Question

McNemar's test, developed in Exercise 56, can also be used when individuals are paired (matched) to yield n pairs and then one member of each pair is given treatment 1 and the other is given treatment 2 . Then \({X_1}\)is the number of pairs in which both treatments were successful, and similarly for\({H_0}\)\({X_2},{X_3}\), and\({X_4}\). The test statistic for testing equal efficacy of the two treatments is given by\(\left( {{X_2} - {X_3}} \right)/\sqrt {\left( {{X_2} + {X_3}} \right)} \), which has approximately a standard normal distribution when \({H_0}\)is true. Use this to test whether the drug ergotamine is effective in the treatment of migraine headaches.

The data is fictitious, but the conclusion agrees with that in the article "Controlled Clinical Trial of Ergotamine Tartrate" (British Med, J., 1970: 325-327).

Short answer

reject the null hypothesis

Step-by-step solution

Step 1: to find the test whether the drug ergotamine is effective in the treatment of migraine headaches

Given values are\({x_1} = 44,{x_2} = 34,{x_3} = 46,{x_4} = 30\). Using this, the test statistic value (given in the exercise) is

\(z = \frac{{34 - 46}}{{\sqrt {34 + 46} }} = - 1.34\)

The hypotheses of interest are\({H_0}:{p_2} = {p_3}\)versus\({H_a}:{p_2} < {p_3}\); thus the P value is the area under the z curve to the left of z

\(P = P(Z < - 1.34) = 0.0901\)

where the value was computed using software (you can use table in the appendix). Because

\(P = 0.0901 > 0.05 > 0.01\)

Reject the null hypothesis

Drug ergotamine is not effective in the treatment of migraine headaches. NOTE: perhaps you could have used two-sided test.

Final proof

Finally we get,

Reject the null hypothesis