A college student organization wants to start a nightclub fo

A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. They expect that about 67% of the student body would respond favorably. (a) What sample size is required to obtain a 95% confidence interval with an approximate margin of error of 0.034? (b) Suppose that 47% of the sample responds favorably. Calculate the margin of error for the 95% confidence interval. A random sample of n = 900 observations from a binomial population produced x = 684 successes. Estimate the binomial proportion p. (Round your answer to three decimal places.) Calculate the 95% margin of error. (Round your answer to three decimal places.) A study of the career paths of hotel general managers sent questionnaires to an SRS of 160 hotels belonging to major U.S. hotel chains. There were 115 responses. The average time these 115 general managers had spent with their current company was 11.4 years. Give a 99% confidence interval for the mean number of years general managers of major-chain hotels have spent with their current company. (Take it as known that the standard deviation of time with the company for all general managers is 4.1 years.)

Solution

(1)

(a) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.96/0.034)^2*0.67*(1-0.67)

=734.7558

Taken n= 735

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(b)

The margin of error = Z*sqrt(p*(1-p)/n)

=1.96*sqrt(0.47*(1-0.47)/735)

=0.03608272

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(2) phat= x/n =684/900 = 0.76

The margin of error = Z*sqrt(p*(1-p)/n)

=1.96*sqrt(0.76*(1-0.76)/900)

=0.028

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(3)Given a=1-0.99=0.01, Z(0.005)=2.58 (from standard normal table)

So the lower bound is

xbar - Z*s/vn = 11.4-2.58*4.1/sqrt(115)=10.4136

So the upper bound is

xbar + Z*s/vn =11.4+2.58*4.1/sqrt(115)=12.3864