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Chapter 7
Estimates and Sample Sizes
7-1 Review and Preview
7-2 Estimating a Population Proportion
7-3 Estimating a Population Mean: σ Known (OMIT)
7-4 Estimating a Population Mean: σ Not Known
7-5 Estimating a Population Variance (OMIT)
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7.1 - 1
Section 7-1
Review and Preview
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7.1 - 2
Review
 Chapters 2 & 3 we used “descriptive
statistics” when we summarized data using
tools such as graphs, and statistics such as
the mean and standard deviation.
 Chapter 6 we introduced critical values:
z denotes the z score with an area of  to
its right.
If  = 0.025, the critical value is z0.025 = 1.96.
That is, the critical value z0.025 = 1.96 has an
area of 0.025 to its right.
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7.1 - 3
Preview
This chapter presents the beginning of
inferential statistics.
 The two major activities of inferential
statistics are (1) to use sample data to
estimate values of a population parameters,
and (2) to test hypotheses or claims made
about population parameters.
 We introduce methods for estimating values
of these important population parameters:
proportions, means, and variances.
 We also present methods for determining
sample sizes necessary to estimate those
parameters.
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7.1 - 4
Section 7-2
Estimating a Population
Proportion
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7.1 - 5
Key Concept
In this section we present methods for using a
sample proportion to estimate the value of a
population proportion.
• The sample proportion is the best point
estimate of the population proportion.
• We can use a sample proportion to construct a
confidence interval to estimate the true value
of a population proportion, and we should
know how to interpret such confidence
intervals.
• We should know how to find the sample size
necessary to estimate a population proportion.
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7.1 - 6
Definition
A point estimate is a single value (or
point) used to approximate a population
parameter.
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7.1 - 7
Definition
ˆ
The sample proportion p is the best
point estimate of the population
proportion p.
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7.1 - 8
Example:
In the Chapter Problem (page 314) we noted
that in a Pew Research Center poll, 70% of
1501 randomly selected adults in the United
States believe in global warming, so the
sample proportion is p̂ = 0.70. Find the best
point estimate of the proportion of all adults in
the United States who believe in global
warming.
Because the sample proportion is the best
point estimate of the population proportion, we
conclude that the best point estimate of p is
0.70. When using the sample results to
estimate the percentage of all adults in the
United States who believe in global warming,
the best estimate is 70%.
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7.1 - 9
Definition
A confidence interval (or interval
estimate) is a range (or an interval) of
values used to estimate the true value
of a population parameter. A
confidence interval is sometimes
abbreviated as CI. Here is an example
of a confidence interval for the
population proportion parameter:
0.677  p  0.723
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7.1 - 10
NOTE
We will learn how to construct
confidence intervals from a sample
statistic later using a formula.
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7.1 - 11
Interpreting a Confidence Interval
We must be careful to interpret confidence intervals
correctly. There is a correct interpretation and many
different and creative incorrect interpretations of the
confidence interval a < p < b
Typically, we interpret the 95% confidence interval as
follows:
“We are 95% confident that the interval from a to b
actually does contain the true value of the population
proportion p.”
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7.1 - 12
Interpreting a Confidence Interval
This means that if we were to select many different
samples of the same size and construct the
corresponding confidence intervals, 95% of them
would actually contain the value of the population
proportion p.
(Note that in this correct interpretation, the level of
95% refers to the success rate of the process being
used to estimate the proportion.)
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7.1 - 13
Interpreting a Confidence Interval
For example, if we calculate the 95%
confidence intervals for 20 different samples
of a population, we expect that 95% of the 20
samples, or 19 samples, would have
confidence intervals that contain the true
value of p.
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7.1 - 14
Interpreting a Confidence Interval
Consider the chapter problem example (global
warming) again. Suppose we know the true
proportion of all adults who believe in global
warming is p=0.75
With 95% confidence interval, if we sample 20
times we may compute a confidence interval
which does not actually contain p=0.75, such
as,
0.677  p  0.723
but, 19 times out of 20 we would find
confidence intervals that do contain p=0.75.
This is illustrated in Figure 7-1.
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7.1 - 15
Interpreting a Confidence Interval
page 319, Figure 7-1
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7.1 - 16
Caution
Know the correct interpretation of a
confidence interval.
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7.1 - 17
NEXT
We will learn now discuss how to
construct confidence intervals from
a sample statistic.
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7.1 - 18
Critical Values
A standard z score can be used to distinguish
between sample statistics that are likely to
occur and those that are unlikely to occur. Such
a z score is called a critical value. Critical values
are based on the following observations:
Under certain conditions, the sampling
distribution of sample proportions can be
approximated by a normal distribution.
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7.1 - 19
Critical Values
DEFINE: A z score of z / 2 associated with a
sample proportion has a probability of /2 of
falling in the right tail. Therefore:
 To find z / 2 , find the z-score
in Table A-2 that corresponds
to an area of 1   / 2
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7.1 - 20
Example
Page 328, problem 7
Find z/2 for  =0.10
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7.1 - 21
Example
Page 328, problem 7
ANSWER:
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7.1 - 22
Definition
A critical value is the number on the
borderline separating sample statistics
that are likely to occur from those that are
unlikely to occur. The number z/2 is a
critical value that is a z score with the
property that it separates an area of /2 in
the right tail of the standard normal
distribution.
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7.1 - 23
Critical Value
Because the standard normal distribution is
symmetric about the value of z=0, the value of
–z/2 is at the vertical boundary for the area of
/2 in the left tail
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7.1 - 24
The Critical Value
z
2
-z/2
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7.1 - 25
Definition
A confidence level is the probability 1 –  (often
expressed as the equivalent percentage value)
that the confidence interval actually does contain
the population parameter, assuming that the
estimation process is repeated a large number of
times. (The confidence level is also called degree
of confidence, or the confidence coefficient.)
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7.1 - 26
Definition
Common choices for confidence levels are:
90% confidence level where  = 10%,
95% confidence level where  = 5%,
99% confidence level where  = 1%,
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7.1 - 27
Confidence Level
For the standard normal distribution a
confidence level of P % corresponds to P
percent of the area between the values –z/2
and z/2
For example, if  = 5% , the confidence level is
1-0.05=0.95=95% which gives the z-score
z / 2  z0.025  1.96
and 95% of the area lies between -1.96 and
1.96
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7.1 - 28
z2 for a 95% Confidence Level
 = 0.05
z2 =  1.96
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7.1 - 29
z2 for a 95% Confidence Level
 = 5%
 2 = 2.5% = .025
z2
-z2
Critical Values
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7.1 - 30
Definition
When data from a simple random sample are
used to estimate a population proportion p, the
margin of error, denoted by E, is the maximum
likely difference (with probability 1 – , such as
0.95) between the observed proportion p̂ and
the true value of the population proportion p.
The margin of error E is also called the
maximum error of the estimate and can be found
by multiplying the critical value and the standard
deviation of the sample proportions:
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7.1 - 31
Margin of Error for
Proportions
E  z 2
NOTE:
ˆˆ
pq
n
qˆ  1  pˆ
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7.1 - 32
Confidence Interval for Estimating
a Population Proportion p
p = population proportion
p̂ = sample proportion
n = number of sample values
E = margin of error
z/2 = z score separating an area of /2 in
the right tail of the standard normal
distribution
7.1 - 33
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Requirements for Using a Confidence
Interval for Estimating a Population
Proportion p
1. The sample is a simple random sample.
2. The conditions for the binomial distribution
are satisfied: there is a fixed number of
trials, the trials are independent, there are
two categories of outcomes, and the
probabilities remain constant for each trial.
3. There are at least 5 successes and 5
failures.
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7.1 - 34
Confidence Interval for Estimating
a Population Proportion p
pˆ – E < p < p̂ + E
where
E  z 2
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ˆˆ
pq
n
7.1 - 35
Confidence Interval for Estimating
a Population Proportion p
ˆ – E < p < pˆ + E
p
ˆ + E
p
ˆ + E)
(pˆ – E, p
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7.1 - 36
Round-Off Rule for
Confidence Interval Estimates of p
Round the confidence interval limits
for p to
three significant digits.
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7.1 - 37
Example
Page 328
Problem 18:
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7.1 - 38
Example
Page 328, problem 18
ANSWER:
compute the critical value z/2
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7.1 - 39
Example
Page 328, problem 18
ANSWER:
compute the sample proportion p̂ and q̂
qˆ  1  pˆ  0.56
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7.1 - 40
Example
Page 328, problem 18
ANSWER:
compute the margin of error E
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7.1 - 41
Calculator Use
Here is what the solution manual suggests
Calculate z/2 first
To use the formula on a TI calculator:
 Calculate p̂ , press the multiply key, the
parentheses key, then 1- ANS (ANS is the 2nd
(-) key on bottom row) , then ENTER. This will
give pˆ qˆ
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7.1 - 42
Example
Press the divide key and input the value of n
then ENTER. This will give pˆ qˆ / n
Press the square root key, then ANS, then
ENTER. This will give pˆ qˆ / n
Finally press the multiply key then input the
value of z/2 then ENTER. This will give E
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7.1 - 43
Procedure for Constructing
a Confidence Interval for p
1. Verify that the required assumptions are satisfied.
(The sample is a simple random sample, the
conditions for the binomial distribution are satisfied,
and the normal distribution can be used to
approximate the distribution of sample proportions
because np  5, and nq  5 are both satisfied.)
2. Refer to Table A-2 and find the critical value z
corresponds to the desired confidence level.
3. Evaluate the margin of error
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/2
that
ˆˆ n
E  z 2 pq
7.1 - 44
Procedure for Constructing
a Confidence Interval for p - cont
4. Using the value of the calculated margin of error, E
and the value of the sample proportion, p, find the
values of p – E and p + E. Substitute those values
in the general format for the confidence interval:
ˆ
ˆ
ˆ
p
ˆ – E < p < pˆ + E
5. Round the resulting confidence interval limits to
three significant digits.
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7.1 - 45
Example
Page 328
Problem 22:
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7.1 - 46
Example
Page 328, problem 22
95% confidence interval gives  = 5%,
compute the critical value z/2
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7.1 - 47
Example
Page 328, problem 22
compute the sample proportion p̂
and q̂
qˆ  1  pˆ  0.8000
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7.1 - 48
Example
Page 328, problem 22
compute the margin of error E
E  z / 2 pˆ qˆ / n
 1.96 (0.2000)(0.8000) / 2000
 0.0175
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7.1 - 49
Example
Page 328, problem 22
compute the upper and lower limits of the
confidence interval
upper limit
pˆ  z / 2 pˆ qˆ / n  0.2000  0.0175
lower limit
pˆ  z / 2 pˆ qˆ / n  0.2000  0.0175
 0.2175
 0.1825
 0.218
 0.183
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7.1 - 50
Example
Page 328, problem 22
ANSWER:
write down the confidence interval using the
upper and lower limits
0.183  p  0.218
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7.1 - 51
Example
Page 328, problem 22
The confidence interval can also be
expressed as
pˆ  E  0.2000  0.0175
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7.1 - 52
Example
Page 328
Problem 10:
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7.1 - 53
Example
Page 328, problem 10
ANSWER:
The sample proportion is the midpoint of the upper
and lower limits of the confidence interval
pˆ 
0.720  0.780
 0.750
2
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7.1 - 54
Example
Page 328, problem 10
ANSWER:
The margin of error is the difference between the
upper limit of the confidence interval and the sample
proportion
E  0.780  0.750  0.030
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7.1 - 55
Example
Page 328, problem 10
The confidence interval can be expressed as
pˆ  E  0.750  0.030
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7.1 - 56
Calculate Confidence Intervals
Directly From Calculator
The TI calculator will compute confidence
intervals as follows:
Press STAT and select TESTS
Select A:1-PropZInt
Enter x,n,C-Level and then calculate
You should be able to calculate a confidence
interval both ways: use the formula as in
previous examples and directly with A:1PropZInt as above
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7.1 - 57
Example
Page 329, problem 30
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7.1 - 58
Example
Page 329, problem 30
a)
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7.1 - 59
Example
Page 329, problem 30
b) 99% confidence interval gives  = 1%,
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7.1 - 60
Example
Page 329, problem 30
b) compute the margin of error E
E  z / 2 pˆ qˆ / n
 2.575 (0.8355)(0.1645) / 152
 0.0774
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7.1 - 61
Example
Page 329, problem 30
b) compute the upper and lower limits of the
confidence interval
upper limit
pˆ  z / 2 pˆ qˆ / n  0.8355  0.0774
lower limit
pˆ  z / 2 pˆ qˆ / n  0.8355  0.0774
 0.9129
 0.7581
 0.913
 0.758
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7.1 - 62
Example
Page 329, problem 30
b) ANSWER:
write down the confidence interval using the
upper and lower limits
0.758  p  0.913
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7.1 - 63
Example
Page 329, problem 30
c) We interpret the answer to part (b) as
follows:
“We are 99% confident that the true value of
the proportion of boys in the population will
be between 0.758 and 0.913.”
Therefore,
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7.1 - 64
Example
Page 329, problem 30
c) if the YSORT method has no effect we
expect the population proportion to be p=0.5
which is not within the 99% confidence
interval from part (b) and we can be 99%
confident that the YSORT method is effective.
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7.1 - 65
Example
Page 329, problem 34
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7.1 - 66
Example
Page 329, problem 34
a)
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7.1 - 67
Example
Page 329, problem 34
b) We interpret the answer to part (a) as
follows:
“We are 99% confident that the true value of
the proportion of people who say they vote in
the population will be between 0.662 and
0.737.”
Therefore,
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7.1 - 68
Example
Page 329, problem 34
c) since the true population proportion is
given as p=0.61 which is not within the 99%
confidence interval from part (b), we can be
99% confident that people do not tell the truth
about their voting record.
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7.1 - 69
Analyzing Polls
When analyzing polls consider:
1. The sample should be a simple random sample,
not an inappropriate sample (such as a voluntary
response sample).
2. The confidence level should be provided. (It is
often 95%, but media reports often neglect to
identify it.)
3. The sample size should be provided. (It is usually
provided by the media, but not always.)
4. Except for relatively rare cases, the quality of the
poll results depends on the sampling method and
the size of the sample, but the size of the
population is usually not a factor.
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7.1 - 70
Caution
Never follow the common misconception
that poll results are unreliable if the
sample size is a small percentage of the
population size. The population size is
usually not a factor in determining the
reliability of a poll.
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7.1 - 71
Sample Size
Suppose we want to collect sample
data in order to estimate some
population proportion. The question is
how many sample items must be
obtained?
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7.1 - 72
Determining Sample Size
z  2
E=
p
ˆ qˆ
n
(solve for n by algebra)
n=
( Z  2)2 p
ˆ ˆq
E2
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7.1 - 73
Sample Size for Estimating
Proportion p
ˆ
When an estimate of p is known:
n=
( z  2 )2 pˆ qˆ
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E2
7.1 - 74
Sample Size for Estimating
Proportion p
ˆ
When no estimate of p is known:
n=
( z  2)2 0.25
E2
NOTE: here we are assuming that
pˆ  qˆ  0.50
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7.1 - 75
Round-Off Rule for Determining
Sample Size
If the computed sample size n is not
a whole number, round the value of n
up to the next larger whole number.
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7.1 - 76
Example
Page 331, problem 42
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7.1 - 77
Example
Page 331, problem 42
a) We are told that the sample percentage is
within 4 percentage points of the true
population percentage. This means that
the margin of error is E=0.04
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7.1 - 78
Example
Page 331, problem 42
a) 90% confidence interval gives  = 10%,
which then gives the values
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7.1 - 79
Example
Page 331, problem 42
a) When sample proportion is unknown use:
pˆ  qˆ  0.50
to get
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7.1 - 80
Example
Page 331, problem 42
b) sample proportion is given
pˆ  0.08
and
qˆ  1  pˆ  0.92
and the same values for the critical value
and margin of error as in part (a) to get
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7.1 - 81
Recap
In this section we have discussed:
 Point estimates.
 Confidence intervals.
 Confidence levels.
 Critical values.
 Margin of error.
 Determining sample sizes.
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7.1 - 82
Section 7-4
Estimating a Population
Mean:  Not Known
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7.1 - 83
Key Concept
This section presents methods for estimating
a population mean when the population
standard deviation is not known. With σ
unknown, we use the Student t distribution
assuming that the relevant requirements are
satisfied.
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7.1 - 84
Sample Mean
The sample mean is the best point
estimate of the population mean.
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7.1 - 85
Notation
 = population mean
x = sample mean
s = sample standard deviation
n = number of sample values
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7.1 - 86
Student t Distribution
If the distribution of a population is essentially
normal, then the distribution of
t =
x-µ
s
n
is a Student t Distribution for all samples of
size n. It is often referred to as a t distribution
and is used to find critical values denoted by
t/2.
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7.1 - 87
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample sizes
(see the following slide, for the cases n = 3 and n = 12).
2. The Student t distribution has the same general symmetric bell
shape as the standard normal distribution but it reflects the
greater variability (with wider distributions) that is expected
with small samples.
3. The Student t distribution has a mean of t = 0 (just as the
standard normal distribution has a mean of z = 0).
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.
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7.1 - 88
Student t Distributions for
n = 3 and n = 12
Figure 7-5
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7.1 - 89
Definition
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. The degree of freedom is often
abbreviated df.
degrees of freedom = n – 1
in this section.
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7.1 - 90
Notation
E = margin of error
t/2 = critical t value separating an area of /2
in the right tail of the t distribution
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7.1 - 91
Margin of Error E for Estimate of 
(With σ Not Known)
Formula 7-6
E = t 
s
2
n
where t2 has n – 1 degrees of freedom.
Table A-3 lists values for tα/2
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7.1 - 92
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7.1 - 93
Confidence Interval for the
Estimate of μ (With σ Not Known)
x–E <µ<x +E
where
E = t/2 s
n
df = n – 1
t/2 found in Table A-3
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7.1 - 94
Example
Use the sample statistics of
n = 49, x = 0.4 and s = 21.0
to construct a 95% confidence interval
estimate of the population mean.
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7.1 - 95
Example
95% confidence level so
 = 5%=0.05
With n = 49, the df = 49 – 1 = 48
Closest df in Table A-3 is 50,
using two tails  = 5%=0.05
using one tail  /2= 2.5%=0.025
t/2 = 2.009
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7.1 - 96
Example
Using t/2 = 2.009, s = 21.0 and n = 49 the
margin of error is:
s
21.0
E  t / 2 
 2.009 
 6.027
n
49
and the confidence interval is
x E    x E
0.4  6.027    0.4  6.027
5.6    6.4
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7.1 - 97
Requirements for Using a Confidence
Interval for Estimating a Population Mean µ
1. The sample is a simple random sample.
2. Either the sample is from a normally
distributed population or n>30
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7.1 - 98
Procedure for Constructing a
Confidence Interval for µ
(With σ Unknown)
1. Verify that the requirements are satisfied.
2. Using n – 1 degrees of freedom, refer to Table A-3 or use
technology to find the critical value t2 that corresponds to
the desired confidence level.
3. Evaluate the margin of error E = t2 • s /
n .
4. Find the values of x  E and x  E. Substitute those
values in the general format for the confidence interval:
x E    x E
5. Round the resulting confidence interval limits.
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7.1 - 99
Round-Off Rule for
Confidence Interval Estimates of µ
If using the original set of data, round to one
more decimal place than is used for the
original data set.
If using summary statistics x ,Sx, n round to
the same number of decimal places used for
the sample mean x
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7.1 - 100
CALCULATOR
You will not be given Table A-3 on the exam
and should be able to use your calculator to
compute the confidence interval for population
mean with σ unknown. This is the method that
will be used for the remaining slides.
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7.1 - 101
Example
Page 354
Problem 14:
NOTE: do part (b) first using the TI calculator
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7.1 - 102
Example
The TI calculator will compute confidence
intervals for population mean with σ
unknown as follows:
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter x ,Sx, n, C-Level and then calculate
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7.1 - 103
Example
Page 354, problem 14
ANSWER to part (b)
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter
x  0.12, Sx  0.04, n  7, C - Level  0.99
Calculate then gives: (0.06395,0.17605)
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7.1 - 104
Example
Page 354, problem 14
ANSWER to part (b)
We must round two decimal places (same
number of places as the sample mean)
0.06    0.18
Units are grams/mile
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7.1 - 105
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 - x
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7.1 - 106
Example
Page 354, problem 14
ANSWER to part (a) uses part (b)
0.06    0.18
0.06  0.18
x
 0.12 grams/mile
2
E  0.18  0.12  0.06 grams/mile
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7.1 - 107
Example
Page 354, problem 18
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7.1 - 108
Example
Page 354, problem 18
a) ANSWER: x  3103 grams
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7.1 - 109
Example
Page 354, problem 18b)
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter
x  3103, Sx  696, n  186, C - Level : 0.95
then calculate to get the confidence interval
(3002.3,3203.7) which rounds to:
3002 g    3204 g
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7.1 - 110
Example
Page 354, problem 18
c) ANSWER:
Yes, since the confidence interval
2608 g    2792 g
for the mean birth weight for mothers who used
cocaine is entirely below the confidence interval
in part (b) for mothers who did not use cocaine,
it appears that cocaine use is associated with
lower birth weights.
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7.1 - 111
Confidence Intervals for
Comparing Data
As in Sections 7-2 and 7-3, confidence
intervals can be used informally to
compare different data sets, but the
overlapping of confidence intervals should
not be used for making formal and final
conclusions about equality of means.
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7.1 - 112
Example
Page 355, problem 22
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7.1 - 113
Example
Page 355, problem 22(a)
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter
x  1.8, Sx  1.4, n  142, C - Level : 0.95
then calculate to get the confidence interval
1.6    2.0 (headaches )
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7.1 - 114
Example
Page 355, problem 22(b)
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter
x  1.6, Sx  1.2, n  80, C - Level : 0.95
then calculate to get the confidence interval
1.3    1.9 (headaches )
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7.1 - 115
Example
Page 355, problem 22(c)
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7.1 - 116
Example
Page 356, problem 26
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7.1 - 117
Example
Page 356, problem 26
Here:
x  ?, Sx  ?, n  7, C - Level : 0.98
and we are not given the mean and
standard deviation. We must determine
these from the given data.
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7.1 - 118
Example
Compute the mean:
x
x =
n
0.06  0.11  0.16  0.15  0.14  0.08  0.15
x
7
0.85

7
 0.121 g/mile
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7.1 - 119
Example
Compute the standard deviation:
s
 (x  x)
2
n 1
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7.1 - 120
Example
x
xx
(x  x)
0.06
-0.061
0.003721
0.11
-0.011
0.000121
0.16
0.039
0.001521
0.15
0.029
0.000841
0.14
0.019
0.000361
0.08
-0.041
0.001681
0.15
0.029
0.000841
Sum of last column:
(x  x)
2
2
 0.009087
0.009087
s
 0.0015145  0.0389 g/mile
7 1
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7.1 - 121
•Calculator:
1) Enter the list of data values into a list using STAT
1:Edit
2) Select 2nd STAT (LIST) and arrow right to choose
MATH option 3:mean(
3) Select 2nd STAT (LIST) and arrow right to choose
MATH option 7:stdDev(
4) Choose the list the data is in
3.1 -
Example
Press STAT and select TESTS
Select 8:Tinterval
Arrow right to select Inpt: Stats
Enter
x  0.121, Sx  0.0389, n  7, C - Level : 0.98
then calculate to get the confidence interval
0.075    0.168 (grams/mil e)
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7.1 - 123
Example
Page 356, problem 26
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7.1 - 124
Recap
In this section we have discussed:
 Student t distribution.
 Degrees of freedom.
 Margin of error.
 Confidence intervals for μ with σ unknown.
 Choosing the appropriate distribution.
 Point estimates.
 Using confidence intervals to compare data.
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7.1 - 125