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

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Discussion Overview

The discussion revolves around solving point estimation problems related to confidence intervals and sample size calculations in statistics. Participants are addressing specific problems involving the estimation of proportions and maximum errors in studies, with a focus on applying statistical formulas correctly.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving estimating the proportion of male electrical engineers who smoke, expressing confusion over the lack of standard deviation (S.D) information needed for calculations.
  • Another participant suggests that the number of smokers follows a binomial distribution, implying that mean and variance can be derived from the observed frequency.
  • A different participant questions how to derive S.D from the given proportion of smokers (16%).
  • One participant critiques the initial approach to the first problem, indicating that the sample size formula used is based on estimating a mean rather than a proportion.
  • Another participant proposes a more direct formula for calculating sample size, suggesting the use of the formula involving the proportion and maximum error.
  • A later reply confirms the proposed formula for sample size calculation, indicating agreement on its appropriateness for the problem at hand.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to solving the problems, with some advocating for specific formulas while others question the assumptions and methods used. No consensus is reached on the best way to proceed with the calculations.

Contextual Notes

Participants note the absence of standard deviation in the problems, which complicates the calculations. There is also a lack of clarity on the definition of "error" in the context of the second problem, which may affect the approach to finding solutions.

mikemike123
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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.
 
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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".
 
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?
 
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.
 
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.
 
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.
 

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