A random sample of 65 flashlight batteries require a sample average of
14 hours with a standard deviation of 6.05 hours. Find a 95%
confidence interval for the average charging time for a battery.

Hi again redcapelliragazo!!


We want that the confidence level be 0.95. So, for a standard normal
variable we need to find Q that:
P(Z > Q) = 0.025. 
Due the symmetry of the normal distribution around 0, if Q is such that 
P(Z > Q) = 0.025 , 
then we will have that
P(Z < -Q) = 0.025. 

Now we can conclude that 
P(-Q < Z < Q) = 1 - P(Z > Q) - P(Z < -Q) =
              = 1 - 0.025 -0.025 =
              = 0.95
This means that the interval (-Q,Q) defines a 95% confidence interval.

Using the table after the link we find that the value 0,025 correspond
to a Z equal to 1.96.
"z-distribution Table":
http://www.math2.org/math/stat/distributions/z-dist.htm

Then for a standard normal variable (-1.96,+1.96) defines a 95%
confidence interval.

Having a large number n of samples (say n > 30) and knowing Mean and
STD (standard deviation) we can calculate the extremes of the 95%
confidence interval by the formula:

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


For more reference about the theory involved in this problem see the
following pages:
"Chapter 6: Confidence Intervals":
http://science.kennesaw.edu/~jdemaio/1107/Chap6.htm

"Confidence Intervals for a Mean":
http://www.ltcconline.net/greenl/courses/201/Estimation/confIntMean.htm

"Confidence Interval for Mean of an Arbitrary Population":
http://www.wku.edu/~david.neal/statistics/confint/zconfint.html

------------------------------------------------------------

Now the problem:

n = 65
Mean = 14
STD = 6.05 

The left extreme of the 95% confidence interval is:

left extreme = 14 - (1.96 * 6.05)/sqrt(65) =
             = 14 - 1.47 =
             = 12.53

and the right extreme is:

right extreme = 14 + (1.96 * 6.05)/sqrt(65) =
              = 14 + 1.47 =
              = 15.47

Then the 95% confidence interval for the average charging time for a battery is:
(12.53,15.47)

------------------------------------------------------------

I hope this helps you. Again if you find something unclear, wrong
and/or incomplete, please let me know by using the clarification
request, I will gladly respond to your requests.

Best regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 07 May 2004 15:54 PDT

I think that the phrase: "Having a large number n of samples (say n >
30) and knowing Mean and STD (standard deviation) we can calculate the
extremes of the 95% confidence interval by the formula" it must be
written as "Having a large number n of SAMPLE SIZE (say n > 30) and
knowing Mean and STD (standard deviation) we can calculate the
extremes of the 95% confidence interval by the formula".

Thank you.

livioflores-ga

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Request for Answer Clarification by redcapelliragazo-ga on 08 May 2004 17:54 PDT

Thanks I am printing it out now and will review it tonight to see if I
have any questions.

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Clarification of Answer by livioflores-ga on 09 May 2004 09:17 PDT

Ok, feel free to request for any clarification needed.

Regards.
livioflores-ga

Oh set sounds good. Everything thing was very clear. Thanks again.