Categorical data HWK1

Question 1

In a crossover trial comparing a new drug to a standard, $\pi$ denotes the probability that the new one is judged better. It is desired to estimate $\pi$ and test $H_0: \pi = 0.5$ against $H_a: \pi \neq 0.5$. In 20 independent observations, the new drug is better each time.

Find and sketch the likelihood function. Give the ML estimate of $\pi$.
Conduct a Wald test and construct a 95% Wald confidence interval for $\pi$. Are these sensible?
Conduct a score test and report the corresponding p-value. Construct a 95% score confidence interval. Interpret.
Conduct a likelihood-ratio test and construct a likelihood-based 95% confidence interval. Interpret.
Construct an exact binomial test and 95% confidence interval. Interpret.
Suppose that researchers wanted a sufficiently large sample to estimate with 95% confidence the probability of preferring the new drug within a margin of 0.05? If the true probability is 0.90, how large should the sample be?

Solutions

n=20

L = function(pi_,x){
  term = (pi_^x)*((1-pi_)^(1-x))
  return(prod(term))
}

x=rep(1,n)
pi_ = seq(0,1,by=0.001)
L_values  = sapply(pi_,L,x=x)
plot(pi_, L_values,type='l',
     ylab = 'Likelihood',xlab = expression(pi),
     main='Likelihood values for different values of pi')

Consider testing $H0:\pi = \pi_0$ for a binomial proportion. $n=20$, $\pi_0=0.5,\hat \pi = 1$.

Wald test \[ \frac{\hat \pi - \pi_0}{\sqrt{\hat\pi(1-\hat\pi)}/n}\sim N(0,1) \] Inverting the Wald test gives the Wald interval \[ \hat \pi\pm Z_{1-\alpha/2}\sqrt{\hat \pi(1-\hat \pi)/n} \]
Score test \[ \frac{\hat \pi - \pi_0}{\sqrt{\pi_0(1-\pi_0)}/n}\sim N(0,1) \] Inverting the Score test yields the Score interval \[ \hat \pi\left(\frac{n}{n+Z_{1-\alpha/2}^2}\right)+\frac{1}{2} \left(\frac{n}{n+Z_{1-\alpha/2}^2}\right)\pm Z_{1-\alpha/2}\sqrt{\frac{1}{n+Z_{1-\alpha/2}}\left[\hat \pi(1-\hat pi)(\frac{n}{n+Z^2_{1-\alpha/2}})+\frac{1}{4}(\frac{Z_{1-\alpha/2}^2}{n+Z_{1-\alpha/2}^2})\right]} \]
Likelihood ratio test \[ -2\left(l(\pi_0)-l(\hat \pi)\right)\sim \chi^2_0 \]
Construct an exact binomial test and 95% confidence interval. Interpret.

Let $X_i$ be that random variable that maps success to 1 and failure to 0, for $i=1,2,,n. Then $X_i$ has a Bernoulli distribution. Let $Y=\sum_{i=1}^{n}X_i$ be the number of success after $n=20$ trials, then \[ P(Y\geq c) =\sum_{y=c}^{20} {20 \choose c}\pi^y(1-\pi)^{20-y} \] Under the null

\[ P(Y\geq c) =\sum_{y=c}^{20} {n \choose c}\pi_0^y(1-\pi_0)^{20-y} \] What is the probability, under the null of getting evidence as extreme or more extreme than we obtained?

\[ P(Y\geq 20) =\sum_{y=20}^{20} {n \choose 20}\pi_0^y(1-\pi_0)^{20-y} = \pi_0^{20} \]

Categorical data HWK1

Question 1

Solutions

Question 2