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Hypothesis Testing. Goodness-of-fit & Independence Chi-Squared Tests. Goodness-of-fit Test. A managed forest has the following distribution of trees:. Mannan & Meslow (1984) made 156 observations of foraging by red-breasted nuthatches and found the following:.
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Hypothesis Testing Goodness-of-fit & Independence Chi-Squared Tests
Goodness-of-fit Test A managed forest has the following distribution of trees: Mannan & Meslow (1984) made 156 observations of foraging by red-breasted nuthatches and found the following: Mannan, R.W., and E.C. Meslow. 1984. “Bird populations and vegetation characteristics in managed and old-growth forests, northeastern Oregon.” J. Wildl. Manage. 48: 1219-1238.
Hypotheses Do the birds forage randomly, without regard to what species of tree they are in? To be true, the observed and expected distributions should be alike. Null: The distributions are alike (good fit, meaning birds forage randomly) Alternate: The distributions are different (lack of fit, or birds prefer certain vegetation)
Expected Values Based on the percentage distribution of trees, the expected counts for each type (out of 156) are: (54% of 156 = 84.24)
Chi-Square Statistic For p-value in Excel, type =CHIDIST(13.593,3), for 3 degrees of freedom (n groups -1)
Conclusion Given the small p-value, we reject the null. These birds are not foraging randomly – they prefer certain types of trees.
Test of Independence Demographic data on 111 students is available. We wish to study gender differences, in this case pertaining to dog ownership. Data Set: Student Variables: Gender, Dog (Yes/No) Are Gender and Dog Ownership independent of each other?
The Hypotheses Null: The two variables are independent of each other (the occurrence of one does not influence the probability of the occurrence of the other.) Alternate: They are not independent(one influences the other)
Chi Square Test Results Tabulated statistics: Gender, Dog Rows: GenderColumns: Dog No Yes All Female 29 23 52 29.98 22.02 52.00 Male 35 24 59 34.02 24.98 59.00 All 64 47 111 64.00 47.00 111.00 Cell Contents: Count Expected count Pearson Chi-Square = 0.143, DF = 1, P-Value = 0.705 Likelihood Ratio Chi-Square = 0.143, DF = 1, P-Value = 0.705