Question

he probability of individuals with blood
types A, B, AB and 0 are 0.45, 013, 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 0. Use goodness of fit test at 5%
level of significance to test whether the
observed frequencies closely correspond to
the theoretical ones.
[You may like to use the values]

Answer 1

Answer on Question #67802 – Math – Statistics and Probability

Question

The probability of individuals with blood types A, B, AB and 0 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 0. Use goodness of fit test at 5% level of significance to test whether the observed frequencies closely correspond to the theoretical ones.

[You may like to use the values]

Solution

Null hypothesis $H_0$: the data are consistent with a specified distribution.

Alternative hypothesis $H_a$: the data are not consistent with a specified distribution.

Test statistic:

$$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} = \frac{(40 - 45)^2}{45} + \frac{(18 - 13)^2}{13} + \frac{(5 - 6)^2}{6} + \frac{(37 - 36)^2}{36} = 2.67.$$

Degrees of freedom:

$$df = 4 - 1 = 3.$$

P-value:

$$p = 0.445.$$

Since P-value is greater than 0.05 we can't reject the null hypothesis.

Answer: the observed frequencies closely correspond to the theoretical ones at 5% level of significance.

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