1. Non-Parametric Statistics Part I: Chi-Square 

2. x 2 Operates on FREQUENCY Data Suppose we have a plot of land on which we hope to harvest wood. Maple is more valuable than Oak and Oak more valuable than pine. We take a sample of the trees (the whole plot is too big) and we ask whether there are significantly unequal amounts of each type (  =.05). We cannot get a mean from these data but there are clear differences between the amounts in each category. This is categorical or nominal data experessed as frequencies. So we use the x 2 PineMapleOak # of trees 145301289

3. What are the null and alternative hypotheses? = (145 + 301 + 289)/3 = 245 PineMapleOak # of trees observed 145301289 # of trees expected x 2 : Homogeneity 245 245 245 = 61.52 H0: The groups have equal frequencies. H1: The groups do not have equal frequencies. Find the critical value: Calculate the obtained statistic: x 2 table (k-1 df = 3-1= 2) = 5.99 Make a decision: Our obtained value is larger than our critical value. Reject the null; the groups do not have equal frequencies.

4. What are the null and alternative hypotheses? = (10 + 15 + 5)/3 = 10 DemocratRepublicanOther # of people observed 10155 x 2 : Homogeneity Example = 5 H0: The groups have equal frequencies. H1: The groups do not have equal frequencies. Find the critical value: Calculate the obtained statistic: x 2 table (k-1 df = 3-1= 2) = 9.21 Make a decision: Our obtained value is smaller than our critical value. Retain the null; the groups have equal frequencies. 10 10 10 # of people expected Is political affiliation distributed equally in our class? (use alpha=.01)

5. PineMapleOak # trees 2009 255115103 Five years ago the tree-lot was also sampled. Has the composition of the lot changed since then (use alpha=.05)? x 2 : Goodness of Fit PineMapleOak # trees 2014 145301289 We need a different expected value based on the previous sample. Notice we’re trying to compare the frequencies from two time points, but the total # of trees categorized in 2014 is different from the 2009 total! Total # 735 473 Pine proportion = 255/473 = 0.54 Maple proportion = 115/473 = 0.24 Oak proportion = 103/473 = 0.22 Pine expected = 0.54(735) = 396.9 Maple expected = 0.24(735) = 176.4 Oak expected = 0.22(735) = 161.7 PineMapleOak # trees expected 396.9176.4161.7

6. PineMapleOak # trees expected 396.9176.4161.7 What are the null and alternative hypotheses? x 2 : Goodness of Fit Example = 348.10 H0: The composition of the lot has not changed. H1: The composition of the lot has changed. Find the critical value: Calculate the obtained statistic: x 2 table (k-1 df = 3-1= 2) = 5.99 Make a decision: Our obtained value is larger than our critical value. Reject the null; the composition of the lot has changed. PineMapleOak # trees 2014 145301289

7. Pine Maple Oak Mirkwood123 234 345 Old Forest233 232 333 H0: Tree type and Forest are independent. H1: Tree type and Forest are not independent. x 2 : Independence

8. What are the null and alternative hypotheses? x 2 : Independence Example (assume alpha=.05) H0: Tree type and forest are independent. H1: Tree type and forest and not independent. Find the critical value: Calculate the obtained statistic: x 2 table (df = 2) = 5.99 df for this test is (r-1)(c-1) We have 2 rows and 3 columns, so (2-1)(3-1) = 2

9. Expected value = (R x C)/ grand total R 702 798 C 356466 678 Grand Total: 1500 Expected Old Forest-Pine = (798 x 356)/1500 = 189.39 Expected Mirkwood-Pine = (702 x 356)/1500 = 166.61 Pine Maple Oak Mirkwood123 234 345 Old Forest233 232 333 x 2 : Independence How to calculate expected values:

10. Observed ValuesExpected Values Pine Maple Oak Mirkwood 123 234 345 Old Forest 233 232 333 Pine Maple Oak Mirkwood 166.8 218.1 317.3 Old Forest 189.4 247.9 360.7   x 2 = E  = 28.18 x 2 : Independence

11. What are the null and alternative hypotheses? x 2 : Independence Example (assume alpha=.05) H0: Tree type and forest are independent. H1: Tree type and forest and not independent. Find the critical value: Calculate the obtained statistic: x 2 table (df = 2) = 5.99 df for this test is (r-1)(c-1) We have 2 rows and 3 columns, so (2-1)(3-1) = 2 28.18 Make a decision: Our obtained value is larger than our critical value. Reject the null; tree type and forest are not independent.