Section 6.1
Confidence Intervals for the Mean
(σ known)
(Large Samples)
Section 6.1 Objectives
• Find a point estimate and a margin of error
• Construct and interpret confidence intervals for the
population mean
• Determine the minimum sample size required when
estimating μ
Point Estimate for Population μ
Point Estimate
• A single value estimate for a population parameter
• Most unbiased point estimate of the population mean
μ is the sample mean x
Estimate Population with Sample
Parameter…
Statistic
x
Mean: μ
Example: Point Estimate for Population μ
A social networking website allows its users to add
friends, send messages, and update their personal
profiles. The following represents a random sample of
the number of friends for 40 users of the website. Find a
point estimate of the population mean, µ.
140 105 130 97 80 165 232 110 214 201 122
98 65 88 154 133 121 82 130 211 153 114
58 77 51 247 236 109 126 132 125 149 122
74 59 218 192 90 117 105
Solution: Point Estimate for Population μ
The sample mean of the data is
x 5232
x

 130.8
n
40
Your point estimate for the mean number of friends for
all users of the website is 130.8 friends.
Interval Estimate
Interval estimate
• An interval, or range of values, used to estimate a
population parameter.
Left endpoint
x  130.8
115.1
(
115
Right endpoint
Point estimate
120
125
130
135
146.5
)
140
145
150
Interval estimate
How confident do we want to be that the interval estimate
contains the population mean μ?
Level of Confidence
Level of confidence c
• The probability that the interval estimate contains the
population parameter.
c is the area under the
c
standard normal curve
between the critical values.
½(1 – c)
½(1 – c)
–zc
z=0
Critical values
z
zc
Use the Standard
Normal Table to find the
corresponding z-scores.
The remaining area in the tails is 1 – c .
Level of Confidence
• If the level of confidence is 90%, this means that we
are 90% confident that the interval contains the
population mean μ.
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
–zc = –1.645
z=0
zc =zc1.645
The corresponding z-scores are ±1.645.
z
Sampling Error
Sampling error
• The difference between the point estimate and the
actual population parameter value.
• For μ:
 the sampling error is the difference x – μ
 μ is generally unknown
 x varies from sample to sample
 If μ =15, and x is 18.5 then, sampling error is 3.5
Margin of Error
Margin of error
• The greatest possible distance between the point
estimate and the value of the parameter it is
estimating for a given level of confidence, c.
• Denoted by E.
E  zcσ x  zc
σ
n
When n ≥ 30, the sample
standard deviation, s, can
be used for σ.
• Sometimes called the maximum error of estimate or
error tolerance.
Example: Finding the Margin of Error
Use the social networking website data and a 95%
confidence level to find the margin of error for the
mean number of friends for all users of the website.
Assume the sample standard deviation is about 53.0.
Solution: Finding the Margin of Error
• First find the critical values
0.95
0.025
0.025
zc
–zc = –1.96
z=0
zczc= 1.96
z
95% of the area under the standard normal curve falls
within 1.96 standard deviations of the mean. (You
can approximate the distribution of the sample means
with a normal curve by the Central Limit Theorem,
because n = 40 ≥ 30.)
Solution: Finding the Margin of Error
E  zc

n
 zc
s
n
You don’t know σ, but
since n ≥ 30, you can
use s in place of σ.
53.0
 1.96 
40
 16.4
You are 95% confident that the margin of error for the
population mean is about 16.4 friends.
Confidence Intervals for the Population
Mean
A c-confidence interval for the population mean μ
sample mean +/- margin of error
𝒙±𝑬
𝒙 ± 𝒔𝒐𝒎𝒆 # 𝒐𝒇 𝒔𝒕𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏𝒔
𝝈
𝒙 ± 𝒛𝒄
𝒏
xE  xE
where E  zc

n
•
The probability that the confidence interval contains μ
is c.
Constructing Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean
(n ≥ 30 or σ known with a normally distributed population)
In Words
1. Find the sample statistics n and
x.
2. Specify σ, if known.
Otherwise, if n ≥ 30, find the
sample standard deviation s and
use it as an estimate for σ.
In Symbols
x
x
n
(x  x )2
s
n 1
Constructing Confidence Intervals for μ
In Words
3. Find the critical value zc that
corresponds to the given
level of confidence.
4. Find the margin of error E.
5. Find the left and right
endpoints and form the
confidence interval.
In Symbols
Use the Standard
Normal Table
E  zc

n
Left endpoint: x  E
Right endpoint: x  E
Interval:
xE  xE
Example: Constructing a Confidence
Interval
Construct a 95% confidence interval for the mean
number of friends for all users of the website.
Solution: Recall x  130.8 and E ≈ 16.4
Left Endpoint:
x E
 130.8  16.4
 114.4
Right Endpoint:
xE
 130.8  16.4
 147.2
114.4 < μ < 147.2
Solution: Constructing a Confidence
Interval
114.4 < μ < 147.2
•
With 95% confidence, you can say that the population
mean number of friends is between 114.4 and 147.2.
Confidence Intervals for the Population
Mean
•
•
•
•
•
mean ±𝐸
3.50 ± .10
3.50 ± .25
3.50 ± .50
3.50 ± 1.00
⇒
⇒
⇒
⇒
3.40, 3.60
3.25, 3.75
3.00, 4.00
2.50, 4.50
The wider the interval, the more confident one is .
Example: Constructing a Confidence
Interval σ Known
A college admissions director wishes to estimate the
mean age of all students currently enrolled. In a random
sample of 20 students, the mean age is found to be 22.9
years. From past studies, the standard deviation is
known to be 1.5 years, and the population is normally
distributed. Construct a 90% confidence interval of the
population mean age.
Solution: Constructing a Confidence
Interval σ Known
• First find the critical values
c = 0.90
½(1 – c) = 0.05
½(1 – c) = 0.05
zc
–zc = –1.645
z=0
zc = 1.645
zc =zc1.645
z
Solution: Constructing a Confidence
Interval σ Known
• Margin of error:

E  zc
n
 1.645 
• Confidence interval:
Left Endpoint:
x E
 22.9  0.6
 22.3
1.5
20
 0.6
Right Endpoint:
xE
 22.9  0.6
 23.5
22.3 < μ < 23.5
Solution: Constructing a Confidence
Interval σ Known
22.3 < μ < 23.5
Point estimate
22.3
(
x E
22.9
23.5
x
xE
•
)
With 90% confidence, you can say that the mean age
of all the students is between 22.3 and 23.5 years.
Interpreting the Results
• μ is a fixed number. It is either in the confidence
interval or not.
• Incorrect: “There is a 90% probability that the actual
mean is in the interval (22.3, 23.5).”
• Correct: “If a large number of samples is collected
and a confidence interval is created for each sample,
approximately 90% of these intervals will contain μ.
Interpreting the Results
The horizontal segments
represent 90% confidence
intervals for different
samples of the same size.
In the long run, 9 of every
10 such intervals will
contain μ.
μ
Sample Size
𝐸 = 𝑧𝑐 ∗
𝜎
𝑛
(multiply by square root of n)
𝑛 ∗ 𝐸 = 𝑧𝑐 *σ (divide by E)
𝑧𝑐 ∗𝜎
𝑛=
𝐸
(square both sides)
 zc 
n

E


2
Sample Size
• Given a c-confidence level and a margin of error E,
the minimum sample size n needed to estimate the
population mean µ is
 zc 
n

E


2
Example: Sample Size
You want to estimate the mean number of friends for all
users of the website. How many users must be included
in the sample if you want to be 95% confident that the
sample mean is within seven friends of the population
mean? Assume the sample standard deviation is about
53.0.
Solution: Sample Size
• First find the critical values
0.95
0.025
0.025
zc
–zc = –1.96
z=0
zc = 1.96
zczc= 1.96
z
Solution: Sample Size
zc = 1.96
σ ≈ s ≈ 53.0
E=7
 zc   1.96  53.0 
n

  220.23

7

 E  
2
2
When necessary, round up to obtain a whole number.
You should include at least 221 users in your sample.
Section 6.1 Summary
• Found a point estimate and a margin of error
• Constructed and interpreted confidence intervals for
the population mean
• Determined the minimum sample size required when
estimating μ