Question

Consider a drug that is used to help prevent blood dots in certain patients. In diamong 5700 patients treated with this drug 150 developed the adverse reaction of nausea. Usea 0.05 granice level to test the claim that of users develop nausea Does nausea appear to be a problema diverse reaction? Identity the nul and alternative hypotheses for this test Choose the correct answer below. OA HD*0.03 H:p<0.00 OBH:p=0.03 Hp000 OC. He pu003 Hp7003 OD HD#0.03 HD 0.03 Identify the best statistic for this hypothesis test The best statistic for this hypothesis tot i wala

Answer 1

Let p denotes the population proportion. Hypotheses: To test null hypothesis H. :p = 0.03 against alternative hypothesis H :P +0.03 Let o denotes the sample proportion and n denotes the sample size. Here, p 150 5790 = 0.026, n = 5790, po = 0.03, standard error(SE) of proportion under null 0.03 * (1 – 0.03) 5790 = 0.002242 The test statistic can be written as: (P - 0.03) 0.03*(1-0.03) n which under Ho follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if Zstat | > 20.025 = 1.959964 0.4 0.3 0.2 0.1 Now, 0.0 - -2 Value of test statistic: The value of the test statistic (0.02591 – 0.03) 2 stat =-1.825839 -1.83 0.03*(1-0.03) V n The critical value = critical = + 1.959964 1.960 P-value = P(|=| > Zstat) = 2* P(Z < -1.825839) = 2 * 0.033937 = 0.0678740.0679 Conclusion: Since p-value > 0.05 and |zobs Zcritical = 1.959964, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that the population proportion is not significantly different from 0.03

The correct conclusion for the hypothesis test - B)

2. low, is not