Question

The white "Spirit" black bear (or Kermode) Ursus americanus kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene (MC1R).

In this population, the gene has two forms (or alleles): the "white" allele b and the "black" allele B. The trait is recessive: white bears have two copies of the white allele of this gene (bb), whereas a bear is black if it has one or two copies of the black allele (Bb or BB). Both color morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If possessing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with BB, Bb, or bb allele combinations in the population will follow a binomial distribution (that is BB- 25%, Bb- 50% and bb- 25%). To investigate, Hedrick and Ritland (2011) sampled and genotyped 87 bears from the northwest coast:

42 were BB

24 were Bb

21 were bb

Assume that this is a random sample.

A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes.

(a)  (null or alternative) hypothesis would be "The frequency distribution of genotypes has a binomial distribution in the population"

whereas "The frequency distribution of genotypes does not have a binomial distribution" is the  (null or alternative) hypothesis.

(b) The degrees of freedom for the test statistic are .

(c) The calculated chi-square value is  (report to one decimal place)

(d) The critical chi-square value at alpha =0.05 is  (report the whole number from the provided chi-square distribution table)

(e) The difference between the observed and expected frequencies is statistically significant.  (Yes or No)

(f) The calculated chi-square value exceeds the critical chi-square value corresponding to = 0.05.  (Yes or No)

(g) The calculated chi-square value exceeds the critical chi-square value corresponding  to  = 0.01.  (Yes or No)

Answer 1

(a)

Null hypothesis would be "The frequency distribution of genotypes has a binomial distribution in the population"

whereas "The frequency distribution of genotypes does not have a binomial distribution" is the alternative hypothesis.

(b)

The degrees of freedom for the test statistic = Number of groups - 1 = 3-1 = 2

(c)

Expected count = npi

Expected count for BB, E1 = 87 * 0.25 = 21.75

Expected count for Bb, E2 = 87 * 0.5 = 43.5

Expected count for bb, E3 = 87 * 0.25 = 21.75

Chi Square test statistic,

= 27.62069

(d) The critical chi-square value at alpha =0.05 and df= 2 is 5.99

(e)

Since the observed Chi Square test statistic is greater than the critical value, the difference between the observed and expected frequencies is statistically significant.  

Yes

(f)

The calculated chi-square value exceeds the critical chi-square value corresponding to = 0.05.  

Yes

(g)

The critical chi-square value at alpha =0.01 and df= 2 is 9.21

The calculated chi-square value exceeds the critical chi-square value corresponding  to  = 0.01.

Yes