Transcript chap 06 sec1
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 Parameter… Mean:
μ
with Sample Statistic
x
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
x n
5232 40 130.8
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 Right endpoint 115.1
x
130.8
( )
146.5
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
c
is the area under the standard normal curve between the
critical values
. ½(1 –
c
) ½(1 –
c
)
z
= 0
z
–
z c z c
Critical values
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
–
z c
=
–
z c z
= 0
z c z c
1.645
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:
μ
is generally unknown
μ
x
varies from sample to sample If
μ =15, x
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
z
σ
c x
z c
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
–
z c
=
z c z
= 0
z c z c
= 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
z c
n
z c
1.96
53.0
40 16.4
s n
You don’t know
σ
, but since
n
≥ 30, you can use
s
in place of
σ
.
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
𝝈 𝒄 𝒏
x
x E
where
E
z c
•
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) 1.
In Words
Find the sample statistics
n x .
and 2.
Specify
σ
, if known. Otherwise, if
n
≥ 30, find the sample standard deviation
s
use it as an estimate for
σ
.
and
In Symbols
x
x n s
x n
x
1 ) 2
Constructing Confidence Intervals for
μ
3.
In Words
Find the critical value
z c
corresponds to the given that level of confidence.
In Symbols
Use the Standard Normal Table 4.
Find the margin of error
E
.
5.
Find the left and right endpoints and form the confidence interval.
E
z c
n
Left endpoint: Right endpoint:
x x
E 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:
x
130.8
E
≈ 16.4
Left Endpoint:
x
E
114.4
Right Endpoint:
x
114.4 < μ < 147.2
E
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
z
–
z c
=
–
z c z
= 0
z c z c
1.645
z c
= 1.645
Solution: Constructing a Confidence Interval σ Known
• • Margin of error:
E
z c
n
1.645
1.5
20 0.6
Confidence interval: Left Endpoint:
x
E
22.3
Right Endpoint:
x
E
23.5
22.3 < μ < 23.5
Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5
x
22.3
(
E
Point estimate 22.9
•
x x
23.5
)
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 = 𝑧 𝑐 ∗ 𝜎 𝑛 (multiply by square root of 𝑛 ∗ 𝐸 = 𝑧 𝑐 *σ (divide by
E
) 𝑛 = 𝑧 𝑐 ∗𝜎 𝐸 (square both sides)
n) n
z c
E
2
Sample Size
• Given a
c
-confidence level and a margin of error
E
, the minimum sample size
n
population mean
µ
is needed to estimate the
n
z c
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
–
z c
=
– z c z
= 0
z c z c
= 1.96
0.025
z c
= 1.96
z
Solution: Sample Size
z c
= 1.96
σ ≈ s ≈
53.0
E
= 7
n
z c
E
2 7 2 220.23
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