Question

In a study of treatments for very painful "cluster" headaches, 148 patients were treated with oxygen and 159 other patients were given a placebo consisting of ordinary ar. Among the 148 patients in the oxygen treatment group, 106 were free from headaches 15 minutes after treatment. Among the 159 patients given the placebo. 37 were free from headaches 15 minutes after treatment. Use a 0.01 significance level to test the claim that the oxygen treatment is effective. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test Consider the first sample to be the sample of patients treated with oxygen and the second sample to be the sample of patients given a placebo. What are the null and alternative hypotheses for the hypothesis test? A. Ho: P1 P2 HP P2 OD. Ho:P SP2 Hy:P*P2 OB. Ho: P = P2 4P1 P2 OE. Ho: P = P2 HP1 P2 Oc. Ho: Pq*P2 HP - P2 F Ho: P2 P2 H: PHP2 - Identify the test statistic. Z (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. parts Clear All as ! 5 remaining

Answer 1

Here we have x =106, n =148, P. = x;/ n =0.7162 X2=37, n =159,92 = x2 / n =0.2327 Hypotheses are: H: P. – P2 = 0 H: P.-P2> 0 Level of significance: a =0.01 Test is right tailed. The pooled proportion is: p=Pini + P212 = (106+37)/(148+159)=0.4658 n + n2 The standard error is: 11 SE = 1 +- = 0.057 nn Test Statistics: _(P -P,)-(P -p,). =8.48 SE Critical value: CV = 2.326 Rejection Region: If z> 2.326, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value =0 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: Excel function for p-value: ENORMSINV(1-0.01) =(1-NORMSDIST(8.48))