Question

In a crossover trial comparing a new drug to a standard, π denotes the probability that the new one is judged better. It is desired to estimate π and test H0 : π = 0.5 against H1 : π =/= 0.5. In 20 independent observations, the new drug is better each time.

(a) Find and PLOT(I can't figure out how to plot it) the likelihood function. Give the ML estimate of π.

(b) Conduct a Wald test and construct a 95% Wald confidence interval for π. Are these sensible? I know the pihat is 1.0, and CI is (1.0,1.0) but I don't know how to arrive at that solution.

(c) Conduct a likelihood ratio test and conduct a likelihood based 95% confidence interval. Interpret.

(d) Suppose that researchers wanted a sufficiently large sample to estimate the probability of preferring the new drug to within 0.05, with confidence 0.95 (i.e., the 95% confidence interval is of length 0.1), If the true probability is 0.80, about how large a sample is needed based on Wald type confidence interval?

Answer 1

ANSWER:

In a crossover trial comparing a new drug to a standard, π denotes the probability that the new one is judged better. It is desired to estimate π and test H0 : π = 0.5 against H1 : π =/= 0.5. In 20 independent observations, the new drug is better each time. (a) Find and PLOT(I can't figure out how to plot it) the likelihood function. Give the ML estimate of π. (b) Conduct a Wald test and construct a 95% Wald confidence interval for π. Are these sensible? I know the pihat is 1.0, and CI is (1.0,1.0) but I don't know how to arrive at that solution. (c) Conduct a likelihood ratio test and conduct a likelihood based 95% confidence interval. Interpret. (d) Suppose that researchers wanted a sufficiently large sample to estimate the probability of preferring the new drug to within 0.05, with confidence 0.95 (i.e., the 95% confidence interval is of length 0.1), If the true probability is 0.80, about how large a sample is needed based on Wald type confidence interval.

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  a) Here 20 independent observati ons, i e., n-20 Observe y = 1 20 20-1 19 = 20π(1-π) =1(a), the likelihood function defined for π between 0 and 1 If t-0, probability is 1 (0) 0 of getting y 1 If π= 0.5, probability is 1(0.5) = 0.00002 of getting y= 1 To satisfies Z- 0(1-70) 72 ±1.96 = 而(1-70) 20 0 is one solution In quadratic shape, 70 = 0.16 is other solution (b) 1 (π) = 20π(1-π)19 maximized at = 0.00002 If π= 0.5, then the ML estimate of π is = 0.00002 95% confidence interval of wald P1.961-P P2002 (1-0) (since P,L-20.0) or 0±1.96 n 20 or 0t1.96 or 0±0 or (0, 0 Now, testing the hypothesis for TT 20 The test statistic is, P-而 P(1-P)
  
  P - value here n - 20 andy 0 20 95% score confidence interval y=0, n= 20, P=0 95% score confidence interval t1.96 而(1-m) 20 1170( 70)=10- 1.96 而 20 0 is one solution. For quadratic, π = 0.16 is other solution. 95% score confidence interval is (0, 0.16) more sensible than wald confidence interval of (0,0) (c) Likelihood confidence interval of 95% (d) Binomial, n 2, observey 1 27(1-7) -1 (π), the likelihood function defined for π between 0 and 1 If r-0 probability is l(0)- 0 of getting y 1 r π= 0.5 probability is 1 (0.5) = 0.5 of getting y = 1