A biologist studying the common morning glory plant ( Ipomoea purpurea ) believe
ID: 3044786 • Letter: A
Question
A biologist studying the common morning glory plant (Ipomoea purpurea) believes that the color of its flowers (which are either purple or pink) is controlled by a single, dominant gene with the purple allele being dominant to the pink one. This would result in a ratio of 3:1 of purple to pink flowers.
Write down the null and alternative hypotheses the biologist would use to test whether this belief is correct.
Null Hypothesis:
Alternative Hypothesis:
(b) He/she then collects 100 flowers at random and observes that 84 are purple and 16 are pink. What are the expected numbers of purple and pink flowers in a sample of 100?
(c) Calculate the chi-square value for these data and provide the degrees of freedom for this test.
(d) What is the critical value of chi-squared in this situation if = 0.05 (use table on p.704-705 of textbook)?
(e) Based on this result, should the null hypothesis be rejected? What conclusion can the biologist draw about flower color in I. purpurea?
Explanation / Answer
null hypothesis: ratio of purple to pink flowers is 3:1.
alternate hypothesis: ratio of purple to pink flowers is differnet from 3:1.
b)
expected number of purple flowers =(3/4)*100 =75
expected number of pink flowers =(1/4)*100 =25
c)
applying chi square test of goodness of fit:
from above chi-square value =4.32
d) for degree of freedom =categories-1=2-1 =1
hence for 1 df and 0.05 level critical value of chi-square =3.841
e)
as test statistic is greater than critical value;we reject null hypothesis
we can conclude that ratio of purple to pink flowers is different from 3:1.
observed Expected residual Chi square x Probability(p) Oi Ei=total*p Ri=(Oi-Ei)/Ei R2i=(Oi-Ei)2/Ei purple 0.7500 84.000 75.0 1.04 1.08 pink 0.2500 16.000 25.0 -1.80 3.24 total 1 100 100 -0.761 4.320Related Questions
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