4.1.4 - Confidence Intervals for a Single Population Proportion

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Page 1 of 2 AP Statistics Assignment: Confidence Intervals for a Single Population Proportion 1. A. Write out the formula for a confidence interval for a population proportion. (1 point) B. Why is the z -statistic used in a confidence interval for a population proportion? (2 points) Because distribution of p-hat is a binomial and can be approximated by a normal distribution. C. What are the conditions a sample needs to meet before you can assume it's binomial and that it approximates a normal distribution? (2 points) Sample is binomial when the proportion is a ration of successes to outcomes; it approximates normal distribution when np-hat>=10, n(1-p-hat)>=10. 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. A. What's the point estimate of p , the population proportion? (1 point) p-hat = X/n = 27/75 = .36 B. 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) The situation is binomial because the students are randomly sampled, and is based on either a need of glasses or no need of glasses, which means its based on success to outcomes. np=75*.36=27>10, and n(1-p)=48>10, so z-distribution can be used to calculate a confidence interval for the proportion of students who need glasses but don’t wear them. C. What's the critical z -value ( z* ) for a 90% confidence interval for the population proportion? (1 point) 1.645 D. What's the margin of error for a 90% confidence interval for the population proportion? (2 points) 1.645(sqrt((.36(1-.36))/75))=.0912 ; margin of error = +/- 9.12% E. Calculate the 90% confidence interval for the population proportion. (1 point) .36+/-.912 = (.269, .451) F. Using your graphing calculator, find a 95% confidence interval for the proportion of students who need to wear glasses but don't. (Show all work,

functions, and inputs on your calculator too. (1 point) 1-PropZInt: x=27 ; n=75 ; C-level=.95  (.2514,.4686) G. 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) sample size of 246 individuals. H. 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) 3. A university wants to renovate a building on campus, and wants to know how many of the 20,000 active members of the alumni association would be willing to contribute funds to this project. However, this is the first time alumni donations would be the sole financial source for such a project, and the university doesn't have an estimate of the proportion who would contribute toward the renovation. A. If the university wanted to estimate, with a 95% confidence interval and a margin of error of 5%, the proportion of alumni who would be willing to donate to this project, what size sample would they need? (Hint: Do you remember how to estimate the minimum sample size when you don't have an estimate for the population proportion?) (1 point) B. The university draws a sample of 385 alumni, and 120 of them say they'd be willing to donate to the building renovation. Construct a 95% confidence interval for the proportion of alumni who would donate to the project. (2 points) p-hat = X/n = 120/385 = .312

.312 +/- 1.645(sqrt((.312(1-.312))/385))= .312 +/ .0463 = (.266, .358) C. Based on the sample size from part b, can you consider this situation binomial? Can you use a normal approximation here? (2 points) np= 385(.312)= 120.12 > 10 ; n(1-p)=385(1-.312) = 264.88>10. Yes, this situation is binomial based on the sample size from part b. I can use normal approximation here. D. The university postpones plans for the building renovations until the following year, when researchers take another sample of 385 alumni. This time, 262 alumni say they'd contribute to the project. Construct a 95% confidence interval for the proportion of alumni who would make donations. (2 points) p-hat = X/n = 262/385 = .681 .681 +/- 1.645(sqrt((.681(1-.681))/385))= .681 +/ .0466 = (.634, .728) E. Use your graphing calculator to calculate a 99% confidence interval for the proportion of alumni who would donate to the building renovations (use n = 385 and x = 262). (Show all work, functions, and inputs on your calculator too. (1 point) 1-PropZInt: x=262 ; n=385 ; C-level=.99  (.62,.742)

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4.1.4 - Confidence Intervals for a Single Population Proportion

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