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3. A Confidence Interval for  When  is Known.
  x  z 2  x You can only use this when you know the population
variance. Don’t forget that there are two formulas for the variance
of the sample mean depending on sample size!
An interval of this type is used in two situations: (i) where the
population variance,  2 , is in fact, known and the sample size is
relatively large; or (ii) where the variance is not known and the sample
variance, s 2 , is used to replace  2 , but the degrees of freedom are so
large that the appropriate value of t n 1 is not very different from z .
The first of these situations is not very realistic, but serves as a good
introduction to confidence intervals. The formula for this type of
confidence interval for the mean is,   x  z  x , where the standard
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deviation of the sample mean, called the standard error is  x  
n
.

N n
n N 1
( n is sample size and N is population size)
Example 1: Assume that a population is Normally distributed with an
unknown mean and a population standard deviation of 36. x ~ N ?,36 
Note: If n  .05 N , use  x 
From a random sample of size n  9 , we get a sample mean of 62.
(Because the population variance is known we can ignore any sample
variance we might compute from the data.). Find a 95% confidence
interval for the mean.
Step 1: State the confidence level and significance level.
The given confidence level of 95% represents the probability that
the interval actually contains the mean and is stated as 1    .95.
The significance level of 5% represents the probability of being
wrong and is   .05 .
Step 2: Find the appropriate value of z. Use the last line of
Table 18 (or Table 17) in the Syllabus Supplement to find (ttable)
z  z.025  1.960 (the bottom number in the .025 column).
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Note that higher confidence levels give larger values of z. , and
thus larger confidence intervals.
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Step 3: Find the standard error.  x  

 12.
n
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Note that larger values of n make the standard error and the confidence interval smaller.
Step 4: Put it together.
  x  z 2 x  62  1.96012  62  23.52 . The last part of this
expression means that the interval extends from 62 – 23.52 = 38.48
to 62 + 23.52 = 85.52. The result can be written
P38 .48    85 .52  .95 .
Example 2: Assume that a population is Normally distributed with
an unknown mean and a population standard deviation of 36.
x ~ N ?,36  From a random sample of size n  9 , we get a sample
mean of 62. (Because the population variance is known we can
ignore any sample variance we might compute from the data.). This
time find a 99% confidence interval for the mean.
Step 1: State the confidence level and significance level.
The given confidence level of 99% represents the probability that
the interval actually contains the mean and is stated as 1    .99. The significance level of 1% represents the probability of being right
and
is   .01.
Step 2: Find the appropriate value of z. Use the last line of
Table 18 (or Table 17) in the Syllabus Supplement to find (ttable)
z  z.005  2.576 (the bottom number in the .005 column).
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Note that higher confidence levels give larger values of z , and
thus larger confidence intervals.
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Step 3: Find the standard error.  x  

 12. No
n
9
change from example 1.
Step 4: Put it together.
  x  z  x  62  2.57612  62  30.91 . The result can be
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written P31 .09    92 .91  .99 . Or make a Normal curve with
62 in the middle and 31.09 and 92.91 on the sides. Label the area
between 31.09 and 92.91 with 99%, the area below 31.09 with 0.5%
and the area above 92.91 with 0.5%.
Definitions:
Note that if we are considering the possibility that the
population mean is 50, we can now say that since this value is
on the confidence interval, and since it is on the interval, we
can say that the mean is not significantly different from 50.
However the mean is significantly different from 20 or 100.
©2002 Roger Even Bove
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