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4. A Confidence Interval for  When  is not known.
  x  tn1 s x This is what you actually use most of the time!
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All that "  unknown" means is that we do not have a value of the
population mean. If you only have the sample standard deviation,
use the t table.
Finding degrees of freedom is easy. In most of the problems we do the
numbers of degrees of freedom is one less than the sample size or n  1 .
The value of t that you need should be in Table 18 in the Syllabus
Supplement. Each row represents the number of degrees of freedom
given by the ‘df’ column. It is a good idea to take a ruler and put a line
across the table after every 10th row. Note that the table skips values
after 100 degrees of freedom, but a good guess is always possible, for
example t.110
05  1.657.
"The variance is not known " implies no previous knowledge or
assumption about the value of  2 . Knowing s 2 is having a guess as
to what the variance is; it is not the same as knowing the variance. If
the population distribution is normal or approximately normal, the
formula for a two-sided confidence interval for the mean is
.
  x  tn1s x , where s x  s
n
2
N n
. Be careful – It is a common
n N 1
error to think that a new population size is actually a sample size.
Note: This is the more common case – if you do not know the
population variance and the sample size is not very large, using z
instead of t is a very bad idea.
Example 1: We have a random sample of 10 homes. The sample mean
of expenditures on maintenance is $838 with a sample standard
deviation of $110. Construct a 95% confidence interval for the mean.
Note: If n  .05 N , use s x 
s
Step1: 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 .
Step2: Find the appropriate value of t Use Table 18 in the
Syllabus Supplement to find t n21  t.9025  2.262 (the number in the
.025 column and the 9th row). Note that higher confidence levels
and lower numbers of degrees of freedom give larger values of t ,
and thus larger confidence intervals.(ttable)
110
Step 3: Find the standard error. s x  s

 34.79 .
n
10
Step 4: Put it together.
  x  tn21 s x  838  2.262 34 .79   838  78 .7 . The result can


be written P 759.3    916.7  .95. Or make a ‘Normal’
curve with 838 in the middle and 759.3 and 916.7 on the sides.
Label the area between 759.3 and 916.7 with 95%, the area below
759.3 with 2.5% and the area above 916.7 with 2.5%.
Example 2: Find a 98% confidence interval for the mean when x  22 ,
n  100 and s x  11.
Step1: Confidence level is 98%, so that the significance level is
  .02 .
Step 2: tn21  t.99
01  2.364 (ttable)
Step 3: s x 
sx
n

11
 1.1
100
Step 4:   x  tn21 s x  22  2.364 1.1  22  2.60 . You
should express this as an interval.
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Example 3: We visit a town of 5000 families. We take a sample of
900 families and find a sample mean of $8536 and a sample standard
deviation of $436. Find a 90% confidence interval for the mean.
x  8536 , s x  436 , n  900 and N  5000 .
Step1: Confidence level is 90%, so that the significance level is
  .10 .
Step 2: Since the degrees of freedom are n  1  899 , we run off
the table. If the degrees of freedom are much over 200, use the
value from the infinity line. t n21  t.899
05  1.645 . (ttable)
Step 3: This is the big change. Since the sample is more than 5%
N n
436

n N 1
900
Note that the smaller the population, the more the finite population
correction will shrink the standard error.
Step 4:   x  tn21 s x  8536  1.645 13 .16   8536  21 .65 .
of the population, use the finite population correction. s x 
sx
4100
 14 .5333 0.9056   13 .16 .
4999
You should express this as an interval.
©2002 Roger Even Bove
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