Question #171761

A new drug cures 9 of 200 patients suffering from a type of cancer, for which the historical cure rate is 2%. Perform a test to check on the significance of this result, at both 5% and 1% levels of significance. Based on your conclusion, comments on the efficacy of the treatment using this new drug.

(15 marks


1
Expert's answer
2021-03-17T12:29:54-0400

H0:p=0.02H1:p0.02N=200p=9200α1=0.05We will use the following test statistic:Z=ppp(1p)nOur text statistic z=9/2000.02(0.02)(0.98)2002.53.Critical value:Φ(zcr)=1α12Φ(zcr)=0.475zcr=1.96(,1.96)(1.96,) is the rejection region.Our test statistic z falls into the rejection region. So we reject H0:p=0.02and accept H1:p0.02.The efficacy of the treatment using this new drug is not equal to 2%.α2=0.01Φ(zcr)=1α22Φ(zcr)=0.495zcr=2.58(,2.58)(2.58,) is the rejection region.Our test statistic z does not fall into the rejection region. So we accept H0:p=0.02.The efficacy of the treatment using this new drug is equal to 2%.H_0: p=0.02\\ H_1: p\neq 0.02\\ N=200\\ \overline{p}=\frac{9}{200}\\ \alpha_1=0.05\\ \text{We will use the following test statistic:}\\ Z=\frac{\overline{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ \text{Our text statistic }z=\frac{9/200-0.02}{\sqrt{\frac{(0.02)(0.98)}{200}}}\approx 2.53.\\ \text{Critical value:}\\ \Phi(z_{cr})=\frac{1-\alpha_1}{2}\\ \Phi(z_{cr})=0.475\\ z_{cr}=1.96\\ (-\infty,-1.96)\cup (1.96,\infty)\text{ is the rejection region}.\\ \text{Our test statistic z falls into the rejection region. So we reject }H_0: p=0.02\\ \text{and accept }H_1: p\neq 0.02.\\ \text{The efficacy of the treatment using this new drug is not equal to 2\%}.\\ \alpha_2=0.01\\ \Phi(z_{cr})=\frac{1-\alpha_2}{2}\\ \Phi(z_{cr})=0.495\\ z_{cr}=2.58\\ (-\infty,-2.58)\cup (2.58,\infty)\text{ is the rejection region}.\\ \text{Our test statistic z does not fall into the rejection region. So we accept }\\ H_0: p=0.02.\\ \text{The efficacy of the treatment using this new drug is equal to 2\%}.


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