To construct a confidence interval for a single unknown population mean μμ where
the population standard deviation is known, we need
x¯
x
as an estimate for μμ and a
margin of error. Here, the margin of error is called the error bound for a
population mean (abbreviated EBM). The margin of error depends on the
confidence level (abbreviated CL). The confidence level is the probability that the
confidence interval produced contains the true population parameter. Most often, it
is the choice of the person constructing the confidence interval to choose a
confidence level of 90% or higher because he wants to be reasonably certain of his
conclusions.
Suppose the sample mean is 7 and the error bound for the mean is 2.5.
x¯
=
x
=
_______ and
EBM
=
EBM=
_______.
x¯
=
x
=
7 and
EBM
=
EBM=
2.5.
The confidence interval is
(
7
-
2.5
,
7
+
2.5
)
(7-2.5,7+2.5).
If the confidence level (CL) is 95%, then we say we are 95% confident that the true
population mean is between 4.5 and 9.5.
A confidence interval for a population mean with a known standard deviation is
based on the fact that the sample means follow an approximately normal
distribution. Suppose we have constructed the 90% confidence interval (5, 15)
where
x¯
=
10
x
=10
and
EBM
=
5
EBM=5. To get a 90% confidence interval, we must include
the central 90% of the sample means. If we include the central 90%, we leave out
a total of 10 % or 5% in each tail of the normal distribution. To capture the central
90% of the sample means, we must go out 1.645 standard deviations on either side
of the calculated sample mean. The 1.645 is the z-score from a standard normal
table that has area to the right equal to 0.05
(5% area in the right tail). The graph shows the general situation.

To summarize, resulting from the Central Limit Theorem,
X¯
X
is normally distributed, that is,
X¯
X
~
N
(
μ
X
,
σ
X
n
)
N(
μ
X
,
σ
X
n
)
Since the population standard deviation, σ, is known, we use a normal curve.
The confidence level, CLCL, is
CL
=
1
-
α
CL=1-α. Each of the tails contains an area
equal to
α
2
α
2
.
The z-score that has area to the right of
α2
α2 is
z
α
2
z
α
2
.
For example, if
α
2
=
0.025
α
2=0.025,
then area to the right
=
0.025
=0.025
and area to the left
=
1
-
0.025
=
0.975
=1-0.025=0.975
and
z
α
2
=
z
0.025
=
1.96
z
α
2
=
z
0.025
=1.96
using a calculator, computer or table. Using the TI83+ or 84 calculator function, invNorm, you can verify this result.
invNorm(.975,0,1)=1.96(.975,0,1)=1.96.
The error bound formula for a single population mean when the variance is known is
EBM
=
z
α
2
⋅
σ
n
EBM=
z
α
2
⋅
σ
n
The confidence interval has the format
(
x¯
−
EBM
,
x¯
+
EBM
)
(
x
−EBM,
x
+EBM).
The graph gives a picture of the entire situation.
CL
+
α
2
+
α
2
=
CL
+
α
=
1
CL+
α
2
+
α
2
=CL+α=1.

Suppose scores on exams in statistics are normally distributed with an
unknown population mean and a population standard deviation of 3 points. A sample
of 36 scores is taken and gives a sample mean (sample average score) of 68. Find a
90% confidence interval for the true (population) mean of statistics exam scores.
- The first solution is step-by-step.
- The second solution uses the TI-83+ and TI-84 calculators.
To find the confidence interval, you need the sample mean,
x¯
x
, and the EBM.
- a.
x¯
=
68
x
=68
- b.
EBM
=
z
α
2
⋅
(
σ
n
)
EBM=
z
α
2
⋅(
σ
n
)
- c.
σ
=
3
σ=3
- d.
n
=
36
n=36
CL = 0.90
CL = 0.90
so
a
=
1
-
CL
=
1
-
0.90
=
0.10
a=1-CL=1-0.90=0.10
Since
α
2
=
0.05
α
2
=0.05, then
z
α
2
=
z
.05
=
1.645
z
α
2
=
z
.05
=1.645
from a calculator, computer or standard normal table
Therefore,
EBM
=
1.645
⋅
(
3
36
)
=
0.8225
EBM=1.645⋅(
3
36
)=0.8225
This gives
x¯
-
EBM
=
68
-
0.8225
=
67.18
x
-EBM=68-0.8225=67.18
and
x¯
+
EBM
=
68
+
0.8225
=
68.82
x
+EBM=68+0.8225=68.82
The 90% confidence interval is (67.18, 68.82).
The TI-83+ and TI-84 caculators simplify this whole procedure. Press
STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press
ENTER. Arrow to Stats and press ENTER. Arrow down and enter 3 for
σσ, 68 for
x¯
x
, 36 for
nn, and .90 for C-level. Arrow down to Calculate and
press ENTER. The confidence interval is (to 3 decimal places) (67.178,
68.822).
We can find the error bound from the confidence interval. From the upper
value, subtract the sample mean or subtract the lower value from the upper value
and divide by two. The result is the error bound for the mean (EBM).
EBM
=
68.822
-
68
=
0.822
EBM=68.822-68=0.822
or
EBM
=
(
68.822
−
67.178
)
2
=
0.822
EBM=
(
68.822
−
67.178
)
2
=0.822
We can interpret the confidence interval in two ways:
- We are 90% confident that the true population mean for statistics exam scores is
between 67.178 and 68.822.
- Ninety percent of all confidence intervals constructed in this way contain the true
average statistics exam score. For example, if we constructed 100 of these
confidence intervals, we would expect 90 of them to contain the true population mean
exam score.
Now for the same problem, find a 95% confidence interval for the true (population)
mean of scores. Draw the graph. The sample mean, standard deviation, and sample
size are:
- a.
x¯
=
x
=
- b.
σ
=
σ=
- c.
n
=
n=
- a.
x¯
=
68
x
= 68
- b.
σ
=
3
σ = 3
- c.
n
=
36
n = 36
The confidence level is CL = 0.95CL= 0.95. Graph:
The confidence interval is (use technology)
(
x¯
-
EBM
,
x¯
+
EBM
)
=
(_______, _______)
(
x
-EBM,
x
+EBM)=(_______, _______).
The error bound EBM =EBM = _______.
(
x¯
-
EBM
,
x¯
+
EBM
)
=
(67.02 , 68.98)
(
x
-EBM,
x
+EBM)=(67.02 , 68.98).
The error bound EBM =EBM = 0.98.
We can say that we are 95 % confident that the true population mean for
statistics exam scores is between 67.02 and 68.98 and that 95% of all
confidence intervals constructed in this way contain the true average
statistics exam score.
Suppose we change the previous problem.
Leave everything the same except the sample size.
- a.
x¯
=
68
x
=68
- b.
σ
=
3
σ=3
- c.
z
α
2
=
1.645
z
α
2
=1.645
If we increase the sample size n n to 100, we decrease the error bound.
EBM
=
z
α
2
⋅
(
σ
n
)
=
1.645
⋅
(
3
100
)
=
0.4935
EBM=
z
α
2
⋅(
σ
n
)=1.645⋅(
3
100
)=0.4935
If we decrease the sample size n n to 25, we increase the error bound.
EBM
=
z
α
2
⋅
(
σ
n
)
=
1.645
⋅
(
3
25
)
=
0.987
EBM=
z
α
2
⋅(
σ
n
)=1.645⋅(
3
25
)=0.987
Leave everything the same except for the confidence level. We increase the confidence level from 0.90 to 0.95.
- a.
x¯
=
68
x
=68
- b.
σ
=
3
σ=3
- c.
z
α
2
=
1.645
z
α
2
=1.645
The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95
is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.
- Confidence Level:
The percent expression for the probability that the confidence interval contains the true population parameter. That is, for ex., if CL=90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
- Error Bound for a Population Mean (EBM):
The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation.
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