Analyze and interpret data. Include a table of descriptive statistics (observed

Question

Analyze and interpret data. Include a table of descriptive statistics (observed and expected values) At five study sites, you have trapped kangaroo rats of three species and you expect to find that Giant kangaroo rats are twice as abundant as either of the other two species. Determine whether the five sites may justifiably be pooled and if the abundance is in the expected 2:1:1 ratio. Study site: A B C D E Giant kangaroo rat 87 92 12 69 54 Heermann's kangaroo rat 45 42 15 39 28 Tipton kangaroo rat 39 51 9 34 29

Explanation / Answer

To test that the expected ratio is 2:1:1. We multiply the observed count of Heermann's kangaroo rat and Tipton kangaroo rat by 2 and test the null hypothesis of Chi-Square Test for Homogeneity to test that the all populations are equal (null hypothesis)

DF = (r - 1) * (c - 1) = (3-1) *(5-1) = 8

Expected frequency counts Er,c = (nr * nc) / n

E(1,1) = 314*255/956 = 83.75

E(1,2) = 314*278/956 = 91.31

E(1,3) = 314*60/956 = 19.71

E(1,4) = 314*195/956 = 64.05

E(1,5) = 314*168/956 = 55.18

E(2,1) = 318*255/956 = 84.82

E(2,2) = 318*278/956 = 92.47

E(2,3) = 318*60/956 = 19.96

E(2,4) = 318*195/956 = 64.86

E(2,5) = 318*168/956 = 55.88

E(3,1) = 324*255/956 = 86.42

E(3,2) = 324*278/956 = 94.22

E(3,3) = 324*60/956 = 20.33

E(3,4) = 324*195/956 = 66.09

E(3,5) = 324*168/956 = 56.93

2 = [ (Or,c - Er,c)2 / Er,c ] = (87-83.75)^2/83.75 + (92-91.31)^2/91.31 + .... + (58-56.93)^2/56.93

= 12.228

p-value for chi-square stat = 12.228 at df = 8 is 0.1413

As, the p-value > 0.05, we fail to reject the null hypothesis and conclude that the expected ratio of Giant Kangaroo Rat, Heermann's kangaroo rat and Tipton kangaroo rat have expected ratio 2:1:1.

Study Site A B C D E Total Giant Kangaroo Rat 87 92 12 69 54 314 Heermann's kangaroo rat 90 84 30 58 56 318 Tipton kangaroo rat 78 102 18 68 58 324 Total 255 278 60 195 168 956

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