Question
The usual confidence interval for estimating a population proportion relies on a Normal approximation. But that approximation fails when the true popu- lation proportion is
The usual confidence interval for estimating a population proportion relies on a Normal approximation. But that approximation fails when the true popu- lation proportion is close to 0 (or 1). The usual rule of thumb is that there needs to be at least both 10 successes and 10 failures in the sample. But what if that’s not true? In particular, how do you construct a confidence interval for a population proportion of success when there are no successes observed in the sample? This problem will show you how, using the connection between hypothesis tests and confidence intervals.
Let X ∼ Binomial(n, p) where p is unknown (but suspected to be close to 0). We will construct a confidence interval for p to use for samples for which X = 0 is observed.
Let p∗ denote the largest value of p0 for which H0 : p = p0 will not be rejected in favor of H1 : p < p0 at level α0 when X = 0 is observed. Find an expression for p∗ in terms of α0 and n.
Explain why the interval [0, p∗] is a 1 − α0 confidence interval for p for samples for which X = 0. Hint: like there is a connection between two-sided tests and confidence intervals, there is a similar connection between one-sided tests and one-sided confidence intervals.
It can be shown that when α0 = 0.05, p∗ ≈ 3/n and so [0, 3/n] is an approximate 95% confidence interval for a population proportion p based on a sample of size n in which 0 successes are observed. Compare p∗ and 3/n for n = 100 and α0 = 0.05.
In NCAA men’s basketball tournaments from 1985 through 2017, no number- 16 seed beat a number-1 seed in 132 games1. Use this data to estimate p, the probability that a number-16 team beats a number-1 seed. You can assume the outcomes of 16/1-seed games are i.i.d. Bernoulli(p).
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