Fingerprint analysis and blood grouping are features that

Adison Palmer

Adison Palmer

Answered question

2022-04-07

Fingerprint analysis and blood grouping are features that do not change through the lifetime of an individual. Fingerprint features appear early in the development of a fetus, and blood types are determined by genetics. Therefore, each is considered an effective tool for identification of individuals. These characteristics are also of interest in the discipline of biological anthropology—a scientific discipline concerned with the biological and behavioral aspects of human beings.
(a) Is the test for an association in this case a chi-square test of independence, or a chi-square test of homogeneity? Justify your choice.
(b) Identify the conditions for the chi-square inference procedure selected in part (a), and indicate whether the conditions are met.
(c) The resulting chi-square test statistic from the appropriate test is approximately 18.930. What are the degrees of freedom and p-value of the test?
(d) Biological anthropology is concerned with the comparative study of human origin, evolution and diversity. Considering the sampling design in this study, to what population is it reasonable for the researchers to generalize their results?
Blood TypeABABOTotalLoops66(71.69)99(112.19)35(32.29)101(84.83)301Whorls51(47.16)91(73.80)15(21.24)41(55.80)198Arches14(12.15)15(19.01)9(5.47)13(14.37)51Total13120559155550

Answer & Explanation

Wernbergbo9d

Wernbergbo9d

Beginner2022-04-08Added 12 answers

(Since it has been not mentioned which question so, as per our protocol we shall only answer the first three part)
a) Here we are testing the association between the blood type and Fingerprint type hence we will go for chi-square test of independence.
The chi-square test of independence is used to test the association ( independence ) of the variables while chi-square test of homogeneity is used for testing if the variables are from the same population.
Therefore, we shall use chi-square test of independence.
b) The data must be independent and one sample should be only assigned to one block, not more than one.
The expected values should be greater than 5 ,which is true as we can see the data ( the numbers in the brackets )
c) The degrees of freedom is
df=(row1)(columns1)=(31)(41)=6
The P-Value is 0.004284

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