Please help..

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Please help..
Message posted by Tom on November 7, 2000 at 12:00 AM (ET)

55 assembly workers are a random sample from population. What is population mean wage of active (mean 28772.25,sd 6302.) and terminated workers (mean 41616.04, sd 4670.) need 95% confidence. Thanks for any suggestions

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Re: Please help..
Message posted by JG on November 8, 2000 at 12:00 AM (ET)

What are you measuring ? Do you really have a random sample ? Is it a simple random sample ? What is the sample size for the terminated workers ? What is the population size ?
This question requires a lot of clarification before a meaningful comment can be made.

Re: Please help..
Message posted by nancy diehl on November 8, 2000 at 12:00 AM (ET)

It sounds like what you are looking for is a 95% Confidence Interval for
the population mean of yearly salary wage for active versus the same for
terminated workers. What is not clear is if the sample size that determined
the avg & stddev. for terminated workers is also 55 (the sample size for
the active workers). The other thing that is unclear is if the population is
known to follow a normal distribution. So I will assume this is the case
for both of my unknowns described above. In which case, this is a
straight-forward application of doing a 95% CI about the mean for a large
sample (n>30) which is: avg. +/- Z*(stdev/sqrt(n)) which for 95% the Z value is
1.96. So for active workers you have $28,772.25 +/- 1.96*(6302/sqrt(55)) - you do
the math from here (assuming you understand +/- means you get an upper and a
lower limit, hence the confidence interval). Apply the same formulat to the
terminated data to obtain its interval.

Re: Please help..
Message posted by JG on November 9, 2000 at 12:00 AM (ET)

If you have a random sample - especially a simple random sample you get normality from the central limit theorem when n is GT than 30.

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