A recent survey of 32 unemployed male executives showed that it took
an average of 20 weeks for them to find another position.  The
standard deviation of the sample was 6.2 weeks.   Construct a 95
percent confidence interval for the population mean.   Is it
reasonable that the population mean is 26 weeks?  Justify your answer.

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Request for Question Clarification by scriptor-ga on 25 Aug 2005 15:59 PDT

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Scriptor

Hi lola5!!


What we want is that the confidence level be 0.95. 
To get an idea on how we can find that confidence level supose that we
have a standard normal variable, we are looking for Q that:
P(Z > Q) = 0.025. 

Due the symmetry of the normal distribution around 0, if P(Z > Q) =
0.025 , then P(Z < -Q) = 0.025.

Working with that probabilities we can conclude that:
P(-Q < Z < Q) = 1 - P(Z > Q) - P(Z < -Q) =
              = 1 - 0.025 -0.025 =
              = 0.95
That confirms the interval (-Q,Q) as a 95% confidence interval.

Use the following table to find the value of Z that correspond to a
probability of 0.025. You will find Z = 1.96
"z-distribution Table":
http://www.math2.org/math/stat/distributions/z-dist.htm

We conclude that, for a standard normal variable, the interval
(-1.96,+1.96) defines a 95% confidence interval.


If you have a large number n of sample size (say n > 30) and know the
Mean and the STD (standard deviation) we can calculate the extremes of
the 95% confidence interval by the formula:

Extremes = Mean +/- (1.96 * STD)/sqrt(n) 


For this problem we have that:

n = 32 (n > 30)
Mean = 20
STD = 6.2

The extremes of the confidence interval are:

left extreme = 20 - (1.96 * 6.2)/sqrt(32) =
             = 20 - 2.15 =
             = 17.85

right extreme = 20 + (1.96 * 6.2)/sqrt(32) =
              = 20 + 2.15 =
              = 22.15

Then the 95% confidence interval is:
(17.85, 22.15)


The sample is sufficiently large , n > 30, to define an interval
within which we can be 95% sure the true population mean will lie,
based on sample estimates of the mean and standard error. Since the
95% confidence interval does not contain the number 26, 26 weeks is
not a reasonable population mean.


See for more references:
"Confident Limits":
http://aerg.canberra.edu.au/envirostats/bm/L3/ff_cflim.htm


Search Strategy:
"sample mean confidence" "population mean"
sample mean confidence



I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards,
livioflores-ga