Section 6.1 Confidence Intervals
for the Mean(Large Samples)
Objective: SWBAT find a point estimate and a margin
of error
And construct and interpret confidence intervals for
the population mean
And determine the minimum sample size required
when estimating p
• In this section you will learn how to use
sample statistics to estimate the value of an
unknown population parameter ц when the
sample size is 30 or when the population is
normally distributed. And the standard
deviation σ is known.
Definition
A point estimate Is a single value estimate for a
population parameter The most unbiased point
estimate of the population mean ц is the sample
mean ҳ
Example 1
Finding a Point Estimate:
Market researchers use the number of sentences per
advertisement as a measure of reliability for magazine
advertisements. The following represents a random
sample of the number of sentences found in 54
advertisements. Find a point estimate of the population
mean ц.
9 20 18 16 9 16 16 9 11 13 22 16 5 18 6 6 5 12
25 17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7
14 6 11 12 11 15 6 12 14 11 4 9 18 12 12 17 11
20
Solution
• The sample mean of the data is
n
X
X
i 1
n
i
671
=
54
 12.4
So your point estimate for the mean length of all magazine
advertisements is
12.4 sentences.
Try it yourself
Another random sample of the number of
sentences found in 30 magazine
advertisements is listed on page 280 use this
sample to find the point estimate for ц.
a. Find the sample mean
b. Estimate the mean sentence length of the
population.
Definition
An interval estimate is an interval or range of values
used to estimate a population
Parameter.
To form an interval estimate use the point
estimate as the center of the interval then add
and subtract a margin of error. For instance if
the point estimate is 12.4 and the margin of
error is 2.1 then an interval estimate would be
given by 12.4 ± 2.1 or 10.3 < ц < 14.5
Before finding an Interval estimate you first
determine how confident you need to be that
your interval estimate contains the
population mean ц.
Definition
The level of confidence c is the probability
that the interval estimate contains the
population parameter.
The level of confidence c is the area under the standard
normal curve between the critical values –zc and zc the
area remaining is 1- c and the area in each tail is ½ (1-c)
Margin of Error
Definition
Given a level of confidence c , the margin of error
sometimes also called the maximum error of estimate
or error tolerance) E is the greatest possible distance
between the point estimate and the value of the
parameter it is estimating.
E  z c  X  zc

n
When n ≥ 30 the sample standard deviation s can be
used in place of σ.
Example 2
Use the data in example 1 and a 95% confidence
level to find the margin of error for the mean
number of sentences in all magazine
advertisements.
Solution : The z score that corresponds to a 95%
confidence level is 1.96 This implies that 95% of
the area under the standard normal curve falls
within 1.96 standard deviations of the mean.
n=54 ≥ 30 We don’t know the population
standard deviation. But because n ≥ 30 you can
use s in place of σ.
Solution (cont.)
Interpretation: You are 95% confident that the
margin of error for the population mean is
about 1.3 sentences.
n
s
(X
i 1
i
 X)
n 1
2
1333.24

 5.0
53
Using the values zc  1.96
  s  5.0 and n=54

E  zc
 1.96
 1.3
n
5.0
54
Try it Yourself
Use the data in try it yourself 1 and a 95%
confidence level to find the margin of error
for the mean number of sentences in a
magazine advertisement.
a.) Identify z, n, and s.
b.) Find E using zc, σ≈ s , and n
c.) State the margin of error.
Confidence Intervals for the
Population Mean
Guidelines
1. Find the sample statistic n and x
n

X
2. Specify σ if known . Otherwise if n≥30
Find the sample standard deviation s and s 
use it as an estimate for σ.
3. Find the critical value Zc that corresponds
To the given level of confidence.
4. Find the margin of error E
i 1
=
Xi
n
n
2
(
X

X
)
 i
i 1
n 1
E  zc

n
5. Find the left and right endpoints Left endpoint x - E
And form the confidence intervals Right endpoint x + E
Interval x – E < ц < x + E
Example 3:
Constructing a Confidence Interval
Construct a 95% confidence interval for the mean
number of sentences in all magazine
advertisements.
Solution in Examples 1 and 2 you found that the mean
= 12.4 and E = 1.3 The confidence interval is as
follows
X – E = 12.4 – 1.3 = 11.1 x+ E = 12.4 +1.3 = 13.7
11.1 < ц < 13.7
Example 5
Constructing a Confidence Interval, σ known
A college admissions director wishes to estimate
the mean age of all students enrolled. In a
random sample of 20 students, the mean age is
found to be 22.9 years. From the past studies
the standard deviation is known to be 1.5 years
and the population tis normally distributed.
Construct a 90% confidence interval of the
population mean age.
Example 5 Cont.
Solution: Using n=20,mean =22.9,σ=1.5 and
Zc = 1.645 the margin of error at the 90%
confidence interval is

1.5
E  zc
n
 1.645
20
 0.6
The 90% confidence interval can be written as
x ± E = 22.9 ± .6 or as follows
22.9 - .6 = 22.3 or 22.9 + .6 = 23.5
22.3 < ц < 23.5
Interpretation: With 90% confidence you can say
that the mean age of all students lies between
22.3 and 23.5 years.
Sample Size
For the sample statistics as the level of
confidence increases, the confidence interval
widens as the interval widens the precision of
the estimate decreases. One way to improve
the precision of an estimate without
decreasing the level of confidence is to
increase the sample size. But how large a
sample size is needed to guarantee a certain
level of confidence for a given margin of error.
Find a Minimum sample size to
estimate ц
Given a confidence level c and a margin of
error E find the minimum sample size the
population mean needed to estimate the
population mean ц.
 zc 
n

 E 
If σ is unknown you can estimate it using s
provided you have a preliminary sample with
at least 30 members.
2
Example 6
Determining a Minimum sample size
You want to estimate the mean number of
sentences in a magazine advertisement. How
many magazine advertisements must be in the
sample to be 95% confident that the sample
mean is within one sentence of the
population mean?
Solution
Using c= 0.95 zc = 1.96 , σ≈ s ≈ 5.0 (from
example 2) and E =1. You can solve for the
Minimum sample size is
 zc 
n

 E 
2
 1.96 5.0 


1


2
 96.04
When necessary round up
to the nearest whole
number. So you could
include at 97 magazine
advertisements in your
sample.
Try it Yourself
• How many magazine advertisements must be
included in the sample if you want to be 95%
confident that the sample mean is within 2
sentences of the population mean
zc = 1.96, E=2, and s ≈ 5.0
 1.96 5.0 


2


 24.01
2
Point Estimate
DEFINITION:
A point estimate is a single value
estimate for a population parameter.
The best point estimate of the
population mean
is the sample mean
Example: Point Estimate
A random sample of 35 airfare prices (in dollars) for a one-way ticket from
Atlanta to Chicago. Find a point estimate for the population mean,
.
99
101
107
102
109
98
105
103
101
105
98
107
104
96
105
95
98
94
100
104
111
114
87
104
108
101
87
103
106
117
94
103
101
The sample mean is
The point estimate for the price of all one way tickets
from Atlanta to Chicago is $101.77.
105
90
Interval Estimates
Point estimate
•
101.77
An interval estimate is an interval or range of values
used to estimate a population parameter.
(
•
101.77
)
The level of confidence, x, is the probability that the
interval estimate contains the population parameter.
Distribution of Sample Means
When the sample size is at least 30, the sampling distribution
for is normal.
Sampling distribution of
For c = 0.95
0.95
0.025
-1.96
0
0.025
1.96
z
95% of all sample means will have standard scores
between z = -1.96 and z = 1.96
Maximum Error of Estimate
The maximum error of estimate E is the greatest possible distance between
the point estimate and the value of the parameter it is, estimating for a given
level of confidence, c.
When n
30, the sample standard deviation, s, can be used for
.
Find E, the maximum error of estimate for the one-way plane fare from Atlanta to Chicago
for a 95% level of confidence given s = 6.69.
Using zc = 1.96, s = 6.69, and n = 35,
You are 95% confident that the maximum error of estimate is $2.22.
Confidence Intervals for
Definition: A c-confidence interval for the population mean is
Find the 95% confidence interval for the one-way plane fare from Atlanta to Chicago.
You found
= 101.77 and E = 2.22
Right endpoint
Left endpoint
(
99.55
•
101.77
)
103.99
With 95% confidence, you can say the mean one-way fare from Atlanta to Chicago is
between $99.55 and $103.99.
Sample Size
Given a c-confidence level and an maximum error of estimate, E, the
minimum sample size n, needed to estimate , the population mean
is
You want to estimate the mean one-way fare from Atlanta to
Chicago. How many fares must be included in your sample if you
want to be 95% confident that the sample mean is within $2 of
the population mean?
You should include at least 43 fares in your sample. Since you
already have 35, you need 8 more.
Homework 1-22 Page 287
Day2: 23-30 all, 31-51 odd pgs 288-290