Question

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Figure 1. Vertical li indicate the 2.5% tails of the alternate distribution. In other words, the red areas (area 0 and 8) sum to 5%, and areas 2, 3, 7 also sums to 5%. dicate the 2.5% tails of the null distribution, while B and D alternate distribution null distribution test statistic Consider a two-sided null hypothesis test with 0.05 as significance level. From the above probability density plot, identify the corresponding areas, or test statistic value (A-D) of: 1. a. Type-I error rate: b. Type-ll error rate: c. Critical value(s): 2. From the figure, how would type-l and type-ll error rates change if the alternate distribution were shifted to the left? 3. A pharmaceutical startup ran a clinical trial to test if drug X cures disease Y. A biostatistician named Z who took BIO 211 was given the task to perform a null hypothesis test. The test hired Z's classmates to evaluate data obtained from each independent trial with the same a. Out of 40 null hypothesis tests, how many tests are expected to reject the null was perfectly done, and the null hypothesis was rejected at the alpha 0.025 significance level. However, unbeknownst to Z, the company ran 39 extra clinical trials and secretly null hypothesis test at the same significance level. hypothesis when the null hypothesis is true? b. What is the probability that at least one (out of 40) clinical trial would reject the null hypothesis when drug X has no effect on curing disease Y? (Hint: each test is an independent coin flip of sorts!) c. If data from one out of 40 of the clinical trials rejected the null hypothesis, and the pharmaceutical company claimed that the drug X has statistically significant effect on curing disease Y. This is highly unethical. Why would it be a wrong conclusion?

Answer 1

$1.~a.~Type~I~error~rate=P(reject~H_0|H_0~is~true)=area ~0+area~8\\ b.~Type~II~error~rate=P(Accept~H_0|H_0~is~false)=1-area~3\\ c.~Critical~values=A,~C.\\ 2.~Type~I~error=area ~0+area~8~(unchanged)\\ Type~II~error~rate=1-area~7.\\ 3.~a.~U=no.~of~tests~are~rejected\sim Binomial(40,0.025)\\ Then~expected~no.~of~tests~are~rejected~null~hypothesis~when~it~is~true\\ =40*0.025=1\\ b.~Required~probability=P(U\geq 1)=1-P(U=0)=1-(1-0.025)^{40}\\=0.6368\\ c.~ P(U=1)=\binom{40}{1}(0.025)(1-0.025)^{39}=0.3725\\ So~probability~that~U=1~when~type~I~error~is~committed~is~high\\ i.e.~>0.05~so~we~can't~reject~null~hypothesis~H_0.$