2. Moderna’s November 16, 2020 press release reported interim results of its phase 3 COVID-19 vaccine trial. The report indicated that there were 95 total cases of COVID-19 illness among their study participants, of which 5 cases were in the treatment group, and concluded that the vaccine is estimated to be 94.5% effective at preventing infection. Approximately equal numbers of participants were in each group.Let \( Y_i = 1\{\text{case } i \text{ was in the treatment group}\} \) be a random variable representing the outcome of the \( i \)th case, and let \( y_i \) denote the observed value of \( Y_i \). Then:- \( Y_1, \ldots, Y_n \overset{\text{iid}}{\sim} \text{Bernoulli}(p) \), where \( p \) represents the conditional probability that a study participant received the vaccine, given that they contracted the virus;- for the interim data, \( n = 95 \) and \( \bar{y} = \frac{5}{95} \approx 0.0526 \).(a) Compute an approximate 95% confidence interval for \( p \).(b) The press release reports a point estimate for efficacy. Efficacy is the relative reduction in case rates, which can be approximated by \( 1 - \frac{p}{1-p} \) because the participant group sizes are close. Find an approximate 95% confidence interval for efficacy by applying this same transformation to the endpoints of your interval in the previous part.(In other words, if your interval was \( (\hat{p}_L, \hat{p}_U) \), compute \( \left(1 - \frac{\hat{p}_L}{1-\hat{p}_L}, 1 - \frac{\hat{p}_U}{1-\hat{p}_U} \right) \). This is valid because efficacy is a monotone function of \( p \) on \( (0, 1) \).)(c) Suppose the local paper summarizes the trial results by concluding that “94.5% of vaccinated adults develop immunity”. Is this interpretation correct?

From the provided information, p⏜=0.05261-α=0.95n=95

Answered: 2. Moderna's November 16, 2020 press…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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