SPSS assignment
Danyah
lecture__11_chi-square_analysis_sp_15.pptxChi-Square Analysis
Chi-Square Statistics
The Chi-square test uses frequency data to generate a statistical results.
Chi-Square analysis is used to:
assess the relationship between two qualitative variables.
test whether a frequency fits a specific pattern (expected frequencies).
Where O is the observed frequency and E is the expected frequency
Types of Chi-Square Tests
Chi-square for goodness-of- fit:
This test is used to test how closely an observed distribution matches an expected distribution.
The null hypothesis: Observed distribution fits an expected distribution
Contingency table analysis –
Chi-square for independence:
This test is used to assess the association between two qualitative variables
The null hypothesis: There is no association between the two variables
Chi-square test for homogeneity:
This test is used to test the claim that different populations have the same proportions of some characteristics.
The null hypothesis: The distribution of the categorical (qualitative) variable is the same across the population
Types of Chi-Square Tests
McNemar’s test :
This test is use to assess the relationship between two paired (related) qualitative variables.
It is similar to paired t-test for paired quantitative variables
The null hypothesis: the proportion of subjects with the specific characteristic (or event) is the same before and after the exposure or the intervention
1. Suppose there are n observations.
2. Each observation falls into a cell (or class).
3. Observed frequencies in each cell: O1, O2, O3, … , Ok.
Sum of the observed frequencies is n.
4. Expected, or theoretical, frequencies: E1, E2, E3, . . . , Ek.
Sum of the expected frequencies is n
Goodness- of- Fit Test
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Example: A group of researchers from the nursing school has suggested that normal births do not take place randomly throughout the day. The researchers observed delivery frequency of 28 deliveries within 24 hours and obtained the following results:
| Observed | |
| 12- 6 am | 12 |
| 6 – 12 am | 5 |
| 12 – 6 pm | 3 |
| 6 – 12 pm | 8 |
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Answer:
H0: Observed number of births do not vary overtime
| Observed | Expected | Test Statistic | ||
| 12- 6 am | 12 | 7 | 3.57 | |
| 6 – 12 am | 5 | 7 | 0.57 | |
| 12 – 6 pm | 3 | 7 | 2.29 | |
| 6 – 12 pm | 8 | 7 | 0.14 | |
| Total | 28 | 28 | 6.57 | |
| Expected value=(12+5+3+8)/4=7 | ||||
| Degree of freedom= n – 1= 4-1=3 Critical value at alpha of 0.05 = 7.815 * n is the number of groups |
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Results:
Decision: Fail to reject H0.
Conclusion: At the 0.05 level of significance, there is no evidence to suggest delivery varies overtime
Test statistic χ 2 = 6.57
Critical value χ2 = 7.815
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Contingency Table Analysis
Test of Independence
Data is sorted into cells, and the observed frequency in each cell is reported in a cross tabulation.
Cross tabulation involves two qualitative variables
Typical question: Are the two variables independent or dependent?
Are the socioeconomic status and levels of physical activity independent?
Is there any association between exercise status at baseline and gender?
The null hypothesis: There is no association (relationship) between the two variables
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H0: Smoking status at baseline is independent of gender
Example: Does smoking status at baseline depend on gender?
| BASELINE SMOKING STATUS * GENDER | ||||
| Gender | Total | |||
| Male | Female | |||
| Smoking Status | Smoker | 18 | 19 | 37 |
| Non-smoker | 149 | 240 | 389 | |
| Total | 167 | 259 | 426 |
Observed Values
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H0: Smoking status at baseline is independent of gender
Example: Does smoking status at baseline depend on gender?
| BASELINE SMOKING STATUS * GENDER | ||||
| Gender | Total | |||
| Male | Female | |||
| Smoking Status | Smoker | (37*167)/426=14.5 | (37*259)/426=22.5 | 37 |
| Non-smoker | (389*167)/426=152.5 | (389*259)/426=236.5 | 389 | |
| Total | 167 | 259 | 426 |
Expected Values
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H0: Smoking status at baseline is independent of gender
Example: Does smoking status at baseline depend on gender?
| BASELINE SMOKING STATUS * GENDER | |||||
| Gender | Total | ||||
| Male | Female | ||||
| Smoking Status | Smoker | Observed | 18 | 19 | 37 |
| Expected | 14.5 | 22.5 | |||
| Non-smoker | Observed | 149 | 240 | 389 | |
| Expected | 152.5 | 236.5 | |||
| Total | 167 | 259 | 426 |
Observed and Expected Values
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| BASELINE SMOKING STATUS * GENDER | |||||
| Gender | Total | ||||
| Male | Female | ||||
| Smoking Status | Smoker | Observed | 18 | 19 | 37 |
| Expected | 14.5 | 22.5 | |||
| Non-smoker | Observed | 149 | 240 | 389 | |
| Expected | 152.5 | 236.5 | |||
| Total | 167 | 259 | 426 |
Chi-square analysis
Test Statistics
=====1.52
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Chi-square test statistic = 1.52
3.84
Degrees of freedom: (rows - 1)(columns - 1) = 1
Critical value from table (α = 0.05) = 3.84
y
x
FTR
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Results
Decision: Fail to reject H0
Conclusion: at the 0.05 level of significance, there is no association between smoking status at baseline and gender
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Contingency Table Analysis
Test of Homogeneity
Data is sorted into cells, and the observed frequency in each cell is reported in a cross tabulation.
Cross tabulation involves two variables
This test is used to test the claim that different populations have the same proportions of some characteristics (to test the equality of proportions in different populations).
The null hypothesis: The distribution of the categorical (qualitative) variable is the same across the populations.
It is computed exactly the same as the chi-square test for independence.
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Example: Is the proportion of the students who drive their own or their parents’ cars the same at all three schools ?
| Drive their own or their parents’ cars * School | |||||
| School | |||||
| A | B | C | Total | ||
| Drive their own or their parents’ cars | Yes | 18 | 22 | 16 | 56 |
| No | 32 | 28 | 34 | 94 | |
| Total | 50 | 50 | 50 | 150 |
Observed Values
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Expected Values
| Drive their own or their parents’ cars * School | |||||
| School | |||||
| A | B | C | Total | ||
| Drive their own or their parents’ cars? | Yes | (56*50)/150=18.67 | (56*50)/150=18.67 | (56*50)/150=18.67 | 56 |
| No | (94*50)/150=31.33 | (94*50)/150=31.33 | (94*50)/150=31.33 | 94 | |
| Total | 50 | 50 | 50 | 150 |
Example: Is the proportion of the students who drive their own or their parents’ cars the same at all three schools ?
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| Drive their own or their parents’ cars * School | ||||||
| School | Total | |||||
| A | B | C | ||||
| Drive their own or their parents’ cars? | Yes | Observed | 18 | 22 | 16 | 56 |
| Expected | 18.67 | 18.67 | 18.67 | |||
| No | Observed | 32 | 28 | 34 | 94 | |
| Expected | 31.33 | 31.33 | 31.33 | |||
| Total | 50 | 50 | 50 | 150 |
Observed and Expected Values
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Chi-square analysis
Test Statistics
| Drive their own or their parents’ cars * School | ||||||
| School | Total | |||||
| A | B | C | ||||
| Drive their own or their parents’ cars | Yes | Observed | 18 | 22 | 16 | 56 |
| Expected | 18.67 | 18.67 | 18.67 | |||
| No | Observed | 32 | 28 | 34 | 94 | |
| Expected | 31.33 | 31.33 | 31.33 | |||
| Total | 50 | 50 | 50 | 150 |
=======1.60
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Chi-square test statistic = 1.596
5.991
Degrees of freedom: (rows - 1)(columns - 1) = 2
Critical value from table (α = 0.05) = 5.991
y
x
FTR
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Results
Decision: Fail to reject H0
Conclusion: at the 0.05 level of significance, the proportion of the students who drive their own or their parents’ cars is the same at all three schools
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General rules to follow in using Chi-square:
No more than 20% of the cells should have an expected frequency less than 5.
No expected frequency should be less than 1.
If the expected frequencies are too small then FISHER’S EXACT TEST should be used in place of the Chi-square.
Observations MUST be considered to be independent. This means that it cannot be used on a ‘before and after’ problem.
If independence cannot be assumed then another non-parametric test called the MCNEMAR test must be used.
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Example: Does baseline exercise level depend on martial status?
H0: Exercise level at baseline is independent of marital status
Assumption not met
We will use Fisher’s Exact test to answer this question because the assumption of Pearson Chi-Square was not met (The minimum expected count is less than 1)
Example: Does the number of smokers change after 6 weeks compared to baseline?
We will use McNemar test to answer this question because it is before and after situation and the variables are categorical.
Example: Does the number of smokers change after 6 weeks compared to baseline?
We will use McNemar test to answer this question because it is before and after situation and the variables are categorical.
Results
Decision: Fail to reject H0
Conclusion: at the 0.05 level of significance, the number of smokers does not change after 6 weeks compared to baseline
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| Observed Frequency | O1 | O2 | O3 | Ok | n | |
| Expected Frequency | E1 | E2 | E3 | Ek | n |
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