2. A nurse in a large university ( N 30,000) is concerned about students\' eye h

Question

2. A nurse in a large university (N 30,000) is concerned about students' eye health. She takes a random sample of 75 students who don't wear glasses and finds 27 that need glasses.

What's the point estimate of p, the population proportion? (1 point)

Is the situation binomial? Also, can the z-distribution be used to calculate a confidence interval for the proportion of students who need glasses but don't wear them? Explain. (2 points)

What's the critical z-value (z*) for a 90% confidence interval for the population proportion? (1 point)

What's the margin of error for a 90% confidence interval for the population proportion? (2 points)

Calculate the 90% confidence interval for the population proportion. (1 point)

Usingyourgraphingcalculator,finda95%confidenceintervalforthe proportion of students who need to wear glasses but don't. (Show all work,

functions, and inputs on your calculator too. (1 point)

The nurse wants to be able to estimate, with a 95% confidence interval and a margin of error of 6%, the proportion of students who need to wear glasses but don't. Find the necessary sample size (n) for this estimate. (2 points)

The following school year, the nurse wants to construct the same 95% confidence interval for the proportion of students who need glasses but don't wear them, but she thinks the proportion has changed since last year. Without using the point estimate of the population proportion from the previous year, find the necessary sample size (n) for a 95% confidence interval for the population proportion with a margin of error (m) of 6%. (2 points)

Explanation / Answer

We can assume the situation is binomial in this case, since N is very large compared to the size of the sample. (i.e, sample size n = 75 is << (0.05)N = 1500)

Also, since the number of "successes" (27) and "failures" (75 - 27 = 48) are both larger than 5, we can assume that the binomial distribution is approximately normal, therefore allowing us to use the z-distribution to calculate a confidence interval for the proportion of students who need glasses

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