Question

there are 200 families of parents heterozygous for albinism, which present normal pigmentation. They all had 4 children, where 60 families with 4 children were found with normal pigmentation, 58 families with 3 children with normal pigmentation, 57 families with 2 children with normal pigmentation, 16 families with 1 child with normal pigmentation and 9 families with 4 albino children. Using the binomial expansion determine the theoretical probability of the possible 5 combinations. Using chi 2, prove that the distribution obtained from people without the condition and with albinism in the 200 families is consistent with what was expected.

Answer 1

Consider X denotes number of children with normal pigmentation.

X	No. of Children (f)	x*f
0	9	0
1	16	16
2	57	114
3	58	174
4	60	240
total	200	544

The random variable X follows binomial distribution with n = 4 and p.

where p =p ( child with normal pigmentation)

for binomial distribution E(X) = mean = np

mean = 544 / 200 = 2.72.

Hence estimated value of p is

phat = 2.72 / 4 = 0.68.

The probability mass function of X is given by

and Expected frequency = N * P(X=x).

Here N = total number of families = 200.

X	P(X=x)	Expected Freq.   = N*P(X=x)
0	0.0105	2.0972
1	0.0891	17.8258
2	0.2841	56.8197
3	0.4025	80.4946
4	0.2138	42.7628
Total	1.0000	200.0000

We have to test the hypothesis that

Whether or not distribution obtained from people without the condition and with albinism in the 200 families is consistent?

i.e. The given frequency distribution is good fitted as binomial distribution or not?

We used Chi-square goodness of fit test.

The value of chi-square test statistic is

Where Oi : Observed Frequency , Ei = Expected frequency and k = number of classes = 5.

Oi	Ei	(Oi-Ei)^2/Ei
9	2.0972	22.7201
16	17.8258	0.1870
57	56.8197	0.0006
58	80.4946	6.2862
60	42.7628	6.9481
200	200.000	36.1420

The value of chi-square test statistic is 36.1420

at 5% level of significance Critical value is

Since calculated value of test statistic is greater than critical value( 36.1420 > 9.488).

We reject Ho at 5% level of significance.

Conclusion : The distribution obtained from people without the condition and with albinism in the 200 families is not consistent.