In order to estimate the average electric usage per month, a sample of
n = 81 houses was selected, and the electric usage was determined.
(a)	Assume a population standard deviation of 450-kilowatt hours. 
Determine the standard error of the mean. (b)	With a probability of
0.95, what is the size of the margin of error? (c) If the sample mean
is 1858-kilowatt hours, what is the 95% confidence interval estimate
of the population mean?

Hi!!


"Standard Error of the Mean: 
If we were to draw all possible samples of size "n" from a given
population, and for each sample calculate the mean, and then make a
frequency distribution of the sample means, then the central limit
theorem states that:
The mean of the distribution is the same as the mean of the population
from which the samples are drawn
The standard deviation of the distribution of the sample means is
equal to the standard deviation of the population divided by the
square root of the sample size "n" and is usually called the Standard
Error. So,

SE = SD(of the population)/sqrt(n) 

For a large sample, the distribution of the sample means is
approximately a normal distribution, even if the population from which
the samples were drawn is not a normal distribution."
From "Standard Error of the Mean" by Richard Reid:
http://www.bjmath.com/bjmath/Stats/serr.htm


Now the question a) can be answered easily because you know that SD(of
the population) is 450 and n is 81:

SE = 450/sqrt(81) = 450/9 = 50

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For the question b) we must find the Z-score for 0.95:

A probability of 0.95 for a standard normal variable Z means:

0.95 = P(-Q < Z < Q) =
     = 1 - P(Z > Q) - P(Z < -Q) =
     = 1 - 2*P(Z < -Q) (due the symmetry of the normal distribution around 0)  
    
Then:
P(Z < -Q) = (1 - 0.95)/2 = 0.025

This means that the area under the standard normal probability curve
(i.e. the probability), below Z is 0.025.
Use the following table to find the value of Z that correspond to a
probability (or area under the curve below Z) of 0.025. You will find
that Z = 1.96
"z-distribution Table":
http://www.math2.org/math/stat/distributions/z-dist.htm

Now we can calculate the margin of error:

E = Z-score(p) * (SD / sqrt(n)) =
  = 1.96 * (450/sqrt(81)) =
  = 1.96 * 50 =
  = 98

Note that the margin of error E results the product of Z-score(p) by
the Standard Error of the mean:

E = Z-score(p)*SE

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We just find the margin of error for a 95% confidence interval estimate
of the population mean (remember that we used 0.95 as the
probability), then if we know that the sample mean is 1,858-kilowatt
hours we can say that the 95% confidence interval estimate of the
population mean is:

           1,858 Kw/h +- 98 Kw/h 
or 
           1,760 Kw/h to 1,956 Kw/h

We are 95% confident that the interval contains the population mean.

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I hope that this helps you. Please request for a clarification if you
need it before rate this answer. I will gladly respond your requests
for further assistance.


Best regards.
livioflores-ga