Question

A geneticist discovered a new species of insect with variable antennae length. He performed the following cross (Bb x bb) and counted the number of offspring with different antennae lengths. He was unsure if his observations matched what he expected to see, so he performed a chi-squared goodness of fit test and calculated ?2 = 0.051. If only two phenotypes were possible, can he reject the null hypothesis? No. P > 0.05 so any difference seen was not due to chance. No. P < 0.05 so any difference seen was due to chance. Yes. P > 0.05 so any difference seen was due to chance. Yes. P < 0.05 so any difference seen was due to chance. Yes. P < 0.05 so any difference seen was not due to chance. No. P < 0.05 so any difference seen was not due to chance. No. P > 0.05 so any difference seen was due to chance. Yes. P > 0.05 so any difference seen was not due to chance.

Answer 1

Our nu hypothesis is that there is no significant difference between results.

Alternate hypothesis is that there are significant difference between results.

As experimental Chi square value is 0.051.

Degree of freedom is = n-1

N= number of phenotype.

Degree of freedom is 2-1= 1.

Chi square value at biologically accepted value is 3.841

As our experimental Chi square value is less than biologically accepted Chi square value. That's why we should accept our result and accept our null hypothesis. The small differences are may be due to by chance.

That's why right option is:

No, P< 0.05 so any differences due to chance.