Solve Point Estimation Problems: 95% Confidence, 3% Error

[mikemike123]

I have 2 problems I am currently stuck on.. one being
In a large sample study according to a report by the U.S. surgeon general, electrical engineers have the lowest smoking rate among all workers surveyed. Only 16% of the male electrical engineers in the sample smoke cigarettes regularly. How many male electrical engineers must be sampled to estimate the proportion of all male electrical engineers who smoke regularly to within 3% of its true value with 95% confidence?

Now for this particular problem I understand they want me to solve for N. In the original equation E=z(alpha/2)* S.D/sqrt(n).. I solved for N and got N=[z(alpha/2)*S.D./E]^2..
So In the problem given, the confidence needed is 95% or .95. Since this is a "large Sample" I assumed using the z table. After looking through it i figured z(alpha/2)=1.96. I would assume E would be .03. This is where i get stuck, there is no S.D given and there is no set of data in the problem so I can not solve for it using the S.D formula. I assume I am interpreting the problem incorrectly. And I am having the same problem with this problem,

In a recent study, 69 of 120 meteorites were observed to enter the Earth's atmosphere with a velocity of less than 26 miles per second.

A) What can you say with 95% confidence about the maximum error?
B) What confidence can we assert that the maximum error of this study is at most 0.055?
C) How large of a sample size is needed if the maximum error of estimate for this study is at most 2.5%

All of these would need a S.D to solve for, or not, I may not be seeing something. Can someone please point me in the right direction.

Thank You.

 

[bpet]

For the first problem, the number of smokers in the sample would follow the binomial distribution, hence the mean and variance of the observed frequency could be worked out.

For the second problem, need to define "error".

 

[mikemike123]

Well in the problem I would assume the mean would be 16% or .16.. how can i solve for a S.D out of this?

 

[statdad]

Your attack of the first problem isn't correct - think about it: you are dealing with percentages, yet you have the sample size formula based on estimating a mean. you could fudge and use the formula you have, but there is a more direct formula.

similar comment(s) apply to the second problem.

 

[mikemike123]

For the more "direct formula", would i use ss = Z^2*(p)*(1-p)/c^2.

So I would have 1.96^2*.16*(1-.16)/.03 which would give me about 574.

 

[statdad]

I haven't checked your numbers but yes, the formula

<br /> \widehat p (1-\widehat p) \left(\frac z c\right)^2<br />

is the one I was directing you to.