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Chapter 8: Estimating With Confidence
8.3 – Estimating a Population Mean
In the previous examples,
we made an unrealistic assumption that the
population standard deviation was known and
could be used to calculate confidence intervals.
Standard Error: When the standard deviation of a
statistic is estimated from the data
s
n
When we know  we can use the Z-table to make
a confidence interval. But, when we don’t know
it, then we have to use something else!
Properties of the t-distribution:
•
σ is unknown
• Degrees of Freedom = n – 1
•
More variable than the normal distribution (it has
fatter tails than the normal curve)
• Approaches the normal distribution when the
degrees of freedom are large (sample size is large).
• Area is found to the right of the t-value
Properties of the t-distribution:
• If n < 15, if population is approx normal, then so
is the sample distribution. If the data are clearly
non-Normal or if outliers are present, don’t use!
• If n > 15, sample distribution is normal, except if
population has outliers or strong skewness
• If n  30, sample distribution is normal, even if
population has outliers or strong skewness
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t > 1.093)
n = 11 10
Picture
1.093
Probability
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t > 1.093)
n = 11 10
Picture
Probability
0.15
1.093
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t < 1.093)
Picture
Probability
0.85
n = 11 10
1.093
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t > 0.685)
Picture
n = 24 23
0.685
Probability
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t > 0.685)
Picture
Probability
0.25
n = 24 23
0.685
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(t > 0.685)
Picture
Probability
0.25
n = 24 23
-0.685
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(0.70<t<1.093) n = 11
Picture
10
0.70 1.093
Probability
Example #1
Determine the degrees of freedom and use the t-table to find
probabilities for each of the following:
DF
P(0.70<t<1.093) n = 11
Picture
10
0.1
0.70 1.093
.25 – .15 = 0.1
Probability
Calculator Tip: Finding P(t)
2nd – Dist – tcdf( lower bound, upper
bound, degrees of freedom)
One-Sample t-interval:
 s 
x  t *n1 

 n
Calculator Tip: One sample t-Interval
Stat – Tests – TInterval
Data: If given actual values
Stats: If given summary of values
Conditions for a t-interval:
1. SRS (should say)
2. Normality (population approx normal and n<15, or
moderate size (15≤ n < 30) with moderate
skewness or outliers, or large sample size
n ≥ 30)
3. Independence (Population 10x sample size)
 N  10n
Robustness:
The probability calculations remain fairly accurate
when a condition for use of the procedure is
violated
The t-distribution is robust for large n values,
mostly because as n increases, the t-distribution
approaches the Z-distribution. And by the CLT, it
is approx normal.
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
n = 10
9
99% CI
n = 20
90% CI
n = 40
95% CI
n = 30
99% CI
t*
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
n = 10
9
99% CI
n = 20
19
90% CI
n = 40
95% CI
n = 30
99% CI
t*
3.250
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
n = 30
99% CI
t*
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
t*
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
2.042
n = 30
29
99% CI
Example #2
Practice finding t*
n
Degrees of
Freedom
Confidence
Interval
t*
n = 10
9
99% CI
3.250
n = 20
19
90% CI
1.729
n = 40
39
95% CI
2.042
n = 30
29
99% CI
2.756
Example #3
As part of your work in an environmental awareness group, you
want to estimate the mean waste generated by American adults.
In a random sample of 20 American adults, you find that the
mean waste generated per person per day is 4.3 pounds with a
standard deviation of 1.2 pounds. Calculate a 99% confidence
interval for  and explain it’s meaning to someone who doesn’t
know statistics.
P: The true mean waste generated per person per
day.
A: SRS: Says randomly selected
Normality: 15<n<30. We must assume the
population doesn’t have strong
skewness. Proceeding with caution!
Independence: It is safe to assume that there
are more than 200 Americans
that create waste.
N: One Sample t-interval
I:
 s 
x  t *n1 

 n
df = 20 – 1 = 19
I:
 s 
x  t *n1 

 n
 1.2 
4.3  2.861

 20 
4.3  0.7677
3.5323, 5.0677
df = 20 – 1 = 19
C: I am 99% confident the true mean waste
generated per person per day is between 3.5323
and 5.0677 pounds.
Matched Pairs t-procedures:
Subjects are matched according to characteristics
that affect the response, and then one member is
randomly assigned to treatment 1 and the other to
treatment 2. Recall that twin studies provide a
natural pairing. Before and after studies are
examples of matched pairs designs, but they require
careful interpretation because random assignment is
not used.
Apply the one-sample t procedures to the differences
Confidence Intervals for Matched Pairs
xd  t
*
n 1
 Sd 


 n
Example #4
Archaeologists use the chemical composition of clay found in
pottery artifacts to determine whether different sites were
populated by the same ancient people. They collected five
random samples from each of two sites in Great Britain and
measured the percentage of aluminum oxide in each. Based on
these data, do you think the same people used these two kiln
sites? Use a 95% confidence interval for the difference in
aluminum oxide content of pottery made at the sites and assume
the population distribution is approximately normal. Can you
say there is no difference between the sites?
New
Forrest
20.8
18
18
15.8
18.3
Ashley
Trails
19.1
14.8
16.7
18.3
17.7
Difference
1.7
3.2
1.3
-2.5
.6
P: μn = New Forrest percentage of aluminum oxide
μa = Ashley Trails percentage of aluminum oxide
μd = μn - μa = Difference in aluminum oxide levels
The true mean difference in aluminum oxide levels
between the New Forrest and Ashley Trails.
A: SRS: Says randomly selected
Normality: Says population is approx normal
Independence: It is safe to assume that there
are more than 50 samples
available
N: Matched Pairs t-interval
I:
xd  t
*
n 1
 Sd 


 n
df = 5 – 1 = 4
I:
xd  t
*
n 1
 Sd 


 n
 2.105469069 
.86  2.776

5


.86  2.613866034
1.754, 3.4743
df = 20 – 1 = 19
C: I am 95% confident the true mean difference in
aluminum oxide levels between the New Forrest
and Ashley Trails is between –1.754 and 3.4743.
Can you say there is no difference between the sites?
Yes, zero is in the confidence interval, so it is safe
to say there is no difference.
Example #5
The National Endowment for the Humanities sponsors summer
institutes to improve the skills of high school language teachers.
One institute hosted 20 Spanish teachers for four weeks. At the
beginning of the period, the teachers took the Modern Language
Association’s listening test of understanding of spoken Spanish.
After four weeks of immersion in Spanish in and out of class,
they took the listening test again. (The actual spoken Spanish in
the two tests was different, so that simply taking the first test
should not improve the score on the second test.) Below is the
pretest and posttest scores. Give a 90% confidence interval for
the mean increase in listening score due to attending the
summer institute. Can you say the program was successful?
Subject
Pretest
Posttest
Subject
Pretest
Posttest
1
30
29
11
30
32
2
28
30
12
29
28
3
31
32
13
31
34
4
26
30
14
29
32
5
20
16
15
34
32
6
30
25
16
20
27
7
34
31
17
26
28
8
15
18
18
25
29
9
28
33
19
31
32
10
20
25
20
29
32
P: μB = Pretest score
μA = Posttest score
μd = μB - μA = Difference in test scores
The true mean difference in test scores between the
Pretest and Posttest
A: SRS: We must assume the 20 teachers are randomly
selected
Normality:
A: SRS: We must assume the 20 teachers are randomly
selected
Normality: 15<n<30 and distribution is
approximately normal, so safe to
assume
Independence: It is safe to assume that there
are more than 200 Spanish
teachers
N: Matched Pairs t-interval
I:
*  Sd 
xd  t n1 

 n
df = 20 – 1 = 19
I:
xd  t
*
n 1
 Sd 


 n
 3.2032 
1.45  1.729 

 20 
1.45 1.2384
 2.689,
 0.2115
df = 20 – 1 = 19
C: I am 90% confident the true mean difference in
test scores between the Pretest and Posttest
is between –2.689 and –0.2115.
Can you say the program was successful?
Yes, zero is not in the confidence interval, so the
pretest score is lower than the posttest score.