Question

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Two human traits controlled by a single gene are the ability to roll one’s tongue and whether one’s ear lobes are free or attached to the neck. Genetic theory says that people will have neither one, one, or both of these traits in the ratio 1:3:3:9 (1/16—attached, noncurling; 3/16—attached, curling; 3/16—free, noncurling; 9/16— free, curling). A biology class of 122 students collected these data:

ear lobes, tongue observed count attached, non-curling 10 attached, curling 22 free, non-curling 31 free, curling 59

Are these data consistent with the genetic theory?

a. State appropriate hypotheses and identify the appropriate test.

Chi-Square Test Independence

b.​​​​​​​ Assuming these data are independent, how do you know if this sample is big enough?

c. ​​​​​​​Calculate the test statistic and p-value.

d.​​​​​​​​​​​​​​ Write a meaningful conclusion.

Answer 1

Expected frequency (E_i)= Expected proprtion * sample size(n)

Therefore in our case the expected proportion for each category are given as :

attached and noncurling : 1/16 =0.0625
attached and curling : 3/16 = 0.1875
free and noncurling :  3/16 = 0.1875
free and curling : 9/16 = 0.5625

The table has observed frequencies and expected frequencies for the sample calculated by the formula on the first line above.

The following table is obtained: Categories Observed Expected attached and noncurling 10 122*0.0625=7.625 attached and curling 22 122*0.1875=22.875 (fo-fe)?lfe (10-7.625)217.625 = 0.74 (22-22.875)2/22.875 = 0.033 (31-22.875)2/22.875 = 2.886 (59-68.625)2/68.625 = 1.35 free and noncurling 31 122*0.1875=22.875 free and curling 59 122*0.5625=68.625 Sum = 122 122 5.009 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho : P1 = 0.0625, P2 = 0.1875, P3 = 0.1875, P4 = 0.5625 Ha Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 4 – 1 = 3, so then the rejection region for this test is R = {x? : x² > 7.815). (3) Test Statistics The Chi-Squared statistic is computed as follows: (O; – E;) E 0.74 + 0.033 + 2.886 +1.35 5.009 i=1 (4) Decision about the null hypothesis Since it is observed that x= 5.009 < xã = 7.815, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the a = 0.05 significance level.