Concept explainers
To perform hypothesis testing.
Answer to Problem T11.13SPT
There is convincing evidence that there is association between education and smoking status.
Explanation of Solution
Given:
Education | ||||
Primary school | Secondary | University | ||
Smoking status | Nonsmoker | 56 | 37 | 53 |
Former | 54 | 43 | 28 | |
Moderate | 41 | 27 | 36 | |
Heavy | 36 | 32 | 16 |
The null and alternative hypotheses:
Using excel,
Primary school | Secondary | University | Total | ||
Nonsmoker | Observed | 56 | 37 | 53 | 146 |
Expected | 59.48 | 44.21 | 42.31 | 146.00 | |
(O - E)² / E | 0.20 | 1.18 | 2.70 | 4.08 | |
Former | Observed | 54 | 43 | 28 | 125 |
Expected | 50.93 | 37.85 | 36.22 | 125.00 | |
(O - E)² / E | 0.19 | 0.70 | 1.87 | 2.75 | |
Moderate | Observed | 41 | 27 | 36 | 104 |
Expected | 42.37 | 31.49 | 30.14 | 104.00 | |
(O - E)² / E | 0.04 | 0.64 | 1.14 | 1.83 | |
Heavy | Observed | 36 | 32 | 16 | 84 |
Expected | 34.22 | 25.44 | 24.34 | 84.00 | |
(O - E)² / E | 0.09 | 1.69 | 2.86 | 4.64 | |
Total | Observed | 187 | 139 | 133 | 459 |
Expected | 187.00 | 139.00 | 133.00 | 459.00 | |
(O - E)² / E | 0.53 | 4.21 | 8.57 | 13.30 | |
13.30 | chi-square | ||||
6 | df | ||||
.0384 | p-value |
Chi-square test statistic = 13.30. The degrees of freedom = 6 and p-value = 0.0384
Decision: P-value < 0.05, reject H0.
Conclusion: There is convincing evidence that there is association between education and smoking status.
Chapter 11 Solutions
PRACTICE OF STATISTICS F/AP EXAM
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