In the same study, if the focus is on the 100 matched pairs consisting of male patients, then theresults are obtained: for 51 matched pairs both drugs are effective; for 36 matched pairs both drugs areineffective; for 3 matched pairs drug A is effective, whereas drug B is ineffective; and for 10 madrug B is effective, whereas drug A is ineffective.(c) How many concordant pairs are there among the male matched pairs?87(d) How many discordant pairs are there among the male matched pairs?13(e) Perform a significance test to assess any differences in effectiveness between the drugs among malepatients. Report a p-value.State the null and alternative hypotheses. (Enter != for # as needed.)p=065Ho:чибA discordant park is a matched pair in which the outcomesdiffer for the members of the banks.H₁:PX 0.5PADRUGADRUG BEffective Ineffective To TAL54EffectiveIneffective 10TOTAL6133639100A concordant pain is a matched paie in which the outcome is the same for eachof the pain51+3Find the test statistic. (If the test does not have a defined test statistic, enter DNE. Round youranswer to two decimal places.) The normal approximation will be valid if hpq = 582DNEMpq = n₂ = 13 = 3₁23,2<5D4Use technology to find the p-value. (Round your answer to four decimal places.)p-value =All answERS are correct.Please explainThanks!!how to use technology to get the P-value,Interpret the results in words.O Reject Ho. We find there is not a significant difference in effectiveness between the twotreatments for male patients.

Given information: Drug B Effective Ineffective Total Drug A Effective a = 51 b = 3 54 Ineffective c = 10 d = 36 46 Total 61 39 100e) The test that can be performed to assess whether there is significant difference in effectiveness between two treatments for male patients is McNemar's test for correlated proportions(exact test). Null hypothesis: H0: Pb=Pc Alternate hypothesis: H0: Pb≠Pc The significance level is α=0.05 and the hypothesis is two tailed test. Test statistics is given by the below formula: Q=b-c2b+c Use the continuity correction because the sample size is small. QC=|b-c|-12b+c=|3-10|-123+10=7-123+10=3613=2.769230769≈2.77The test statistics Q approximate the chi square distribution with 1 degree of freedom. Use the excel formula, "=CHISQ.DIST.RT(2.77,1)" for determining the P-value from the chi-square statistic having degrees of freedom 1. P-value=2×Pχ2>2.77=2×0.0960=0.192 Since the P-value (0.192) is greater than the significance level (0.05), we failed to reject the null hypothesis. Failed to reject the H0. We find there is not a significant difference in effectiveness between the two treatments for male patients.