Question

The probability of individuals with blood types A, B, AB and O are 0.45, 0.13,
0.06 and 0.36, respectively. A geneticist tested 100 individual blood types and
found that 40 had type A, 18 had type B, 5 had type AB and 37 had type O. Use
goodness of fit test at 5% level of significance to test whether the observed
frequencies closely correspond to the theoretical ones.

Answer 1

Chi-Square Distribution.

H0 : The observed frequencies closely correspond to the theoretical ones (“no difference” situation).

H1 ; H0 is false, alpha=0.05.

Test statistics is

$$\chi^2=\Sigma(O-E)^2/E$$

O = observed count.

E = expected count.

χ2 = chi-square test value.

χ2_crit = chi-square value from the table.

df = degrees of freedom.

$$\chi^2=(40-45)^2/45+(18-13)^2/13+(5-6)^2/6+(37-36)^2/36=2.673.$$

The critical value for df=4-1=3 degrees of freedom and 5%  level of significance is 

$$\chi crit^2=7.82.$$

Since

$$\chi^2<\chi crit^2 ,$$

there is not sufficient evidence at alpha=0.05 to reject the null hypothesis.

Thus the observed frequencies closely correspond to the theoretical ones.