  with  12 degrees  of freedom  and  a level  of  
significance  of 0.05, the  chi­square  value  from  the  table  is 21.0261.  The  computed  value  
of the test statistics is 25.1687. This will lead  us  to reject the  null  hypothesis.
ANSWER: T
 Chi-Squared Tests 82

61. In  a goodness­ of­fit test,  the  null  hypothesis  states  that  the  data  came  from  a  normally  
distributed  population.   The  researcher  estimated  the  population  mean  and  population  
standard  deviation  from  a sample  of 200 observations.  In addition,  the  researcher  used  5  
standardized  intervals  to test for normality.  Using  a 10% level of significance, the  critical  
value  for this test is 4.60517.
ANSWER: T

62. In chi­square  tests, the  conventional  and  conservative  rule  – known  as the  Rule of Five – 
is to require  that  difference  between  the  observed  and  expected  frequency  for  each  cell  
be at least five.
ANSWER: F

63. Whenever  the expected  frequency  of a cell is less than  5, one  remedy  for this condition  is  

to decrease  the significance level.
ANSWER: F

64. The  area  to  the  right  of a chi­square  value  is 0.01.  For  8 degrees  of freedom,  the  table  
value  is 1.64648.
ANSWER: F

65. A multinomial  experiment,  where  the  outcome  of each  trial  can  be classified  into  one  of  
two  categories, is identical to the  binomial  experiment.
ANSWER: T

66. The chi­squared  goodness­ of­fit test  is usually  used  as a test  of multinomial  parameters,  

but  it can also be used  to determine  whether  data  were  drawn  from  any  distribution.
ANSWER: T

67. The   chi­squared   test   of   a   contingency   table   is   used   to   determine   if   there   is   enough  
evidence  to  infer  that  two  nominal  variables  are  related,  and  to  infer  that  differences  
exist among  two  or more  populations  of nominal  variables.
ANSWER: T

68. The number  of degrees  of freedom  for a contingency  table  with  r rows  and  c columns  is 
ν  = rc ,  provided  that  both  r and  c are greater  than  or equal  to 2.
ANSWER: F

69. When  the  problem  objective  is  to  describe  a  population  of  nominal  data  with  exactly  
two  categories,  we  can  employ  either  the  z­test  of population  proportion  p, or  the  chi­
squared  goodness­ of­fit test.
ANSWER: T

70. If we  want  to perform  a one­tail test  of a population  proportion  p, we  can  employ  either  
the z­test of p, or the chi­squared  goodness­ of­fit test.
ANSWER: F
83 Chapter Sixteen

71. If we  want  to perform  a two­ tail test  of a population  proportion  p, we  must  employ  the  
z­test  of p.
ANSWER: F

72. If we  want  to test  for differences  between  two  populations  of nominal  data  with  exactly  
two  categories,  we  can  employ  either  the  z­test  of   p1 − p2 , or  the  chi­squared  test  of a  
contingency  table (squaring  the  value  of the  z statistic yields  the  value  of  χ 2  statistics).
ANSWER: T

73. If   we   want   to   perform   a   two­ tail   test   for   differences   between   two   populations   of  
nominal  data  with  exactly  two  categories,  we  can  employ  either  the  z­test  of  p1 − p2 , or 
the chi­squared  test  of a contingency  table (squaring  the  value  of the  z statistic yields  the  
value  of  χ 2  statistics).
ANSWER: T

74. If we  want  to perform  a one­tail test for differences  between  two  populations  of nominal  

data  with  exactly two  categories, we  must  employ  the  z­test of  p1 − p2 .
ANSWER: T

75. The  number  of degrees  of freedom  associated  with  the  chi­squared  test  for  normality  is  
the number  of intervals  used  minus  the  number  of parameters  estimated  from  the  data.
ANSWER: F
 Chi-Squared Tests 84

TEST QUESTIONS

76. The following  data  are believed  to have  come  from  a normal  probability  distribution.

26 21 25 20 21 29 26 23 22 24
24 30 23 32 26 24 32 16 36 26
21 31 26 23 32 35 40 30 14 26
46 27 33 25 27 21 26 18 29 36

The  mean  of this  sample  equals  26.80, and  the  standard  deviation  equals  6.378. Use  the  
goodness­ of­fit test at the 5% significance level to test  this claim.

ANSWER:
H 0 :  The population  has  a normal  probability  distribution  
              H 1 :  The population  does  not  have  a normal  probability  distribution
Since the sample  size is less than  80, we  employ  the  minimum  number  of intervals  4.

Selected Standardized   Probability Observed   Expected   ( f − e ) 2

i i
Intervals Intervals pi Frequency Frequency  
ei
fi ei

X  ≤  20.422 Z  ≤  ­1 0.1587 4 6.348 0.8685

20.422  ≤ X  ≤  26.80 ­1  ≤  Z  ≤  0 0.3413 20 13.652 2.9517
26.80  ≤ X  ≤ 33.178 0  ≤  Z  ≤  1 0.3413 11 13.652 0.5152
X > 33.178 Z > 1 0.1587 5 6.348 0.2862

 Rejection  region:  χ > χ 0.05,1 = 3.8415

2 2

    Test statistic:  χ 2 = 4.6216
 p­value  = 0.0316
 Conclusion:  Reject the null hypothesis; concluding  that  the  population  does  not  have     
                                     a  normal  probability  distribution  

77. Conduct  a  test  to  determine  whether  the  two  classifications  A  and  B are  independent,  
using  the data  in the accompanying  table and   α = 0.05  

B1 B2 B3
A1 35 25 20
A2 25 20 25
ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 2.121,  and   p­value  = 0.3464.

 Conclusion: Don’t reject the null  hypothesis. Yes
85 Chapter Sixteen

78. The  personnel  manager  of  a  consumer  product  company  asked  a  random  sample  of  
employees  how  they  felt  about  the  work  they  were  doing.  The  following  table  gives  a  
breakdown  of  their  responses  by  gender.  Do  the  data  provide  sufficient  evidence  to  
conclude  that  the level of job satisfaction  is related  to gender?  Use  α = 0.10.

     Response

Gender Very Interesting Fairly Interesting Not  Interesting

Male 70 41 9
Female 35 34 11

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.10,2 = 4.605
2 2

 Test statistic:  χ 2 = 4.708
  p­value  = 0.095
 Conclusion: Reject the null  hypothesis. Yes

79. The  personnel  manager  of  a  consumer  products  company  asked  a  random  sample  of  
employees  how  they  felt  about  the  work  they  were  doing.  The  following  table  gives  a  
breakdown  of their  responses  by  age.  Is there  sufficient  evidence  to  conclude  that  the  
level of job satisfaction  is related  to age? Use  α = 0.10.

     Response

Age Very   Fairly Interesting Not  Interesting

Interesting
Under  30 31 24 13
Between  30 and  50 42 30 4
Over  50 32 21 3

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.10,4 = 7.779
2 2

 Test statistic:  χ 2 = 9.692
  p­value  = 0.046
 Conclusion: Reject the null  hypothesis. Yes
 Chi-Squared Tests 86

80. A  firm  has  been  accused  of  engaging  in  prejudicial  hiring  practices.  According  to  the  
most   recent   census,   the   percentages   of   whites,   blacks,   and   Hispanics   in   a   certain  
community  are 72%, 10%, and  18%, respectively.  A random  sample  of 200 employees  of  
the  firm  revealed  that  165 were  white,  14 were  black, and  21 were  Hispanic. Do the  data  
provide  sufficient  evidence  to conclude  at  the  5% level  of significance  that  the  firm  has  
been  engaged  in prejudicial hiring  practices?

ANSWER:
H 0 : p1 = 0.72 ,  p2 = 0.10,   p3 = 0.18  
H 1 :  At least two  proportions  differ from  their  specified  values    
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 11.113
 p­value  = 0.0039
              Conclusion:  Reject the null hypothesis. Yes

81. Five brands  of orange  juice are  displayed  side  by side  in several  supermarkets  in a large  
city.   It was  noted  that  in  one  day,  180 customers  purchased  orange  juice. Of  these,  30  
picked  Brand  A,  40 picked  Brand  B, 25 picked  Brand  C,    35 picked  Brand  D,  and  50  
picked  brand  E.  In this  city, can  you  conclude  at the  5% significance  level that  there  is a  
preferred  brand  of orange  juice?

ANSWER:
H 0 : p1 = p 2 = p3 = p 4 = p 5  
H 1 :  At least two  proportions  differ from  their  specified  values
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

Test statistic:  χ 2 = 10.278
p­value  = 0.036
Conclusion:  Reject the null  hypothesis. Yes

82. A sport  preference  poll showed  the  following  data  for men  and  women:

          Favorite Sport

Gender Baseball Basketball Football Golf Tennis

Male 24 17 30 18 22
Female 21 20 22 12 28

Use the 5% level of significance  and  test to determine  whether  sport  preferences  depend  

on gender.
87 Chapter Sixteen

ANSWER:
H 0 : The two  variables  (gender  and  favorite  sport)  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

 Test statistic:  χ 2 = 3.30
  p­value  = 0.509
 Conclusion:  Don’t reject the  null hypothesis. No

83. Last year, Brand  A microwaves  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 appliance  stores  shows  that  of 10,000 microwaves  sold, 4350 were  
Brand  A, 3450 were  Brand  B, and  2200 were  Brand  C.  Has  the  market  changed?   Test at  
α = 0.01.

ANSWER:
H 0 : p1 = 0.45 ,  p2 = 0.35,   p3 = 0.20  
H 1 :  At least two  proportions  differ from  their  specified  values    
 Rejection  region:  χ > χ 0.01,2 = 9.210
2 2

 Test statistic:  χ 2 = 25.714
 p­value  = 0.0
 Conclusion: Reject the null  hypothesis. Yes

84. A study  of educational  levels of 500 voters  and  their  political party  affiliations  in a  

particular  state  in the USA showed  the  following  results:

                    Party Affiliation

Educational level Democrat Republican Independent

Didn’t Complete  High  School 40 20 80
High  School Diploma 70 30 60
Has  College Degree 90 50 60

Use  the  1% level  of significance  and  test  to see  if party  affiliation  is independent  of the  
educational  level of the voters.

ANSWER:
H 0 : The two  variables  (educational  level and  political party  affiliation)  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.01,4 = 13.277
2 2

 Test statistic:  χ 2 = 26.830
 p­value  = 0.0
 Conclusion:  Reject the null hypothesis. No
 Chi-Squared Tests 88

85. 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  

H 0 : p1 = 0.45 ,  p2 = 0.30,   p3 = 0.25

H 1 :  At least two  proportions  differ from  their  specified  values    

ANSWER:
Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

Test statistic:  χ 2 =  2.082
             p­value  = 0.3531
Conclusion:  Don’t reject the null  hypothesis.

86. In  2000, the  student  body  of  a  state  university  in  Michigan  consists  of  30% freshmen,  
25% sophomores,  27% juniors,  and  18% seniors.   A  sample  of 400 students  taken  from  
the  2001 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.

ANSWER:
H 0 : p1 = 0.30 ,  p2 = 0.25,   p3 = 0.27,   p4 = 0.18  
H 1 :  At least two  proportions  differ from  their  specified  values    
 Rejection  region:  χ > χ 0.05,3 = 7.815
2 2

 Test statistic:  χ 2 = 6.844
 p­value  = 0.077
 Conclusion: Don’t reject the null  hypothesis. No

87. 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  

H 0 : p1 = p 2 = p3 = p 4  
H 1 :  At least two  proportions  differ from  their  specified  values
89 Chapter Sixteen

ANSWER:
Rejection  region:  χ > χ 0.10,3 = 6.251
2 2

Test statistic:  χ 2 = 7.450
p­value  = 0.0589
Conclusion:  Reject the null  hypothesis. No

88. A   statistics   professor   posted   the   following   grade   distribution   guidelines   for   his  
elementary  statistics  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.

ANSWER:
H 0 : p1 = 0.08 ,  p2 = 0.35,   p3 = 0.40,   p4 = 0.12,   p4 = 0.05
H 1 :  At least two  proportions  differ from  their  specified  values    
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

Test statistic:  χ 2 = 5.889
p­value  = 0.2076
Conclusion:  Don’t reject the null  hypothesis. No

89. Conduct  a  test  to  determine  whether  the  two  classifications  A  and  B are  independent,  
using  the data  in the accompanying  table and   α = .01  

B1 B2
A1 42 28
A2 23 57

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.01,1 = 6.635
2 2

 Test statistic:  χ 2 = 14.847
  p­value  = 0.0001
 Conclusion: Reject the null  hypothesis. No

90. Explain  what  is meant  by the rule of five.

ANSWER:
The  rule  of five  requires  that  the  expected  frequency  for  each  cell be  at  least  5. Where  
necessary,  cells should  be combined  in order  to satisfy this  condition.  The choice of cells  
to be combined  should  be  made  in  such  a way  that  meaningful  categories  (cells) result  
from  the combination.
 Chi-Squared Tests 90

91. In 1996, computers  of Brand  A controlled  25% of the  market,  Brand  B 20%, Brand  C 10%, 

and  brand  D  45%.   In  2000, sample  data  was  collected  from  many  randomly  selected  
stores  throughout  the country.   Of the  1200 computers  sold,  280 were  Brand  A, 270 were  
Brand  B, 90 were  Brand  C, and  560 were  Brand  D.  Has  the  market  changed  since  1996?  
Test at the 1% significance level.

ANSWER:
H 0 : p1 = 0.25 ,  p2 = 0.20,   p3 = 0.10,   p4 = 0.45
H 1 :  At least two  proportions  differ from  their  specified  values    
 Rejection  region:  χ > χ 0.01,3 = 11.345
2 2

 Test statistic:  χ 2 = 13.324
 p­value  = 0.004
 Conclusion: Reject the null  hypothesis. Yes

92. A major insurance  firm  interviewed  a random  sample  of 1200 college  students  to find  

out  the  type  of life insurance  preferred,  if any.  The results  follow:

               Insurance Preference

Gender Term Whole  Life No  Insurance

Female 100 80 325
Male 160 60 475

Is there  evidence  that  life insurance  preference  of male  students  is different  than  that  of  
female students?   Test using  the  5% level of significance.

ANSWER:
H 0 : The two  variables  (gender  and  insurance  preference) are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 15.124
  p­value  = 0.0005
 Conclusion:  Reject the null hypothesis. Yes

93. The number  of cars sold  by three  salespersons  over  a 3­month  period  are shown  below:

                      Brand of Car

Salesperson Brand  A Brand  B Brand  C

David  7 2 6
Edward 11 4 8
Frank   8 5 3

Use the 5% level of significance  and  test for the  independence  of salesperson  and  type  of  

product  sold
91 Chapter Sixteen

ANSWER:
H 0 : The two  variables  (salesperson  and  brand  of car) are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

 Test statistic:  χ 2 = 2.662
 p­value  = 0.6158
 Conclusion:  Don’t reject the  null hypothesis. Yes

94. A telephone  company  prepared  four  versions  of a set  of instructions  for  placing  collect  
calls.  The  company  asked  a  sample  of  1600 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?

ANSWER:
H 0 : p1 = p 2 = p3 = p 4  
H 1 :  At least two  proportions  differ from  their  specified  values
Rejection  region:  χ > χ 0.05,3 = 7.815
2 2

Test statistic:  χ 2 = 4.25
p­value  = 0.2357
Conclusion:  Don’t reject the null  hypothesis. No

95. Suppose  that  a random  sample  of 60 observations  was  drawn  from  a population.   After  
calculating  the  mean  and  standard  deviation,  each  observation  was  standardized  and  
the  number  of observations  in each  of the  intervals  below  was  counted.   Can  we  infer  at  
the 10% significance level that  the  data  were  drawn  from  a normal  population?

Intervals Frequency
Z   ≤  ­1 8
­1 < Z   ≤  0 30
0 < Z   ≤  1 17
Z  > 1 5

ANSWER:
H 0 : p1 = 0.1587 ,  p2 = 0.3413,   p3 = 0.3413,   p4 = 0.1587
                     (The population  is normal)
H 1 :  At least two  proportions  differ from  their  specified  values  
        (The population  is not  normal)   
 Rejection  region:  χ > χ 0.10,1 = 2.7055
2 2

 Test statistic:  χ 2 =  4.479
 p­value  = 0.0343
 Conclusion: Reject the null  hypothesis. No
 Chi-Squared Tests 92

96. Suppose  that  a random  sample  of 150 observations  was  drawn  from  a population.   After  
calculating  the  mean  and  standard  deviation,  each  observation  was  standardized  and  
the  number  of observations  in each  of the  intervals  below  was  counted.   Can  we  infer  at  
the 5% significance level that  the  data  were  drawn  from  a normal  population?

Intervals Frequency
Z   ≤  ­1.5 15
­1.5 < Z   ≤  ­.5 32
­.5  ≤  Z   ≤  .5 65
.5 < Z   ≤  1.5 25
Z  > 1.5 13

ANSWER:
H 0 : p1 = 0.0668 ,  p2 = 0.2417,   p3 = 0.3830,   p4 = 0.2417 ,  p5 = 0.0668  
       (The population  is normal)
H 1 :  At least two  proportions  differ from  their  specified  values
       (The population  is not  normal)   
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 =  8.347
  p­value  = 0.0154
 Conclusion: Reject the null  hypothesis. No

97. The   president   of   a   large   university   has   been   studying   the   relationship   between  
male /female  supervisory  structures  in  his  institution  and  the  level  of  employees’  job  
satisfaction.   The results  of a recent  survey  are  shown  in the  table  below.   Conduct  a test  
at the 5% significance level to determine  whether  the  level of job satisfaction  depends  on  
the boss /em ployee  gender  relationship.

                                          Boss/Employee

Level of Satisfaction Male /  Female Female / Male Male /Male Female /Female

Satisfied 60 15 50 15
Neutral 27 45 48 50
Dissatisfied 13 32 12 55

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,6 = 12.592
2 2

    Test statistic:  χ 2 = 92.709
 p­value  = 0.0
 Conclusion: Reject the null  hypothesis. Yes
93 Chapter Sixteen

98. Consumer  panel  preferences  for three  proposed  fast food  restaurants  are as follows:

           

Restaurant  A Restaurant  B Restaurant  C

48 62 40

Use  0.05 level  of  significance  and  test  to  see  if there  is  a  preference  among  the  three  
restaurants.

ANSWER:
H 0 : p1 = p 2 = p3  
H 1 :  At least two  proportions  differ from  their  specified  values
Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

Test statistic:  χ 2 = 4.96
 p­value  = 0.0837
Conclusion:  Don’t reject the null  hypothesis. No

99. A   cafeteria   proposes   to   serve   4   main   entrees.     For   planning   purposes,   the   manager  
expects  that  the proportions  of each  that  will be selected  by his customers  will be:

Selection Proportion
Chicken 0.50
Roast  Beef 0.20
Steak 0.10
Fish 0.20

Of the first 100 customers,  44 selected  chicken,  24 selected  roast  beef, 13 select steak, and  

10 selected  fish.  Should  the manager  revise  his estimates?   Use  α = 0.01.

ANSWER:
H 0 : p1 = 0.50 ,  p2 = 0.20,   p3 = 0.10,   p4 = 0.20
H 1 :  At least two  proportions  differ from  their  specified  values    
 Rejection  region:  χ > χ 0.01,3 = 11.345
2 2

 Test statistic:  χ 2 = 7.264
 p­value  = 0.064
 Conclusion: Don’t reject the null  hypothesis. No
 Chi-Squared Tests 94

100. A  large  carpet  store  wishes  to  determine  if the  brand  of carpet  purchased  is related  to  
the  purchaser’s  family  income.    As  a  sampling  frame,  they  mailed  a  survey  to  people  
who  have  a  store  credit  card.    Five  hundred  customers  returned  the  survey  and  the  
results  follow:

             Brand of Carpet

Family Income Brand  A Brand  B Brand  C

High  Income 65 32 32
Middle  Income 80 68 104
Low  Income 25 35 59

At  the  5% level  of significance,  can  you  conclude  that  the  brand  of carpet  purchased  is  
related  to the purchaser’s family  income?

ANSWER:
H 0 : The two  variables  (family  income  and  brand  of carpet)  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

 Test statistic:  χ 2 = 27.372
 p­value  = 0.0
 Conclusion:  Reject the null hypothesis. Yes

101. 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 there  sufficient  evidence  at  the  5% significance  level to allow  you  
to conclude  that  the coin is not  fair?

ANSWER:
H 0 : p1 = 0.50,  p 2 = 0.50   (the coin is fair)
H 1 : At least one   pi is not  equal  to its specified  value   (the coin is not  fair)
 Rejection  region:  χ > χ 0.05,1 = 3.841
2 2

 Test statistic:  χ 2 = 2.88
 p­value  = 0.0897
 Conclusion: Don’t reject the null  hypothesis. Yes
95 Chapter Sixteen

QUESTIONS 102 THROUGH  105 ARE BASED ON  THE FOLLOWING  INFORMATION:

Consider  a  multinomial  experiment  involving  n  =  200 trials  and  k  =  5 cells.    The  observed  
frequencies  resulting  from  the experiment  are shown  in the  following  table:

Cell 1 2 3 4 5
Frequency 16 44 56 48 36

The null hypothesis  to be tested  is as follows.

H 0 : p1 = 0.10 ,  p2 = 0.25,   p3 = 0.30,   p4 = 0.20,   p5 = 0.15 p5 = 0.15

102. Test the hypothesis  at the 5% level of significance

ANSWER:
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

Test statistic:  χ 2 = 4.587
p­value  = 0.3324
Conclusion:  Don’t reject the null  hypothesis.

103. Re­do  Question  102 with  the following  frequencies:

Cell 1 2 3 4 5
Frequency 8 22 28 24 18

ANSWER:
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

Test statistic:  χ 2 = 2.293
p­value  = 0.6820
Conclusion:  Don’t reject the null  hypothesis.

104. Re­do  Question  102 with  the following  frequencies:

Cell 1 2 3 4 5
Frequency 4 11 14 12 9

ANSWER:
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2

Test statistic:  χ 2 = 1.147
p­value  = 0.8868
Conclusion:  Don’t reject the null  hypothesis.
 Chi-Squared Tests 96

105. Review  the  results  of Questions  102 to 104.  What  is the  effect of decreasing  the  sample  
size?

ANSWER:
As the  sample   size  decreased   by  50%,  the   value  of the  test statistic also decreased  by 
50% and  the p­value  became  much  larger. 

QUESTIONS 106 THROUGH  109 ARE BASED ON  THE FOLLOWING  INFORMATION:

Consider  the data  in the  accompanying  table with  classifications  A  and  B:

B1 B2
A1 40 80
A2 56 48

106. Conduct  a  test  to  determine  whether  the  two  classifications   A   and  B  are  independent,  
using   α = 0.05

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,1 = 3.841
2 2

 Test statistic:  χ 2 = 3.923
 p­value  = 0.0476
 Conclusion: Reject the null  hypothesis. No

107. Re­do  Question  106 using  the following  table:

B1 B2
A1 20 30
A2 28 24

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,1 = 3.841
2 2

    Test statistic:  χ 2 = 1.962
 p­value  = 0.1613
 Conclusion: Don’t reject the null  hypothesis. Yes
97 Chapter Sixteen

108. Re­do  Question  106 using  the following  table:

B1 B2
A1 10 15
A2 14 12

ANSWER:
H 0 : The two  variables  are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,1 = 3.841
2 2

 Test statistic:  χ 2 = 0.981
 p­value  = 0.322
 Conclusion: Don’t reject the null  hypothesis. Yes

109. Review  the  results  of Questions  106 to  108. What  is the  effect  of decreasing  the  sample  
size?

ANSWER:
As the  sample   size  decreased   by  50%,  the   value  of the  test statistic also decreased  by  
50 % and  the p­value  became  much  larger. 

QUESTIONS 110 THROUGH  114 ARE BASED ON  THE FOLLOWING  INFORMATION:

A biology  professor  claimed  that  the proportions  of grades  in his classes  are the  same. A sample  

of 100 students  showed  the following  frequencies:

Grade A B C D F
Frequency 18 20 28 23 11

110. State the null and  alternative  hypotheses  to be tested.

ANSWER:
H 0 : p1 = 0.20 ,  p2 = 0.20,   p3 = 0.20,   p4 = 0.20, p5 = 0.20  
H 1 :  At least two  proportions  differ from  their  specified  values    

111. Determine  the rejection  region  at the  5% significance level.

ANSWER:
Rejection  region:  χ > χ 0.05,4 = 9.488
2 2
 Chi-Squared Tests 98

112. Compute  the  value  of the test statistics.

ANSWER:
Test statistic:  χ 2 = 7.90

113. Use statistical software  to compute  the  p­value  for this test.

ANSWER:
p­value  = 0.0953

114. Do the data  provide  enough  evidence  to support  the  professor’s claim?

ANSWER:
Don’t  reject  the  null  hypothesis.  Yes, the  data  provide  enough  evidence  to  support  the  
professor’s claim.

 QUESTIONS 115 AND  116 ARE BASED ON  THE FOLLOWING  INFORMATION   :

A  salesperson  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 0.50.

115. 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.

ANSWER:

x fi pi ei
0 10 0.0313 6.26
1 38 0.1562 31.24
2 69 0.3125 62.50
3 63 0.3125 62.50
4 18 0.1562 31.24
5 2 0.0313 6.26
99 Chapter Sixteen

116. Should   the   assumption   of   a   binomial   distribution   be   rejected   at   the   5%   significance  

level?

ANSWER:
H 0 :  The population  has  a binomial  probability  distribution  
              H 1 :  The population  does  not  have  a binomial  probability  distribution
Rejection  region:  χ > χ 0.05,5 = 11.07
2 2

   Test statistic:  χ 2 = 12.888
p­value  = 0.0245
Conclusion:  Reject the null  hypothesis. Yes

QUESTIONS 117 THROUGH  120 ARE BASED ON  THE FOLLOWING  INFORMATION:

A tire manufacturer  operates  a plant  in New  York and  another  plant  in New  Jersey. Employees  

at  each  plant  have  been  evenly  divided  among  three  issues  (wages,  working  conditions,  and  
pension  benefits)  in terms  of which  one  they  feel should  be the  primary  issue  in the  upcoming  
contract  negotiations.  The  president  of the  union  has  recently  circulated  pamphlets  among  the  
employees,  attempting  to  convince  them  that  pension  benefits  should  be  the  primary  issue.  A  
subsequent  survey  revealed  the  following  breakdown  of the  employees  according  to  the  plant  
at which  they  worked  and  the issue  that  they  felt should  be supported  as the  primary  one.

                Issues

Plant Location Very Interesting Fairly Interesting Not  Interesting

New  York 60 62 78
New  Jersey 70 56 74

117. Can   you   infer   at   the   5%   significance   level   that   the   proportional   support   by   the  
employees  at both  plants  for the  issues  has  changed  since the  pamphlet  was  circulated?

ANSWER:
H 0 : p1 = p 2 = p3  
H 1 :  At least two  proportions  differ from  their  specified  values
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 4.460
 p­value  = 0.1075
 Conclusion: Don’t reject the null  hypothesis. No
 Chi-Squared Tests 100

118. Can  you  infer  at  the  5% significance  level  that  the  proportional  support  by  the  New  
York employees  for the  three  issues  has  changed  since the  pamphlet  was  circulated?  

ANSWER:
H 0 : p1 = p 2 = p3  
H 1 :  At least two  proportions  differ from  their  specified  values
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 2.92
 p­value  = 0.2322
 Conclusion: Don’t reject the null  hypothesis. No

119. Can  you  infer  at  the  5% significance  level  that  the  proportional  support  by  the  New  
Jersey employees  for the three  issues  has  changed  since the  pamphlet  was  circulated?  

ANSWER:
H 0 : p1 = p 2 = p3  
H 1 :  At least two  proportions  differ from  their  specified  values
Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

Test statistic:  χ 2 = 2.68
p­value  = 0.2618
Conclusion:  Don’t reject the null  hypothesis. No

120. Do  the  data  indicate  at  the  5% significant  level  that  there  are  differences  between  the  
two  plants  regarding  which  issue  should  be the  primary  one? 

ANSWER:
H 0 : The two  variables  (plant  location  and  issues) are independent
H 1 : The two  variables  are dependent
 Rejection  region:  χ > χ 0.05,2 = 5.991
2 2

 Test statistic:  χ 2 = 1.18
 p­value  = 0.5544
 Conclusion: Don’t reject the null  hypothesis. No