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.
- Reject the null hypothesis.
- p-value < alpha
- 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
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 |
- The data fit the distribution.
- The data does not fit the distribution.
- 3
- chi-square distribution with df = 3
- 19.27
- 0.0002
- Check student’s solution.
- Alpha = 0.05
- Decision: Reject null
- Reason for decision: p-value < alpha
- 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.
77.
- H0: The actual college majors of graduating females fit the distribution of their expected majors
- Ha: The actual college majors of graduating females do not fit the distribution of their expected majors
- df = 10
- chi-square distribution with df = 10
- test statistic = 11.48
- p-value = 0.3211
- Check student’s solution.
- Alpha = 0.05
- Decision: Do not reject null when a = 0.05 and a = 0.01
- Reason for decision: p-value > alpha
- 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.
- H0: Surveyed obese fit the distribution of expected obese
- Ha: Surveyed obese do not fit the distribution of expected obese
- df = 4
- chi-square distribution with df = 4
- test statistic = 54.01
- p-value = 0
- Check student’s solution.
- Alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: p-value < alpha
- 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.
- H0: Car size is independent of family size.
- Ha: Car size is dependent on family size.
- df = 9
- chi-square distribution with df = 9
- test statistic = 15.8284
- p-value = 0.0706
- Check student’s solution.
- Alpha: 0.05
- Decision: Do not reject the null hypothesis.
- Reason for decision: p-value > alpha
- Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.
89.
- H0: Honeymoon locations are independent of the bride’s age.
- Ha: Honeymoon locations are dependent on the bride’s age.
- df = 9
- chi-square distribution with df = 9
- test statistic = 15.7027
- p-value = 0.0734
- Check student’s solution.
- Alpha: 0.05
- Decision: Do not reject the null hypothesis.
- Reason for decision: p-value > alpha
- Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.
91.
- H0: The types of fries sold are independent of the location.
- Ha: The types of fries sold are dependent on the location.
- df = 6
- chi-square distribution with df = 6
- test statistic =18.8369
- p-value = 0.0044
- Check student’s solution.
- Alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: p-value < alpha
- Conclusion: At the 5% significance level, there is sufficient evidence that types of fries and location are dependent.
93.
- H0: Salary is independent of level of education.
- Ha: Salary is dependent on level of education.
- df = 12
- chi-square distribution with df = 12
- test statistic = 255.7704
- p-value = 0
- Check student’s solution.
-
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.
- H0: Age is independent of the youngest online entrepreneurs’ net worth.
- Ha: Age is dependent on the net worth of the youngest online entrepreneurs.
- df = 2
- chi-square distribution with df = 2
- test statistic = 1.76
- p-value 0.4144
- Check student’s solution.
- Alpha: 0.05
- Decision: Do not reject the null hypothesis.
- Reason for decision: p-value > alpha
- 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.
- H0: The distribution for personality types is the same for both majors
- Ha: The distribution for personality types is not the same for both majors
- df = 4
- chi-square with df = 4
- test statistic = 3.01
- p-value = 0.5568
- Check student’s solution.
- Alpha: 0.05
- Decision: Do not reject the null hypothesis.
- Reason for decision: p-value > alpha
- Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.
103.
- H0: The distribution of fish caught is the same in Green Valley Lake and in Echo Lake.
- Ha: The distribution of fish caught is not the same in Green Valley Lake and in Echo Lake.
- 3
- chi-square with df = 3
- 11.75
- p-value = 0.0083
- Check student’s solution.
- Alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: p-value < alpha
- Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake
105.
- H0: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.
- Ha: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.
- df = 4
- chi-square with df = 4
- test statistic = 2.7434
- p-value = 0.7395
- Check student’s solution.
- Alpha: 0.05
- Decision: Do not reject the null hypothesis.
- Reason for decision: p-value > alpha
- 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.
- H0: The distribution for technology use is the same for community college students and university students.
- Ha: The distribution for technology use is not the same for community college students and university students.
- 2
- chi-square with df = 2
- 7.05
- p-value = 0.0294
- Check student’s solution.
- Alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: p-value < alpha
- 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.
- H0: σ = 15
- Ha: σ > 15
- df = 42
- chi-square with df = 42
- test statistic = 26.88
- p-value = 0.9663
- Check student’s solution.
- Alpha = 0.05
- Decision: Do not reject null hypothesis.
- Reason for decision: p-value > alpha
- Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
124.
- H0: σ ≤ 3
- Ha: σ > 3
- df = 17
- chi-square distribution with df = 17
- test statistic = 28.73
- p-value = 0.0371
- Check student’s solution.
- Alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: p-value < alpha
- Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
126.
- H0: σ = 2
- Ha: σ ≠ 2
- df = 14
- chi-square distiribution with df = 14
- chi-square test statistic = 5.2094
- p-value = 0.0346
- Check student’s solution.
- Alpha = 0.05
- Decision: Reject the null hypothesis
- Reason for decision: p-value < alpha
- 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.
- 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.
- Testing to see if the data fits the distribution “too well” or is too perfect.