Answers to Selected Exercises

Facts About the Chi-Square Distribution – Practice

1. mean = 25 and standard deviation = 7.0711

3. when the number of degrees of freedom is greater than 90

5. df = 2

Goodness-of-Fit Test – Practice

7. a goodness-of-fit test

9. 3

11. 2.04

13. We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

15. H0: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.

17. right-tailed

19. 2016.136

21. Graph: Check student’s solution. Decision: Reject the null hypothesis. Reason for the Decision: p-value < alpha Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.

Test of Independence – Practice

23. a test of independence

25. a test of independence

27. 8

29. 6.6

31. 0.0435

33.

Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White Totals
1-10 9,886 2,745 12,831 8,378 7,650 41,490
11-20 6,514 3,062 4,932 10,680 9,877 35,065
21-30 1,671 1,419 1,406 4,715 6,062 15,273
31+ 759 788 800 2,305 3,970 8,622
Totals 18,830 8,014 19,969 26,078 27,559 10,0450

 

 

35.

Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White
1-10 7777.57 3310.11 8248.02 10771.29 11383.01
11-20 6573.16 2797.52 6970.76 9103.29 9620.27
21-30 2863.02 1218.49 3036.20 3965.05 4190.23
31+ 1616.25 687.87 1714.01 2238.37 2365.49

37. 10,301.8

39. right

41.

  1. Reject the null hypothesis.
  2. p-value < alpha
  3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.

Test for Homogeneity – Practice

43. test for homogeneity

45. test for homogeneity

47. All values in the table must be greater than or equal to five.

49. 3

51. 0.00005

Comparison of the Chi-Square Tests – Practice

53. a goodness-of-fit test

55. a test for independence

57. Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way, [latex]\sum{ }_{(ij)}\frac{{{({O}-{E})}^{2}}}{{E}}[/latex] In addition, all values must be greater than or equal to five.

Test of a Single Variance – Practice

59. a test of a single variance

61. a left-tailed test

63.

  • H0: σ2 = 0.812;
  • Ha: σ2 > 0.812

65. a test of a single variance

67. 0.0542

Facts About the Chi-Square Distribution – Homework

69. true

71. false

Goodness-of-Fit Test – Homework

73.

Marital Status Percent Expected Frequency
never married 31.3 125.2
married 56.1 224.4
widowed 2.5 10
divorced/separated 10.1 40.4
  1. The data fit the distribution.
  2. The data does not fit the distribution.
  3. 3
  4. chi-square distribution with df = 3
  5. 19.27
  6. 0.0002
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Reject null
    3. Reason for decision: p-value < alpha
    4. Conclusion: Data does not fit the distribution.

75.
H0: The local results follow the distribution of the U.S. AP examinee population
Ha: The local results do not follow the distribution of the U.S. AP examinee population
df = 5
chi-square distribution with df = 5
chi-square test statistic = 13.4
p-value = 0.0199
Check student’s solution.

Alpha = 0.05
Decision: Reject null when a = 0.05
Reason for Decision: p-value < alpha
Conclusion: Local data do not fit the AP Examinee Distribution.
Decision: Do not reject null when a = 0.01
Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.

77.

  1. H0: The actual college majors of graduating females fit the distribution of their expected majors
  2. Ha: The actual college majors of graduating females do not fit the distribution of their expected majors
  3. df = 10
  4. chi-square distribution with df = 10
  5. test statistic = 11.48
  6. p-value = 0.3211
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Do not reject null when a = 0.05 and a = 0.01
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.

79. true

81. true

83. false

85.

  1. H0: Surveyed obese fit the distribution of expected obese
  2. Ha: Surveyed obese do not fit the distribution of expected obese
  3. df = 4
  4. chi-square distribution with df = 4
  5. test statistic = 54.01
  6. p-value = 0
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

Test of Independence – Homework

87.

  1. H0: Car size is independent of family size.
  2. Ha: Car size is dependent on family size.
  3. df = 9
  4. chi-square distribution with df = 9
  5. test statistic = 15.8284
  6. p-value = 0.0706
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.

89.

  1. H0: Honeymoon locations are independent of the bride’s age.
  2. Ha: Honeymoon locations are dependent on the bride’s age.
  3. df = 9
  4. chi-square distribution with df = 9
  5. test statistic = 15.7027
  6. p-value = 0.0734
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.

91.

  1. H0: The types of fries sold are independent of the location.
  2. Ha: The types of fries sold are dependent on the location.
  3. df = 6
  4. chi-square distribution with df = 6
  5. test statistic =18.8369
  6. p-value = 0.0044
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence that types of fries and location are dependent.

93.

  1. H0: Salary is independent of level of education.
  2. Ha: Salary is dependent on level of education.
  3. df = 12
  4. chi-square distribution with df = 12
  5. test statistic = 255.7704
  6. p-value = 0
  7. Check student’s solution.

  8. Alpha: 0.05

    Decision: Reject the null hypothesis.

    Reason for decision: p-value < alpha

    Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.

95. true

97. true

99.

  1. H0: Age is independent of the youngest online entrepreneurs’ net worth.
  2. Ha: Age is dependent on the net worth of the youngest online entrepreneurs.
  3. df = 2
  4. chi-square distribution with df = 2
  5. test statistic = 1.76
  6. p-value 0.4144
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the age and net worth of the youngest online entrepreneurs are dependent.

Test for Homogeneity – Homework

101.

  1. H0: The distribution for personality types is the same for both majors
  2. Ha: The distribution for personality types is not the same for both majors
  3. df = 4
  4. chi-square with df = 4
  5. test statistic = 3.01
  6. p-value = 0.5568
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.

103.

  1. H0: The distribution of fish caught is the same in Green Valley Lake and in Echo Lake.
  2. Ha: The distribution of fish caught is not the same in Green Valley Lake and in Echo Lake.
  3. 3
  4. chi-square with df = 3
  5. 11.75
  6. p-value = 0.0083
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake

105.

  1. H0: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.
  2. Ha: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.
  3. df = 4
  4. chi-square with df = 4
  5. test statistic = 2.7434
  6. p-value = 0.7395
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.

Comparison of the Chi-Square Tests – Homework

107.

  1. H0: The distribution for technology use is the same for community college students and university students.
  2. Ha: The distribution for technology use is not the same for community college students and university students.
  3. 2
  4. chi-square with df = 2
  5. 7.05
  6. p-value = 0.0294
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.

Test of a Single Variance – Homework

110. 225

112. H0: σ2 ≤ 150

114. 36

118. The claim is that the variance is no more than 150 minutes.

120. a Student’s t– or normal distribution

122.

  1. H0: σ = 15
  2. Ha: σ > 15
  3. df = 42
  4. chi-square with df = 42
  5. test statistic = 26.88
  6. p-value = 0.9663
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Do not reject null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.

124.

  1. H0: σ ≤ 3
  2. Ha: σ > 3
  3. df = 17
  4. chi-square distribution with df = 17
  5. test statistic = 28.73
  6. p-value = 0.0371
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.

126.

  1. H0: σ = 2
  2. Ha: σ ≠ 2
  3. df = 14
  4. chi-square distiribution with df = 14
  5. chi-square test statistic = 5.2094
  6. p-value = 0.0346
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Reject the null hypothesis
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.

128.

The sample standard deviation is $34.29.

H0 : σ2 = 252

Ha : σ2 > 252

df = n – 1 = 7.

test statistic:[latex]{x}^{2}={x}^{2}_{7}=\frac{{{(n-1)}{s}^{2}}}{{{25}^{2}}}=\frac{{{(8-1)}{34.29}^{2}}}{{{25}^{2}}}={13.169}[/latex]

p-value: P(x27>13.169)=1–P(x27 ≤13.169)=0.0681

Alpha: 0.05

Decision: Do not reject the null hypothesis.

Reason for decision: p-value > alpha

Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.

Bringing It Together: Homework

130.

 

  1. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.
  2. Testing to see if the data fits the distribution “too well” or is too perfect.