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Close Window Save and Submi &Click; Submit to complete this assessment Question 5 Question S A genetioist discovered a new species lengths. He was unsure if his observations matched what he expected to see, so he performed a chi-squared of insect with variable antennae length. He performed the following cross (Bb x bb) and counted the number of offspring with different antennae 1 points Save Ansa goodness of ft tet and calcuated x2-403. only two phenotypes were Table34 Critical values of the x' distribution possible, can he reject the null hypothesis?9 .75a1 0.000 0,000 0016 0455 2.706 341 .02445 879 0.0100.051 21113 4605 .991 37 210 10.597 3 0.072021654 2.36 6251 75 34811.345 12.83 .207 484 1.064 3.357 7.779 9408 11.143 13277 460 O No. P 0.05 so any difference seen was due to chance. O Yes. P 0.05 so any difference seen was not due to chance O No. Pc0.05 so any difference seen was due to chance. o Yes. Pc0.05 so any difference seen was due to chance. O No. P>0.05 so any difference seen was not due to chance o Yes.P>0.05 so any difference seen was due to chance. O Yes.P xa.os so any difference seen was not due to chance. Question 5 ofs > cck submit to complete this assessment. Close Window

Explanation / Answer

For a chi square test, the degrees of freedom are (n-1) where n is the number of classes.

We reject the null hypothesis when p-value < ?

Here Calculated chi square = 4.03

and Tabulated chi square for 1 df and (?=.05) = 3.81

Hence (Calculated chi square) > (Tabulated chi square) which implies that (p-value for calculated chi square) < .05

Hence we reject the null hypothesis and conclude that any difference seen was not due to chance.

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