Question

Gregor Mendel established the concept of dominant and recessive genes and characteristics. According to Mendel’s law when crossing two inbred lines, 75% of the offspring should show dominant and 25% recessive characteristics. In his 19th century work studying inheritance of seed form in peas, Mendel observed 7,324 individuals, 5,474 of which had dominant characteristics.

a. Does this support or contradict Mendel’s law? Explain.

b. If the observation would be 5,566 (out of 7,324), would it be likely that his conjecture is true, i.e., 75% percent would show dominant characteristics? Explain.

Answer 1

H0: p =0.75 (claim) (Mendel’s law)

H1: p 0.75 (two-tailed test)

[H0: Null Hypothesis; H1: Alternative Hypothesis; p =Population proportion of dominant characteristics; Hypothesised population proportion, p₀ =0.75].

(a)

Sample proportion, =x/n =5,474/7,324 =0.7474

Standard Error, SE = = =0.005077

Test statistic, Z = =(0.7474 - 0.75)/0.005077 = -0.5121

Let the significance level be 5% =0.05

For a two-tailed test, at 0.05 significance level, critical value of Z is: Z_crit =1.96

p-value for Z statistic of -0.5121, for a two-tailed test is: p-value =0.6086

Decision criteria: Reject H0 if Z < -1.96 or Z > 1.96 (or reject H0 if p-value < significance level of 0.05)

Conclusion: Since Z of -0.5121 > -1.96, we failed to reject the null hypothesis, H0 at 5% significance level (or failed to reject H0 because the p-value: 0.6086 > 0.05).

Thus, it supports Mendel’s law because we do not have a sufficient statistical evidence to reject Mendel’s law (H0) and p-value is very high supporting H0.

(b)

=5566/7324 =0.76

SE =sqrt[0.76*0.24/7324] =0.0049904

Z =(0.76 - 0.75)/0.0049904 =2.0038

p-value (for a two-tailed test, Z =2.0038) =0.0451

Since p-value of 0.0451 < 0.05 significance level, we reject the null hypothesis, H0 at 5% significance level.

Thus, it would not be likely that his conjecture is true because we have a sufficient statistical evidence to reject Mendel’s law (H0).