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find confidence interval for proportion

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I need help coing up with solutions for the following problem:

For many of us a person aged 80 years is considered "old." However, for legal purposes involving the establishment of age discrimination, 40 is typically considered the magic number. In fact, cases are won or lost based on the proportion of workers employed who are "under 40" and "over 40."

a) Suppose you wish to obtain a 95% confidence interval estimate for the proportion of the people of Gator Hollow who are over 40, and you want the confidence interval to be no larger than plus or minus 0.06. You are willing to assume that this proportion is no larger than 0.4 (but you want a more refined estimate). What sample size should you use?

The Buena Vista Social Club is located in Gator Hollow, Florida. The Club has been sued for age discrimination by a group of former, "older" employees of the Club. To prove their case, the group has argued that 35 percent of the people in the labor pool in the Club's region are over 40 years old. But of the 84 people currently employed by the Club, only 23 are over 40 years old. Assume these 84 people represent a random sample of the type of people the Club has been employing for some time now.

b) Develop a 95 percent confidence interval for the proportion of all employees employed over the years by the Club who are over 40 years old.

c) Does the group have a legitimate claim? Is the proportion of Buena Vista's employees over 40 years old significantly less than the proportion in the labor pool? If ? is the proportion of all of Buena Vista's employees over 40 years old, the group should test
H0: ? = 0.35 versus
Ha: ? < 0.35
Test at the 5 percent level of significance. Draw a picture to illustrate the test. Show the test statistic and prob-value for the test, and give your conclusion in terms of the statement of the problem.

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Solution Summary

The solution provides detailed explanation how to find out the confidence interval for the proportion.

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a) 95% confidence interval is 1.96.
margin of error=critical value*standard deviation/sqrt(n)
The proportion is around 0.40-0.06=0.34.
0.06=1.96*sqrt(0.34*(1-0.34)/n).
N=239.45=240.
The sample size is 240.

b). the mean proportion =23/84=0.2738
the critical value for 95% confidence ...

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