Question

Test the null hypothesis at the .01 level of significance that the distribution of blood types for college students complies with the proportions described in the blood bank bulletin, namely, .44 for O, .41 for A, .10 for B, and .05 for AB. Now, however, assume that the results are available for a random sample of only 60 students. The results are as follows: 27 for O, 26 for A, 4 for B, and 3 for AB. NOTE: The expected frequency for AB, (.05)(60) = 3, is less than 5, the smallest permissible expected frequency. Create a sufficiently large expected frequency by combining B and AB blood types.

Answer 1

The B and AB groups are combined here to have sufficiently large expected frequencies.

The observed and expected frequencies are then computed as:

Group	O	A	B or AB
Observed Frequency	27	26	4 + 3 = 7
Expected Frequency	0.44*60 = 26.4	0.41*60 = 24.6	0.15*60 = 9

Now the chi square test statistic for the test here is computed as:

Now for n - 1 = 2 degrees of freedom, the p-value for this test for goodness of fit is computed form the chi square distribution tables as:

As the p-value here is 0.7642 > > 0.01 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence that the given data has different proportions.