An experiment was conducted to test the effect of a new drug on a viral infectio

Question

An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection. The second group received the drug. After a 30-day period, the proportions of survivors, p_1^and p_2^, in the two groups were found to be 0.36 and 0.60, respectively. a. Is there sufficient evidence to indicate that the drug is effective in treat- ing the viral infection? Test at 5% significance level. (Make sure to state your null and alternative hypotheses.) b. Use a 95% confidence interval to estimate the actual difference in the cure rates. i.e. p_1 - p_2, for the treatment versus the control groups. Based on this confidence interval can you conclude that the drug is effective? Why?

Explanation / Answer

Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2

p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.36* 50) + (0.60 * 50)] / (50 + 50)= 0.48

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = sqrt [ 0.48 * 0.52 * ( 1/50 + 1/50 ) ] = sqrt [0.001] = 0.0316

z = (p1 - p2) / SE = (0.36 - 0.60)/0.0316 = -7.60

p-value is

we reject null hypothesis .

SE*Tcrtical = 0.0316*1.96 = 0.0620

p1-p2 = 0.36-0.60 = -0.24

CI 95% = ( -0.24-0.0620,-0.24+0.0620)

1.48065E-14 which is less than significance level 0.05

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