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Because of the popularity of movies as an entertainment medium for adolescents, an entrepreneur plans to...

Because of the popularity of movies as an entertainment medium for adolescents, an entrepreneur plans to do a national study of the average cost of a movie ticket. If you assume that s = $0.50, what sample size would the entrepreneur have to take to be 95% confident that the estimate was within $0.25 of the true mean ticket prices?

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we have standard deviation s = 0.50

margin of error = ME = 0.25

z critical value for 95% confidence level is 1.96 (using z distribution table)

We have to find the value of sample size n

Formula for the sample size n is given as

n = ((z*s)/ME)^2

setting the given values, we get

n = ((1.96 * 0.50)/0.25)2-(1.96 * 2)2-3.922-15.3664

rounding it off to next whole number, we get n = 16

So, required sample size is n = 16

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