What's your interpretation? |
March and April, 1997 |
This comparison shows the farrowing rates of two groups of breeding females within a single herd. One group was mated using natural service (NS), and artificial insemination (AI) was used in the other. Breeding females were assigned at random at the time of mating to either treatment. The statistical question of interest is: "What is the probability that the observed difference in farrowing rates (76% versus 58%) could have occurred by chance?" Although this question is similar to that posed regarding average daily gain in the last issue, we cannot use the two-sample t-test to compare rates or proportions such as farrowing rates. What is the appropriate statistical test to use in this case? What information will we need to perform such a test?
n this experiment, 143 breeding females were randomly assigned at mating to either natural service (NS) or artificial insemination (AI). A success was recorded as each female consequently farrowed, and a failure recorded if the female returned to service, was culled, or failed to farrow for any reason. The appropriate test statistic to compare the two groups is the Chi-square test.
Step 1
To perform the Chi-square test, we cannot start with the two farrowing rates as presented in the graphic. Rather, the raw data (counts) are arranged in a 2 * 2 table (Table 1).
The essence of the Chi-square test is to compare the difference between the observed values in each cell with the values which we would expect to find if there was no difference between treatments.
Step 2
For each cell, we calculate an expected value (EV). The computation is quite easy -- simply divide the product of the margin totals by the grand total:
EV_{a} = 84 * 94 / 143 = 55.22
EV_{b} = 84 * 49 / 143 = 28.78
EV_{c} = 59 * 94 / 143 = 38.78
EV_{d} = 59 * 49 / 143 = 20.22
Step 3
Calculate the c^{2} value for each cell using the formula:
c^{2} = (observed - expected)^{2} / expected
c^{2}_{a} = (49 - 55.22)^{2} / 55.22 = 0.70
c^{2}_{b} = (35 - 28.78)^{2} / 28.78 = 1.34
c^{2}_{c} = (45 - 38.78)^{2} / 38.78 = 1.00
c^{2}_{d} = (14 - 20.22)^{2} / 20.22 = 1.91
(Table 2)
Overall c^{2} = (0.70 + 1.34 + 1.00 + 1.91) = 4.95
Step 4
Calculate the degrees of freedom (d.f.):
d.f. = (Rows - 1) * (Cols - 1) = (2 - 1) * (2 - 1) = 1
Looking up c^{2} = 4.95, 1 d.f. in statistical tables, we find P = .0261.
Our interpretation is that the probability of our observed difference of 18% (58 vs. 76) occurring by chance is about 2.6% (26 times in 1000.) Therefore we can conclude that the observed difference in farrowing rate is statistically (and also economically!) significant. Clearly, some effort and investment in improving AI techniques in this herd is warranted.