Green Parakeet Crosses I

  1. α = 0.1.
  2. H0: “The distribution of parakeets will follow the 9 green to 3 blue to 3 yellow to 1 white theoretical distribution” and HA: “The distribution of parakeets will NOT follow the 9 green to 3 blue to 3 yellow to 1 white theoretical distribution”.
  • The hypotheses for a goodness-of-fit test will usually be “wordy” like this.
  • The H0 will always have the format of the “distribution of INDIVIDUALS into the LEVELS OF THE RESPONSE VARIABLE will follow the THEORETICAL DISTRIBUTION.” Thus, the H0 above simply has the capitalized items replaced with their specific meanings in this example. That is, the individuals are parakeets, the levels of the response variable are the colors (i.e. phenotypes), and the theoretical distribution is the 9:3:3:1 genetic theory.
  • The HA will be exactly like H0 except that it will read “NOT follow.”
  1. A Goodness-of-Fit test will be used because
    1. one group/population considered (this cross of parakeets) and
    2. a categorical variable (phenotypic color) was recorded.
  • These are the two characteristics that separate a Goodness-of-Fit test from other tests we will see this semester.
  • Please include some sort of description that shows that the characteristics are met and that you are not just listing the characteristics – in other words, include the parts in parentheses.
  1. This study is experimental because the researcher performed the crosses.
  2. The assumptions are met because the expected table has more than five individuals in all six cells
Green Blue Yellow White
84.375 28.125 28.125 9.375


  • Each expected value is found by multiplying the sample size (n=150) by the expected proportion in each group. Thus, the expected values for “Green” is \(150\frac{9}{16}\), for blue and yellow is \(150\frac{3}{16}\), and for white is \(150\frac{1}{16}\).
  • Expected values are expected counts of individuals, but we traditionally keep at least one decimal place so that our calculations below will be more precise.
  1. The statistic for a Goodness-of-Fit test is the observed frequency table.
Green Blue Yellow White Total
95 27 22 6 150


  • I usually add the marginal totals to my observed tables in preparation for making the expected tables.
  1. The χ2 test statistic is 3.932, as computed from the observed and expected tables as shown below. This test statistic has 3 df.

\[ \chi^{2} = \sum\frac{(obs-exp)^{2}}{exp} = \] \[ \frac{(95-84.375)^{2}}{84.375} + \frac{(27-28.125)^{2}}{28.125} + \frac{(22-28.125)^{2}}{28.125} + \frac{(6-9.375)^{2}}{9.375} = \] \[ 1.338 + 0.045 + 1.334 + 1.215 = \] \[ 3.932 \]

  • I urge you to write out each of the parts of the test statistic and the intermediate computation for each part. This will help you solve problems if something goes sideways.
  • The df is the number of groups (4) minus 1.
  1. p-value0.2689 as computed with distrib(3.932,distrib="chisq",df=3,lower.tail=FALSE).
  • When computing the p-value on a chi-square distribution it will always be a “right-of” and, thus, will always include lower.tail=FALSE.
  1. Do not reject H0 because the p-value > α.
  2. It appears that the phenotypic colors of the parakeets follows the 9:3:3:1 ratio suggested by the genetic theory. Thus, the genetic theory is upheld by these data.
  • There is not usually a Step 11 for a goodness-of-fit test. We will not ever do a Step 11 in this class.