Confidence Intervals for the Mean (Large Samples) Larson/Farber 4th ed 1 Section 6.1.

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Transcript of Confidence Intervals for the Mean (Large Samples) Larson/Farber 4th ed 1 Section 6.1.
Confidence Intervals for the Mean (Large Samples)
Larson/Farber 4th ed1
Section 6.1
Section 6.1 Objectives
Larson/Farber 4th ed2
Find a point estimate and a margin of errorConstruct and interpret confidence
intervals for the population meanDetermine the minimum sample size
required when estimating μ
Point Estimate for Population μ
Larson/Farber 4th ed3
Point EstimateA single value estimate for a population
parameterMost unbiased point estimate of the
population mean μ is the sample meanx
Estimate Population Parameter…
with Sample Statistic
Mean: μx
Example: Point Estimate for Population μ
Larson/Farber 4th ed4
Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. The following represents a random sample of the number of sentences found in 50 advertisements. Find a point estimate of the population mean, . (Source: Journal of Advertising Research)
9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 2517 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 714 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20
Solution: Point Estimate for Population μ
Larson/Farber 4th ed5
The sample mean of the data is 620
12.450
xx
n
Your point estimate for the mean length of all magazine advertisements is 12.4 sentences.
Interval Estimate
Larson/Farber 4th ed6
Interval estimate An interval, or range of values, used to estimate a
population parameter.
Point estimate
• 12.4
How confident do we want to be that the interval estimate contains the population mean μ?
( )
Interval estimate
Level of Confidence
Larson/Farber 4th ed7
Level of confidence c The probability that the interval estimate
contains the population parameter.
zz = 0zc zc
Critical values
½(1 – c) ½(1 – c)
c is the area under the standard normal curve between the critical values.
The remaining area in the tails is 1 – c .
c
Use the Standard Normal Table to find the corresponding zscores.
zc
Level of Confidence
Larson/Farber 4th ed8
If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ.
zz = 0 zc
The corresponding zscores are +1.645.
c = 0.90
½(1 – c) = 0.05½(1 – c) = 0.05
zc = 1.645 zc = 1.645
Sampling Error
Larson/Farber 4th ed9
Sampling error The difference between the point estimate
and the actual population parameter value.For μ:
the sampling error is the difference – μμ is generally unknown varies from sample to sample
x
x
Margin of Error
Larson/Farber 4th ed10
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.
Sometimes called the maximum error of estimate or error tolerance.
c x cE z zn
σσ When n 30, the sample standard deviation, s, can be used for .
Example: Finding the Margin of Error
Larson/Farber 4th ed11
Use the magazine advertisement data and a 95% confidence level to find the margin of error for the mean number of sentences in all magazine advertisements. Assume the sample standard deviation is about 5.0.
zc
Solution: Finding the Margin of Error
Larson/Farber 4th ed12
First find the critical values
zzcz = 0
0.95
0.0250.025
zc = 1.96
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 ≥ 30.)
zc = 1.96
Solution: Finding the Margin of Error
Larson/Farber 4th ed13
5.01.96
501.4
c c
sE z z
n 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 1.4 sentences.
Confidence Intervals for the Population Mean
Larson/Farber 4th ed14
A cconfidence interval for the population mean μ
The probability that the confidence interval contains μ is c.
where cx E x E E zn
Constructing Confidence Intervals for μ
Larson/Farber 4th ed15
Finding a Confidence Interval for a Population Mean (n 30 or σ known with a normally distributed population)
In Words In Symbols
1. Find the sample statistics n and .
2. Specify , if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for .
xxn
2( )1
x xsn
x
Constructing Confidence Intervals for μ
Larson/Farber 4th ed16
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.
Use the Standard Normal Table.
Left endpoint: Right endpoint: Interval:
cE zn
x Ex E
x E x E
In Words In Symbols
Example: Constructing a Confidence Interval
Larson/Farber 4th ed17
Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements.
Solution: Recall and E = 1.4
12.4x
12.4 1.4
11.0
x E
12.4 1.4
13.8
x E
11.0 < μ < 13.8
Left Endpoint:
Right Endpoint:
( )
Solution: Constructing a Confidence Interval
Larson/Farber 4th ed18
11.0 < μ < 13.8
• 12.411.0 13.8
With 95% confidence, you can say that the population mean number of sentences is between 11.0 and 13.8.
Example: Constructing a Confidence Interval σ Known
Larson/Farber 4th ed19
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.
zc
Solution: Constructing a Confidence Interval σ Known
Larson/Farber 4th ed20
First find the critical values
zz = 0 zc
c = 0.90
½(1 – c) = 0.05½(1 – c) = 0.05
zc = 1.645 zc = 1.645
zc = 1.645
Solution: Constructing a Confidence Interval σ Known
Larson/Farber 4th ed21
Margin of error:
Confidence interval:
1.51.645 0.6
20cE z
n
22.9 0.6
22.3
x E
22.9 0.6
23.5
x E
Left Endpoint:
Right Endpoint:
22.3 < μ < 23.5
Solution: Constructing a Confidence Interval σ Known
Larson/Farber 4th ed22
22.3 < μ < 23.5
( )
• 22.922.3 23.5
With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years.
Point estimate
xx E x E
Interpreting the Results
Larson/Farber 4th ed23
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
Larson/Farber 4th ed24
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
Larson/Farber 4th ed25
Given a cconfidence level and a margin of error E, the minimum sample size n needed to estimate the population mean is
If is unknown, you can estimate it using s provided you have a preliminary sample with at least 30 members.
2
czn
E
Example: Sample Size
Larson/Farber 4th ed26
You want to estimate the mean number of sentences in a magazine advertisement. How many magazine advertisements must be included in the sample if you want to be 95% confident that the sample mean is within one sentence of the population mean? Assume the sample standard deviation is about 5.0.
zc
Solution: Sample Size
Larson/Farber 4th ed27
First find the critical values
zc = 1.96
zz = 0 zc
0.95
0.0250.025
zc = 1.96 zc = 1.96
Solution: Sample Size
Larson/Farber 4th ed28
zc = 1.96 s = 5.0 E = 1
221.96 5.0
96.041
czn
E
When necessary, round up to obtain a whole number.
You should include at least 97 magazine advertisements in your sample.
Section 6.1 Summary
Larson/Farber 4th ed29
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