Transcript Section 5-1 and 5-2 Notes - Pendleton County Schools

Sta220 - Statistics
Mr. Smith
Room 310
Class #16
Section 5-1 and 5-2 Notes
Our goal in this chapter is to estimate the value
of an unknown population parameter, such as
the population mean.
Example
• The mean gas mileage for a new car model
• The average expected life of a flat-screen
computer monitor.
The unknown population parameter (e.g., mean
or proportion) that we are interested in
estimating is called the target parameter.
Procedure
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A point estimator of a population parameter is a
rule or formula that tells us how to use the sample
data to calculate a single number that can be used
as an estimate of the target parameter.
An interval estimator (or confidence interval) is a
formula that tells us how to use the sample data to
calculate an interval that estimates the target
parameter.
Example:
We’ll use the sample mean 𝑥 to estimate the
population mean 𝜇. Consequently, 𝑥, is a point
estimator. We then attach a measure of
reliability, similar to a standard deviation, to our
estimate by obtaining an interval estimator, also
called confidence interval.
5-2: Confidence Interval for a
Population Mean: Normal (z)
Statistic
Suppose a large hospital wants to estimate the
average length of time patients remain in the
hospital.
• The hospital’s target parameter is the population
mean 𝜇.
• Hospital administrators plan to randomly sample
100 of all previous patients’ records.
• The sample mean 𝑥 of the lengths of stay to
estimate 𝜇, the mean of all patients’ visits. So the
sample mean 𝑥 represents a point estimator.
According to the Central Limit Theorem, the sampling
distribution of the sample mean is approximately normal
for a large samples.
(n > 30)
Now we can calculate the interval estimator:
𝑥 ± 1.96 𝜎𝑥
This means we form an interval from 1.96 standard
deviations (probability = .95) below the sample mean to
1.96 standard deviations above the mean.
Sampling distribution of x
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Example 5.1:
Consider the large hospital that wants to estimate
the average length of stay of its patients, 𝜇. The
hospital randomly samples n = 100 of its patients
and finds that the sample mean length of stay is
𝑥 = 4.5 days. Also, suppose it is known that the
standard deviation of the length of stay for all
hospital patients is 𝜎 = 4 days. Use the interval
estimator 𝑥 ± 1.96𝜎𝑥 to calculate a confidence
interval for the target parameter, 𝜇.
Solution
Substitution 𝑥 = 4.5 and 𝜎 = 4 into the interval
estimator formula, we obtain:
𝑥 ± 1.96 𝜎𝑥
= 𝑥 ± 1.96
= 𝑥 ± 1.96
𝜎
𝑛
4
100
= 4.5 ± .78
or (3.72, 5.28).
The confidence coefficient is the probability that an
interval estimator encloses the population
parameter – that is, the relative frequency with
which the interval estimator encloses the
population parameter when the estimator is used
repeatedly a very large number of times. The
confidence level is the confidence coefficient
expressed as a percentage.
𝑥 ± 𝑧𝛼/2 𝜎𝑥
Confidence intervals for m: 10 samples
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reserved.
Locating za/2 on the standard normal curve
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The value 𝑧𝛼 is defined as the value of the
standard normal random variable z such that the
area 𝛼 will lie to its right. In other words,
𝑃(𝑧 > 𝑧𝛼 ) = 𝛼.
Example 5-2:
Find 𝑧𝛼/2 for 𝛼 = .80.
Solution
To illustrate, for confidence coefficient of .80, we
have (1 − 𝛼) = .80, 𝛼 = .20 and 𝛼/2 = .10.
𝑧.10 is z value that locates area .10 in the upper tail
of the sampling distribution. Since the total area to
the right of the mean is .10, we find the z value
corresponding to an area of .5- .1 = .4 to the right of
the mean.
This z value is 𝑧.10 = 1.28
Table 7.2
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reserved.
Procedure
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reserved.
Definition
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reserved.
Example 5-3
Unoccupied seats on flights cause airlines to lose
revenue. Suppose a large airline wants to estimate its
average number of unoccupied seats per flight over the
past year. To accomplish this, the records of 225 flights
are randomly selected, and the number of unoccupied
seats is noted for each of the sampled flights. Descriptive
statistics for the data are displayed in the MINITAB
printout below.
Estimate μ, the mean number of unoccupied seats per
flight during the past year, using 90% confidence interval.
Solution
The form of a large-sample 90% confidence
interval for a population mean is:
𝑥 ± 𝑧𝛼/2 𝜎𝑥
= 𝑥 ± 𝑧.05 𝜎𝑥
= 𝑥 ± 1.645
𝜎
𝑛
= 11.6 ± 1.645
4.1
225
= 11.6 ± .45
or the interval 11.15 to 12.05. That is, at the 90%
confident level, we estimate the mean number of
unoccupied seats per flight to be between 11.15
and 12.5 during the sampled year.
Example 5-4
Many middle schools have initiated a program that provides
every student with a free laptop (notebook) computer.
Student usage of laptops at a middle school that participates
in the initiative was investigated in American Secondary
Education (fall 2009). In a sample of 106 students, the
researchers reported the following statistics on how many
minutes per day each student used his or her laptop for taking
notes: 𝑥 = 13.2 and s = 19.5. Now the researchers want to
estimate the average amount of time per day laptops are used
for taking notes for all middle school students across the
country.
a. Calculate a 90% confidence interval for the
target parameter. Interpret the results.
b. Explain what the phrase “90% confidence”
implies in part a.
Solution
a. For confidence coefficient .90, 𝛼 = .10 and 𝛼/2 = .05.
From the table, 𝑧.05 = 1.645.
The confidence interval is:
𝑥 ± 𝑧𝛼/2 𝜎𝑥
= 𝑥 ± 1.645
19.5
106
= 13.2 ± 3.116
(10.084, 16. 316)
We are 90% confident that the true average
amount of time per day laptops are used for
taking notes for all middle school students
across the country is between 10.084 and
16.316 minutes.
b. “90% confidence” means that in a repeated
sampling, 90% of all confidence intervals
constructed in this manner will contain the true
mean.
Sometime, the we produce a confidence interval that is
too wide. In this case, we want to reduce the width of the
interval to obtain a more precise estimate of 𝜇.
Go back to the hospital example. A 90% confidence
interval for 𝜇 is (3.92, 5.14). The interval is narrower than
the previously calculated 95% confidence interval (3.81,
5.25).
However, we also have “less confidence” in the 90%
confidence interval.
5-2 Homework Due Friday (also 5-3 and 5-5 will
be due Friday, so I encourage you to work
tomorrow).