Suppose in health-science, D denotes the diseased and D denotes the non-diseased
ID: 3218723 • Letter: S
Question
Suppose in health-science, D denotes the diseased and D denotes the non-diseased (healthy) states. In medical diagnostics, there are two possible errors that may occur: Type I error means that a non-diseased (healthy) condition is wrongly) diagnosed as diseased and medical treatment is pursued (denoted as the state, F); and, Type II error means that a diseased condition is wrongly diagnosed as non-diseased and medical treatment is not administered (denoting the state F (a) Indicate Type I and Type II errors on the following diagram: (b) Assume PD = 0.01 so that P(D) = 0.99: Probability (Type I error) = 0.02 and Probability (Type II error) = 0.05 Label all the probabilities between (D, D) and (F, F) as appropriate. Hence determine (i) P(F); (ii) P(D/F); (iii) P(D/F); and (vi) P(D/F).Explanation / Answer
(a) Type I error is when null hypothesis is true but rejected so D-- -> F is the case of type I error when non deceased is treated and medical treatment is persued.
Type I error is when null hypothesis is false but erronously fail to reject so D-> F-- is the case of type II error when deceased is untreated and medical treatment is not persued.
(b) so all probabilities
P ( D -> F) = 0.01 * 0.95 = 0.0095
P(D - > F-) = 0.01 * 0.05 = 0.0005
P(D-- --> F) = 0.99 * 0.02 = 0.0198
P (D-- --> F--) = 0.99 * 0.98 = 0.9702
(i) P (F--) = P(D-- --> F--) + P(D--> , F-- ) = 0.99 * 0.98 + 0.01 * 0.05 = 0.9702 + 0.0005 = 0.9707
(ii) P( D/F) = (0.01 * 0.95) / ( 0.01 * 0.95 + 0.99 * 0.02 ) = 0.3242
(iii) P(D--/F) = 1 - 0.3242 = 0.6758
(iv) P(D--/F--) = (0.99 * 0.98)/ (0.99 * 0.98 + 0.01 * 0.05) = 0.9995