Download 8.1

Section 8.1 - Estimating Population
Means (When  is Known)
• A point estimate is a single-number estimate of a
population parameter (such as the population mean,
the population standard deviation, the population
variance). In 8.1 we estimate population means.
An unbiased estimator is a point estimate that does
not consistently underestimate or overestimate the
population parameter.
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𝑥 is the best point estimate of 𝜇
Usually, we can’t find the exact population mean 𝜇 by
surveying the entire population (time, expense, or
sheer impossibility).
We can take a sample and find the 𝑥.
That’s as good as we can do.
But we want to express some uncertainty.
Being uncertain in a formal mathematical way.
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Confidence Intervals
math courseware specialists
8.1 Introduction to Estimating
Population Means
Confidence Interval for Population Means:
E is the
Margin of Error
We claim that μ is between these values.
We claim that μ is in this (low,high) interval
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Example 8.1: Finding a Point Estimate for a
Population Mean
Find the best point estimate for the population mean of test
scores on a standardized biology final exam. The following is a
simple random sample taken from the population of test scores.
45 68 72 91 100 71
SOLUTION:
xi
69 83 86 55 89 97
x
 81.6.
n
76 68 92 75 84 70
We take a sample and compute its
81 90 85 74 88 99
mean. The sample mean, 𝑥, is 81.6.
76 91 93 85 96 100

So 81.6 is our Point Estimate of the
population mean.
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Estimating Population Means
A point estimate is a single-number estimate of a
population parameter.
“We estimate that the population mean is 81.6.”
taking the
estimation idea
one step further
“We are 90% confident that the population
mean is between 77.5 and 85.8.”
An interval estimate is a range of possible values for a
population parameter.
The level of confidence is the probability that the interval
estimate contains the population parameter.
A confidence interval is an interval estimate associated with a
certain level of confidence.
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Estimating Population Means
The margin of error, or maximum error of estimate, E,
is the largest possible distance from the point estimate
that a confidence interval will cover.
(The sample mean
plus or minus some
“margin of error”.)
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Example 8.2: Constructing a Confidence Interval
with a Given Margin of Error
A college student researching study habits collects data
from a random sample of 250 college students on her
campus and calculates that the sample mean is x  15.7
hours per week. If the margin of error for her data
using a 95% level of confidence is E = 0.6 hours,
construct a 95% confidence interval for her data.
Interpret your results.
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Example 8.2: Constructing a Confidence Interval
with a Given Margin of Error (cont.)
Thus, the lower endpoint is calculated as follows.
Lower Endpoint of the Interval:
𝑥 − 𝐸 = ______ - _______ = _______ hours per week
Upper Endpoint of the Interval:
𝑥 + 𝐸 = ______ + _______ = _______ hours per week
Write it as an inequality:
_______ < μ < ________
Write it as an interval:
( ______ , ______ )
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Example 8.2: Constructing a Confidence Interval
with a Given Margin of Error (cont.)
The interpretation of our confidence interval is that we
are ____% confident that the true population mean for
the number of hours per week that students on this
campus spend studying is between _____ and _____
hours.
But this was too easy. They handed us the value of E.
We did minus on one side of the mean.
We did plus on the other side of the mean.
We really want to know “Where does that E value
come from?”
As the old saying goes, “Give a man the value of E and he will calculate one confidence
interval; teach a man how to find E and he will enjoy statistics for a lifetime.”
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Using the Standard Normal Distribution to
Estimate a Population Mean
Margin of Error of a Confidence Interval for a
Population Mean ( Known)
When the population standard deviation is known, the
sample taken is a simple random sample, and either
the sample size is at least 30 or the population
distribution is approximately normal, the margin of
error of a confidence interval for a population mean is
given by
  
E  z 2   x   z 2 
 n 
 
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 
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E = margin of error
The parts of the formula
c = the level of confidence,
  
E  z 2   x   z 2 
such as c = .90 for 90%
 n 
alpha, α = 1 – c, such as 0.10
𝑧𝛼/2 = the critical value, the z value such that
• the area to the right of 𝑧𝛼/2 is 𝛼/2
• the area to the left of −𝑧𝛼/2 is 𝛼/2
• the area in the middle is c
σ = the population standard deviation
n = the sample size
 
 
Practice/Review – Find the Area
Example: Find the critical value 𝑧𝛼/2
to use in calculating a 90%
confidence interval.
1. Sketch: 90% in middle.
2. Leftover ____ in 2 tails.
3. = _____ in each 1 tail
4. What negative z value
has that area its left?
(Use table or use invNorm).
5. By symmetry, the positive z is _____; that‘s 𝑧𝛼/2 .
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A convenient table for common critical values
Level of Confidence
0.80 (or 80%)
0.85 (or 85%)
0.90 (or 90%)
0.95 (or 95%)
0.98 (or 98%)
0.99 (or 99%)
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zα/2
1.28
1.44
1.645
1.96
2.33
2.575
So you can use this
table with the
special values.
Or you can reason it
out like you did on
the previous slide
where we
calculated the
critical value for a
90% confidence
interval.
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EXAMPLE: A sample of 32 airport shuttle
vans’ records shows that the mean annual
mileage was 75,118. Assume that the
standard deviation of the population of all
such vans is 10,000 miles. Construct the
90% confidence interval of the annual
mileage. Construct the 95% confidence
interval, too.
• The next slide shows the final result of
the by-hand solution.
• The next slide after that shows the final
result of the TI-84 ZInterval solution.
• To view the entire step-by-step solution,
go to http://2205.drscompany.com,
click on Examples,
click on Chapter 8,
click on Section 8.1,
then click “Confidence interval with z”.
A sample of 32 …mean annual mileage
was 75,118…. standard deviation of
the population of all such vans is
10000 miles. Construct the 90% and
the 95% confidence intervals…
Always: Read the problem
and make notes of important
quantities.
By-Hand:
Find the critical value, 𝑧𝛼/2 .
Find the margin of error,
𝑧𝑎/2 ∙𝜎
𝐸= 𝑛 .
Find the confidence
interval, 𝑥 − 𝐸, 𝑥 + 𝐸 .
TI-84 Method:
STAT, TESTS, 7:ZInterval
You may be required to give
some of the by-hand
method’s values, too.
stats/confint/z/01-basic
19
x = 75118
σ = 10000
n = 32
Critical Value: zα/2:
zα/2 =1.645 for 90% C.I.
zα/2 = 1.96 for 95% C.I.
Margin of Error, E:
For 90% C.I.:
For 95% C.I.:
E=
(1.645)(10000)
√32
E = 2908.0 miles
x – E = 75118 – 2908.0
= 72210.0
x + E = 75118 + 2908.0
= 78026.0
90% C.I. is
(72210.0, 78026.0) miles
or 72210.0 < μ < 78026.0
E=
(1.96)(10000)
√32
E = 3464.8 miles
x – E = 75118 – 3464.8
= 71653.2
x + E = 75118 + 3464.8
= 78582.8
95% C.I. is
(71653.2, 78582.8) miles
or 71653.2 < μ < 78582.8
A sample of 32 …mean annual mileage
was 75,118…. standard deviation of
the population of all such vans is
10000 miles. Construct the 90% and
the 95% confidence intervals…
20
x = 75118
σ = 10000
n = 32
Always: Read the problem
and make notes of important
quantities.
By-Hand:
Find the critical value, 𝑧𝛼/2 .
Find the margin of error,
𝑧𝑎/2 ∙𝜎
𝐸= 𝑛 .
Find the confidence
interval, 𝑥 − 𝐸, 𝑥 + 𝐸 .
TI-84 Method:
STAT, TESTS, 7:ZInterval
You may be required to give
some of the by-hand
method’s values, too.
stats/confint/z/01-basic
95% C.I. is
90% C.I. is
(72210.0, 78026.0) miles (71653, 78583) miles
or 71653 < μ < 78583
or 72210 < μ < 78026
TRY THIS: Suppose that on Thursday afternoon we had
40 applicants to the college take a placement exam. The
mean score was a 38. The standard deviation of that
exam score is 5.5 according to a recent study.
Construct the 90% confidence interval for the population
mean score of all applicants taking that exam.
Make notes of values and variables:
40 = _______
38 = _______
5.5 = ______
0.90 = ______
Compute the margin of error using the primitive formula:
 
 
E  z 2   x   z 2
  


n
Then assemble the Confidence Interval:
Lower Endpoint of the Interval:
𝑥 − 𝐸 = ______ - _______ = _______ points on the exam
Upper Endpoint of the Interval:
𝑥 + 𝐸 = ______ + _______ = _______ points on the exam
Write it as an inequality:
_______ < μ < ________
Write it as an interval:
( ______ , ______ )
Do it again using TI-84 ZInterval
Inpt:
Stats
σ: ______________
x: ______________
n: ______________
C-Level:
Calculate
_______
What does the results
screen tell you?
Further words about ZInterval
•
•
•
If you’re asked for a confidence interval,
• Use ZInterval for a normal distribution situation.
• It’s easier than using the primitive formula
• The calculator keeps more decimal precision
If the problem asks for a critical value of z, too,
• Then you have to use invNorm( or the printed
table to answer that question.
Make the right choice between
• Stats, if you’re given the mean, etc.
• Data, if you’re given a list of raw data
Excel Function
=CONFIDENCE.NORM( α, σ, n )
• α = level of significance = 1 – level of confidence
• σ = population standard deviation
• n = sample size
(These values came
from some other
example)
Minimum Sample Size for Estimating a
Population Mean
Minimum Sample Size for Estimating a Population
Mean
The minimum sample size required for estimating a
population mean at a given level of confidence with a
particular margin of error is given by
 z 2   
n
 E 
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What is the Minimum Sample Size I need?
2
The Situation:
 z 2   
n
• I’m going to calculate
 E 
a confidence interval.
• I want to be within E = . . . . of the true mean.
• I want c = . . . . % confidence level.
My Big Question: “What’s the minimum sample size I
need to achieve this level of accuracy?”
Why do I care?
• It costs time and money to collect the sample data.
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What is the Minimum Sample Size I need?
The Formula:
2
 z 2   
• I want to be within E = . . . .
n

E

of the true mean. It goes right in.
• I want c = . . . .% confidence level. Just like before, c
leads to α leads to α/2 leads to zα/2.
• I have σ available from somewhere, probably other
people’s previous studies.
My Big Question: “What’s the minimum sample size I
need to achieve this level of accuracy?” The formula
tells me n. Always bump up (don’t round, bump up.)
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EXAMPLE: The number of samples needed
to estimate the price of something within
$200. Go to http://2205.drscompany.com
Click on Chapter 8
Click on Section 8.1
Click on What sample size is needed?
The final results screen is shown in the next
slide.
95% confidence interval …The
standard deviation … assumed to be
$1,000. If the maximum allowable
margin of error is $200, how many
prices need[ed] … for the sample?
Recognize the problem
type.
Write the formula.
Plug into the formula.
Calculator.
Always bump up to the next
higher whole number. Do
not round off in this
calculation; always bump
up.
30
n=
n=
za/2 • σ
(
(
E
)
(1.96)(1000)
200
.
2
)
2
n = 97 is the minimum sample size needed
to obtain a 95% confidence interval with a
margin of error of no more than $200.
stats/confint/z/05-sample-size-needed
Example 8.8: Finding the Minimum Sample Size Needed for
a Confidence Interval for a Population Mean
Determine the minimum sample size needed if we wish
to be 90% confident that the sample mean is within
two units of the population mean. An estimate for the
population standard deviation of 8.4 is available from a
previous study.
Make notes as you read:
90% = ______
2 = _____,
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8.4 = ______.
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Example 8.8: Finding the Minimum Sample Size Needed for
a Confidence Interval for a Population Mean (cont.)
 z 2   
n
 E 
2
𝑧𝛼/2 =________, σ = ________,
E = ________
Example 8.8: Finding the Minimum Sample Size Needed for
a Confidence Interval for a Population Mean (cont.)
So the size of the sample that we need to construct a
90% confidence interval for the population mean with
the desired margin of error is at least _______.
What’s good about this?
• We have some assurance about the results we get,
knowing how many we need in our sample.
• We know not to over-sample
• Saves time
• Saves effort
• Saves money!
Drawback? Need to know σ. Or rely on some claim about σ.
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Example 8.8 with Excel
• Line 1: typing the formula with all numbers.
• Line 2: wrap it in the CEILING(value, multiple of)
function to bump up to next highest integer.
• Lines 3 and 4: same, but we use
–NORM.S.INV(0.05) instead of the table lookup.
One more kind of problem you’ll see
Example: A survey showed that the 95% confidence
interval for the price of homes in Pine City is
($101000, $145000).
(a) What was the sample mean?
(b) What is the point estimate for the population
mean?
(c) What is the margin of error?
(d) What is the standard deviation?
http://2205.drscompany.com, Chapter 8, Section 8.1,
“Working backward from the interval”
(a) The sample mean is the midpoint between the low
endpoint and high endpoint of the interval. Recall the
formula: 𝑥 − 𝐸, 𝑥 + 𝐸 Calculate it here:
(b) The point estimate is the same as the sample mean.
$____________. That’s where this whole thing started!
(c) Margin of Error: how far is it between an endpoint
(either endpoint) and the sample mean? Calculate it
here.
(d) What is the standard deviation? That’s trickier. Plug
𝑧𝛼/2 ∙𝜎
into formula 𝐸 =
and solve for the unknown σ.
𝑛