Question

A local statistician is interested in the proportion of high school students that drink coffee. He takes a random sample of 90 high school students and finds that 18 of these students drink coffee. He wants to construct a 99

1) z = (pÌ‚ - p) / √(p(1-p)/n)
2) z = (pÌ‚ - p) / √(p(1-p)/n)
3) z = (pÌ‚ - p) / √(p(1-p)/n)
4) z = (pÌ‚ - p) / √(p(1-p)/n)

Answer 1

The local statistician can construct a 99% confidence interval for the high school students' coffee drinking proportion using the provided formulas, a binomial distribution, and the central limit theorem conditions.

Step-by-step explanation:

The local statistician is interested in constructing a confidence interval to estimate the proportion of high school students that drink coffee. In a random sample of 90 high school students, 18 were found to drink coffee.

To construct a 99% confidence interval for the proportion, the statistician should use the formula for the confidence interval of a single population proportion p. The sample proportion (p̂) is 18/90, which would be needed to calculate the z-score.

Given the conditions - a certain number of independent trials, two possible outcomes (success or failure), and having the same probability of success on each trial - this scenario could be modeled by a binomial distribution. If np and nq (where q = 1 - p) are both greater than five, the binomial distribution can be approximated by a normal distribution.

Since the sample size is reasonably large, the statistician could use the z-distribution for the hypothesis test and confidence interval estimation as long as np and nq are both greater than five.