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Non-Parametric Statistics Part I: Chi-Square

Non-Parametric Statistics Part I: Chi-Square. c 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

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Non-Parametric Statistics Part I: Chi-Square

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  1. Non-Parametric Statistics Part I:Chi-Square c2

  2. x2 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 (a=.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 x2

  3. x2 :Homogeneity What are the null and alternative hypotheses? H0: The groups do not have different frequencies. H1: The groups do have different frequencies. Find the critical value: x2 table (k-1 df = 3-1= 2) = 5.99 Calculate the obtained statistic: = (145 + 301 + 289)/3 = 245 245 245 245 = 61.52 Make a decision: Our obtained value is larger than our critical value. Reject the null; the groups do have different frequencies.

  4. x2 :Homogeneity Example Is political affiliation distributed equally in our class? (use alpha=.01) What are the null and alternative hypotheses? H0: The groups do not have different frequencies. H1: The groups have different frequencies. Find the critical value: x2 table (k-1 df = 3-1= 2) = 9.21 Calculate the obtained statistic: = (10 + 15 + 5)/3 = 10 # of people expected 10 10 10 = 5 Make a decision: Our obtained value is smaller than our critical value. Retain the null; the groups do not have different frequencies.

  5. x2 :Goodness of Fit We need a different expected value based on the previous sample. Five years ago the tree-lot was also sampled. Has the composition of the lot changed since then (use alpha=.05)? Total # 735 473 Pine proportion = 255/473 = 0.54 Maple proportion = 115/473 = 0.24 Oak proportion = 103/473 = 0.22 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! Pine expected = 0.54(735) = 396.9 Maple expected = 0.24(735) = 176.4 Oak expected = 0.22(735) = 161.7

  6. x2 :Goodness of Fit Example What are the null and alternative hypotheses? H0: The composition of the lot has not changed. H1: The composition of the lot has changed. Find the critical value: x2 table (k-1 df = 3-1= 2) = 5.99 Calculate the obtained statistic: = 348.10 Make a decision: Our obtained value is larger than our critical value. Reject the null; the composition of the lot has changed.

  7. x2 :Independence Pine Maple Oak Mirkwood 123 234 345 Old Forest 233 232 333 H0: Tree type and Forest are independent. H1: Tree type and Forest are not independent.

  8. x2 :Independence Example (assume alpha=.05) What are the null and alternative hypotheses? H0: Tree type and forest are independent. H1: Tree type and forest and not independent. Find the critical value: df for this test is (r-1)(c-1) We have 2 rows and 3 columns, so (2-1)(3-1) = 2 x2 table (df = 2) = 5.99 Calculate the obtained statistic:

  9. x2 :Independence How to calculate expected values: Pine Maple Oak Mirkwood 123 234 345 Old Forest 233 232 333 R 702 798 356 C 466 678 Grand Total: 1500 Expected value = (R x C)/ grand total Expected Mirkwood-Pine = (702 x 356)/1500 = 166.61 Expected Old Forest-Pine = (798 x 356)/1500 = 189.39

  10. ( O - E)2 x2= E S x2 :Independence Observed Values Expected Values Pine Maple Oak Mirkwood166.8 218.1 317.3 Old Forest 189.4 247.9 360.7 Pine Maple Oak Mirkwood 123 234 345 Old Forest 233 232 333 = 28.18

  11. x2 :Independence Example (assume alpha=.05) What are the null and alternative hypotheses? H0: Tree type and forest are independent. H1: Tree type and forest and not independent. Find the critical value: df for this test is (r-1)(c-1) We have 2 rows and 3 columns, so (2-1)(3-1) = 2 x2 table (df = 2) = 5.99 28.18 Calculate the obtained statistic: Make a decision: Our obtained value is larger than our critical value. Reject the null; tree type and forest are not independent.

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