1.) A simple random sample of 60 items resulted in a sample mean of 86. The popu

Question

1.) A simple random sample of 60 items resulted in a sample mean of 86. The population standard deviation is 15.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

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b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

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2.) A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $69.50.

a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals).

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b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association?

3.) In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a 95% level of confidence? How large a sample should be taken for a 99% level of confidence? Use a planning value for the population standard deviation of 7 minutes.

95% Confidence (to the nearest whole number):

99% Confidence (to the nearest whole number):

4.) The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .32.

a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence.

b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the sample proportion of smokers (to 4 decimals)?

c. Based on the answer in part (b), what is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

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Explanation / Answer

We have  Confidence interval for population mean (Mu) =sample mean +/- confidence coefficient*standard error of mean.

1.)n=60,S.D=15,z(confidence coefficient)=1..96 for 95% confidence interval.

So Confidence interval(Mu)=86 +/- 1.96*15/sqrt(60).

lower boundary=86 - 1.96*15/sqrt(60)=82.2

upperboundary=86 + 1.96*15/sqrt(60)=89.8

so the CI is [82.2,89.8].

2.)252.45 +/- 1.96*69.50/sqrt 64
252.45 +/- 18.25
loweer boundary is 252.45 - 17.0275 = 235.42
upper boundary is 252.45 + 17.0275 = 269.48
The CI is ($235.42, $269.48)

b) Since the mean reported by the AMA ($215.60) is NOT falling within the confidence interval, it can be inferred that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the AMA.

3.)The desired margin of error is 2 mins at 95% confidence interval .Given S.D=7mins.

n=[1.96*7/2]2=47.(sample size).

for 99% confidence interval z=2.5758

so n=[2.5758*7/2]2=81(sample size).

4.)The formula for finding the sample size needed to estimate a population proportion is

n = phat * (1-phat) * (z/E)^2, where

phat is the sample proportion, which according to the preliminary estimate is 0.32

z is the multiplier based on the confidence level. Since this is a 95% confidence interval, then you would use 1.96, and

E is the desired margin of error, which is 0.02.So

a.)n = 0.32 * 0.7 * (1.96/0.02)^2 = 2016.84, which should be rounded up to 2017

b).If there are 520 smokers out of these 2017, then the point estimate, which is you best guess, is

phat = 520/2017 = 0.2578.

c.)The 95% confidence interval for the population proportion is found by computing

phat +- z*sqrt{phat*(1-phat)/n}

0.258 +- 1.96 * sqrt{0.258*(1-0.258)/2017}
0.258 +- 0.019

So I am 95% confident that the proportion of smokers in the population is between 0.258-0.019 = 0.239 and 0.258+0.019 = 0.277.

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