1 Hypothesis Testing Goodness-of-fit & Independence Chi-Squared Tests

2 Goodness-of-fit Test A managed forest has the following distribution of trees: Douglas Fir54% Ponderosa Pine40% Grand Fir5% Western Larch1% Mannan & Meslow (1984) made 156 observations of foraging by red-breasted nuthatches and found the following: Mannan, R.W., and E.C. Meslow “Bird populations and vegetation characteristics in managed and old-growth forests, northeastern Oregon.” J. Wildl. Manage. 48: Douglas Fir70 Ponderosa Pine 79 Grand Fir3 Western Larch4

3 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)

4 Expected Values Based on the percentage distribution of trees, the expected counts for each type (out of 156) are: Douglas Fir84.24 Ponderosa Pine62.40 Grand Fir7.80 Western Larch1.56 (54% of 156 = 84.24)

5 Chi-Square Statistic ExpectedObservedo-e(o-e)Sq(o-e)Sq/e Douglas Fir Ponderosa Pine Grand Fir Western Larch Chi-square = p-value = For p-value in Excel, type =CHIDIST(13.593,3), for 3 degrees of freedom (n groups -1)

6 Conclusion Given the small p-value, we reject the null. These birds are not foraging randomly – they prefer certain types of trees.

7 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? No Dog Have Dog Female2923 Male3524

8 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)

9 Chi Square Test Results Tabulated statistics: Gender, Dog Rows: Gender Columns: Dog No Yes All Female Male All Cell Contents: Count Expected count Pearson Chi-Square = 0.143, DF = 1, P-Value = Likelihood Ratio Chi-Square = 0.143, DF = 1, P-Value = 0.705