Question

Colebrook administered Prontosil to many women suffering from perpeural fever following childbirth. This disease typically carried a high mortality rate of about 20%, and at the hospital where he worked, 42 out of the previous 210 patients died. After Colebrook began using the drug on his women patients suffering from this illness, only 3 out of his 38 original patients died. Although the treated patients and the un-treated patients were not randomized in a clinical trial, we can apply the methods of the likelihood ratio test to these data to see if this difference in mortality rates would have achieved statistical significance. Using the data below, answer the questions in this quiz.

Question 1 (1 point) Under the Null hypothesis of no difference in mortality between the treated and un-treated groups of patients, what is the Likelihood of seeing the data that Colebrook compiled?

Not Totals Treated treated Survived 35 168 203 42 45 3 Died Totals38 210

Q2

Under the Alternative hypothesis, where there is a difference in mortality between the treated and un-treated groups of patients, what is the Likelihood of seeing the data that Colebrook compiled?

Q3

Calculate the value of G for the ratio of these Likelihoods.

Answer 1

Q1.

From the given table total number women both treated and not treated N = 248

Under the Null hypothesis of no difference in mortality between the treated and un-treated groups of patients, the Likelihood is calculated as

No.ofdeathswhentreated + No.ofdeathsw hennottreated 248 しnull =

Given that disease typically carried a high mortality rate of about 20%. Hence under null hypothesis when there is no difference in mortatily between treated and un-treated

No. of deaths when treated = (38 * 20)/100 = 7.6

No. of deaths when treated = (210 * 20)/100 = 42

7.6+4249.6 Lnull 248 248

Q2.

Under the Alternate hypothesis of there is a difference in mortality between the treated and un-treated groups of patients, the Likelihood is calculated as

No.ofdeathswhentreated + No.ofdeathsw hennottreated 248 alt

3+42 45 alt 248 24801815

Q3.

The value of G is calculated as

0.2 G = null. 0.2 _ = 1.1019 t 0.1815