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Section 7-4
Estimating a
Population Mean:
 Not Known
 unknown
This section presents methods for
finding a confidence interval estimate
of a population mean when the
population standard deviation is not
known. With
, we will use
the Student t distribution assuming
that certain requirements are
satisfied.
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different
sample sizes (see Figure 7-5, following, for the cases n
= 3 and n = 12). SEE GRAPH
2. The Student t distribution has the same general
symmetric bell shape as the standard normal
distribution
3. The Student t distribution has a mean of t = 0 (just as
the standard normal distribution has a mean of z = 0).
Important Properties of the
Student t Distribution
4. The standard deviation of the Student t distribution
varies with the sample size and is greater than 1
(unlike the standard normal distribution, which has a
 = 1).
5. As the sample size n gets larger, the Student t
distribution gets closer to the normal distribution.
Using the Different t-distributions
The number of degrees of freedom for a collection of
sample data is the number of sample values that can
vary after certain restrictions have been imposed on
all data values.
degrees of freedom = n – 1
in this section.
We will use degrees of freedom to identify
critical values.
Student t Distributions for
n = 3 and n = 12
Figure 7-5
Using table A-3 Find the Critical
Values (t / 2 )
1) n= 12, CL: 90%,  unknown
t / 2 = 1.796
2) n= 22, CL: 95%,  unknown
t / 2 = 2.080
3) n= 35, CL: 99%,  unknown
t / 2 = 2.728
4) n= 28, CL: 98%,  unknown
t / 2 = 2.473
Confidence Intervals for Mean (σ unKnown)
SAME THREE STEPS:
1.Check Assumptions.
2. Perform Calculations.
3. State Conclusion.
Assumptions/Requirements with σ
Unknown to Use Student t distribution
1) The sample is a simple random sample.
2) Normality of the sampling distribution:
a) the sample is from a normally distributed
population
or
b)n > 30.
or
c) the graph of the sample appears
approximately normal.
Calculations:
Confidence Interval for the Estimate of μ
(With σ Not Known)
x–E <µ<x +E
Margin of Error →
Formula 7-6
E = t/
s
2
n
Table A-3 lists values for tα/2
where t/2 has n – 1 degrees of freedom.
State Conclusion
Interpret the interval in context!! Same conclusion!
We are ____% confident
that the true mean of
(CONTEXT) is between
____ and ____.
Summary: Procedures for Constructing a
Confidence Interval for µ (With σ Unknown)
1. Verify that the requirements are satisfied.
• Given/assume SRS
• Normal?
n≥30 OR Normally Distributed Population OR Graph the data
2. Find the values of x - E and x + E. (CALC: T-Interval)
𝑥±
𝑠
𝑡𝛼 2
𝑛
3. Conclusion: Interpret the interval in context!!
We are ____% confident that the true mean of (CONTEXT)
is between ____ and ____.
Example: A study found the body
temperatures of 106 healthy
adults. The sample mean was 98.2
degrees and the sample standard
deviation was 0.62 degrees. Find
the margin of error E and the 95%
confidence interval for µ.
Example: A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval for µ.
Assumptions: Assume SRS of body temperatures
and n=106 > 30 therefore, population can be
assumed approximately normal
n = 106
E = t / • s = 1.984 • 0.62 = 0.1195
2
98.20o
x=
s = 0.62o
 = 0.05
/2 = 0.025
n
106
x–E << x +E
98.08o <
 < 98.32o
t / 2 = 1.984
We are 95% confident that the true mean body
temperature is between 98.08o and 98.32o.
Example 2: Flesch ease of reading
scores for 12 different pages randomly
selected from J.K. Rowling’s Harry Potter
and the Sorcerer’s Stone. Find the 95%
interval estimate of , the mean Flesch
ease of reading score. (The 12 pages’
distribution appears to be bell-shaped
with x = 80.75 and s = 4.68.)
Example: Flesch ease of reading scores for 12 different pages
randomly selected from J.K. Rowling’s Harry Potter and the
Sorcerer’s Stone. Find the 95% interval estimate of , the mean
Flesch ease of reading score. (The 12 pages’ distribution appears
to be bell-shaped with x = 80.75 and s = 4.68.)
Assumptions: Given SRS of pages and sample appears normal therefore,
population can be assumed approximately normal
x = 80.75
s = 4.68
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t/2
s = (2.201)(4.68) = 2.97355
n
12
x–E<µ<x+E
80.75 – 2.97355 < µ < 80.75 + 2.97355
77.77645 <  < 83.72355
77.78 <  < 83.72
We are 95% confident that this interval contains the mean Flesch ease of
reading score for all pages.
Minimal Answer
Assumptions: Given SRS of pages and sample
appears normal therefore, population can be
assumed approximately normal
Calculations:
T-Interval
(77.78, 83.72)
Conclusion: We are 95% confident that the true
mean Flesch ease of reading score is between
77.78 and 83.72.
Finding the Point Estimate
and E from a Confidence Interval
Point estimate of µ:
x = (upper confidence limit) + (lower confidence limit)
2
Margin of Error:
E = (upper confidence limit) – (lower confidence limit)
2
Graphing Example:
The following is a list of temperatures taken
on random days in February. Create a 90%
confidence interval to estimate the average
temperature for a certain city in February.
57 54 55 51 56 48 52 51 59 59
a) Check Assumptions for the 90% CI.
b) Calculate the interval.
c) Interpret the interval found in part b.
Solution
a) Given random sample of days
graph appears approximately normal
(YOU MUST SKETCH THE GRAPH!)
b)
c) We are 90% confident that the true mean
temperature in February is between 52.069 and
56.331.