Consider the Bernoulli distribution Ber(p) = Bin(1, p), answer the following que

Question

Consider the Bernoulli distribution Ber(p) = Bin(1, p), answer the following questions. 1. Take a sample X of size 50 from Ber(p_0) with p_0 = 0.2. 2. Calculate the sample proportion p* of 1's in X. 3. Is p* reasonably good enough as an estimator of p_0? Explain why. 4. Pretend that you forgot p_0 you used to generate the above sample. Your guess now is 0.4. You want to figure out whether it is 0.4. Below are two methods for your choice. Which one do you prefer? Why? (a) Method one. Compare the sample proportion you got in the above with your guess 0.4. If they are reasonably close, you probably will adopt your guess 0.4. Think about how you could address the close-ness (how close is close enough?). (b) Method two. The logic behind this is that if p_0 = 0.4 indeed, what will happen? If something unusual happens, then you may doubt your guess; if everything happens reasonably under p_0 = 0.4, then it seems no reason for you to reject your guess. Below is how we implement this. Generate N = 10,000 samples of size n = 50 from Ber(p_0 = 0.4). Calculate the sample proportion for each of N = 10,000 samples. Denote them by P_50 = {pk, k = 1, 2 ..., N = 10,000}. Plot the histogram of P_50 to see the distribution of the proportions P_50 under the assumption that p_0 = 0.4. If the observed p* is not in one of the two extreme regions (tails) of the distribution, you probably will adopt your guess 0.4. Calculate the proportion that pk

Explanation / Answer

1., 2., and 3. We simulate a sample of size 50 from Ber (p0=0.2). We run an R program for the simulation:

> set.seed(123)
> bs <- rbinom(50,1,0.2)
> table(bs)
bs
0 1
39 11
> phat <- sum(bs)/1
> phat <- sum(bs)/50
> phat
[1] 0.22
The sample proportion is a reasonable estimate of the p0. This is because the sample proportion also turns out to be MLE, unbiased estimator, etc.

4. The option (a) is reasonable and leads towards the UMPU test.

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