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A heterozygous white-fruited squash plant is crossed with a yellow-fruited plant, yielding 200 seeds. Does the data support or not support the hypothesis?

Hypothesis: White is autosomal dominant to yellow. Of these, 110 produce white-fruited plants while only 90 produce yellow-fruited plants.

1 Answer
Dec 1, 2017


The data "supports" the hypothesis. (The data fails to reject the hypothesis.)


For this problem, I will use W to refer to an allele for white fruit, and w to refer to an allele for yellow fruit. Per our hypothesis, we're assuming that W (white) is dominant to w (yellow). Thus, using a standard Punnett square, we can get a feeling for what a single crossing of a heterozygous white-fruited plant (Ww) should be with a yellow-fruited plant (ww, since we're assuming W is dominant):

#{: (, , underline(" W "), , underline(" w "),), ("w", "|", underline(" Ww "), "|", underline(" ww "), "|"), ("w", "|", underline(" Ww "), "|", underline(" ww "), "|") :}#

From this, we would expect there to be #1/2# white-fruited and #1/2# yellow-fruited offspring. Since there were 200 seeds, that means we'd expect there to be 100 white and 100 yellow.

To perform the #chi^2# Goodness of Fit test in this case, recall that the general formula for a #chi^2# test is:

#chi^2 = sum_{i=1}^n (O_i - E_i)^2/E_i#

where #O_1, O_2, ... O_n# are the observed counts for each category, while #E_1, E_2, ... E_n# are the expected counts for each. A table shows our calculations for this problem:

#{: ("Color", O_i, E_i, (O_i-E_i), (O_i-E_i)^2, (O_i-E_i)^2/E_i), ("White", 110, 100, 10, 100, 1), ("Yellow", 90, 100, -10, 100, 1), (,,,,bar(bb("Sum")), bar(bb(2))) :}#

#chi^2 = 2#. To determine how this relates to our hypothesis, we have to choose an #alpha# level, which is often set at 5% (.05) for many such tests, and then we must reference a #chi^2# table to find the critical value for our test. Since there are only 2 categories (white and yellow), we have 1 (one) degree of freedom.

A #chi^2# table for 1 df (degree of freedom) at an #alpha = .05# level shows a critical value of 3.841. This means we have enough statistical weight to reject the hypothesis if our #chi^2# value is greater than 3.841.

Since our #chi^2# was only 2, which is not greater than 3.841, we do not have enough evidence to reject our hypothesis. (Put another way, the p-value for our test is about .157 per Excel, which is clearly not less than .05.)

Therefore, we can say that the data seem to support the hypothesis. (Actually, I'd feel more comfortable saying that the data do not support rejecting the hypothesis, but that's just me.)