Question

In order to study the genetic history of three different large groups of Americans, a random sample of 50 persons is drawn from group 1, a random sample of 100 persons is drawn from group 2, and a random sample of 200 persons is drawn from group 3. The blood type of each person in the samples is classified as A, B, AB, or O, and the results are as given in Table 10.47. Test the hypothesis that the distribution of blood types is the same in all three groups at the level of significance 0.1.

Short answer

There is not sufficient evidence to reject the claim that the distribution of blood types is the same in all three groups.

Step-by-step solution

Given Information

The number of rows is 3 (r).

The number of columns is 4(c).

The significance level is 0.1\(\left( \alpha \right)\)

Compute the row and column totals with observed values

The given information is tabulated as follows,

Group/blood types	1	2	3	Total
A	24	43	69	136
B	6	24	47	77
AB	5	7	22	34
O	15	26	62	103
Total	50	100	200	350

The hypothesis is formulated that the distribution of blood types is the same in all three groups as follows:

\({H_0}\): Distribution of blood types is the same in all three groups.

\({H_1}\): Distribution of blood types is not the same in all three groups.

Compute the expected frequencies

The expected frequency E is the product of the column and row total, divided by the grand total.

Mathematically,

\({E_{ij}} = \frac{{{r_i} \times {c_j}}}{G}\)

Thus, the expected values are:

Group/ blood types	1	2	3	Total
A	19.43	38.86	77.71	136
B	11	22	44	77
AB	4.86	9.71	19.43	34
O	14.71	29.43	58.86	103
Total	50	100	200	350

Compute the chi-square test statistic

The test statistic is formulated using the following formula;

\(\begin{array}{c}{\chi ^2} = \sum\limits_{i,j = 1}^n {\frac{{{{\left( {{O_{ij}} - {E_{ij}}} \right)}^2}}}{{{E_{ij}}}}} \\ = \frac{{{{(24 - 19.43)}^2}}}{{19.43}} + \frac{{{{(6 - 11.00)}^2}}}{{11.00}} + ... + \frac{{{{(62 - 58.86)}^2}}}{{58.86}}\end{array}\)

The chi-square value is the sum of the following values;

Group/ blood types	1	2	3	Total
A	1.0756	0.4417	0.9771	136
B	2.2727	0.1818	0.2045	77
AB	0.0042	0.7584	0.3403	34
O	0.0055	0.3994	0.1678	103
Total	50	100	200	350

Thus, the chi-square test statistic is 6.829.

The degree of freedom is,

\(\begin{array}{c}df = \left( {r - 1} \right)\left( {c - 1} \right)\\ = \left( {3 - 1} \right)\left( {4 - 1} \right)\\ = 6\end{array}\)

The critical value is obtained at the intersection of row 6 and column with \(\alpha = 0.1\) in the chi-square distribution table as follows,

\(\chi * = 10.645\)

Obtain the result and conclusion

The decision rule is determined as the rejection region then contains all values greater than 10.64.

If the value of the test statistic is within the rejection region, then the null hypothesis is rejected:

Since,\(6.8293 < 10.64\left( {\chi * } \right)\)

Thus, the hypothesis is failed to be rejected at 0.1 level of significance.

Thus, there is not sufficient evidence to reject the claim that the distribution of blood types is the same in all three groups.