Two drugs (A, B) are compared for the medical treatment of duodenal ulcer. For t

Question

Two drugs (A, B) are compared for the medical treatment of duodenal ulcer. For this purpose, patients are carefully matched with regard to age, sex, and clinical condition. The treatment results based on 200 matched pairs show that 89 matched pairs are effective; for 90 matched pairs both treatments are ineffective; for 5 matched pairs drug A is effective whereas B is ineffective; and for 16 matched pairs drug B is effective whereas drug A is ineffective. A) What test procedure can be used to assess the results? B) Perform the test and report p-value. In the same study, if the focus is on the 100 matched pairs consisting of male patients, then the following results are obtained: for 52 matched pairs both drugs are effective; for 35 matched pairs both drugs are ineffective; for 4 matched pairs drug A is effective whereas drug B is ineffective; and for 9 matched pairs drug B is effective whereas the drug A is ineffective. A) How many concordant pairs are there among the male matched pairs? B) How many discordant pairs are there among the male matched pairs? C) Perform a significant test to assess any differences in effectiveness between the drugs among males.

Explanation / Answer

part (A)

Here we go for McNemar’s test;

let p=probability of discordant pair of type A

H0: p = 0.5 vs H1: P0.5

here total 21 discordant pairs of which there are 5 type A discordant pairs and 16 type B discordant pairs. we go for normal theory and chi-squre test statistic will be as

chi-square=(|5-21/2|-0.5)2/(21/4)=25/5.25=4.76

we compare it with tabulate chi-square with 1 df.

the p-value =0.029 ,so at alpha=5% there is significant difference in effectiveness between A and B.

Part(B)

(a) 87 concordant matched male pairs

(b) 13 discordant matched male pairs

(c)

Here we go for McNemar’s test;

let p=probability of discordant pair of type A

H0: p = 0.5 vs H1: P0.5

here total 13 discordant pairs of which there are 4 type A discordant pairs and 9 type B discordant pairs. we go for normal theory and chi-squre test statistic will be as

chi-square=(|4-13/2|-0.5)2/(13/4)=4/3.25=1.23

we compare it with tabulate chi-square with 1 df.

the p-value =0.26 ,so at alpha=5% there is insignificant difference in effectiveness between A and B.

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