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Stat 155, Section 2, Last Time• Binomial Distribution
– Normal Approximation– Continuity Correction– Proportions (different scale from “counts”)
• Distribution of Sample Means– Law of Averages, Part 1 – Normal Data Normal Mean– Law of Averages, Part 2:
Everything (averaged) Normal
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Reading In Textbook
Approximate Reading for Today’s Material:
Pages 382-396, 400-416
Approximate Reading for Next Class:
Pages 425-428, 431-439
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Chapter 6: Statistical Inference
Main Idea:
Form conclusions by
quantifying uncertainty
(will study several approaches,
first is…)
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Section 6.1: Confidence Intervals
Background:
The sample mean, , is an “estimate”
of the population mean,
How accurate?
(there is “variability”, how
much?)
X
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Confidence Intervals
Recall the Sampling Distribution:
(maybe an approximation)
nNX
,~
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Confidence Intervals
Thus understand error as:
How to explain to untrained consumers?
(who don’t know randomness,
distributions, normal curves)
ndistX 'n
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Confidence Intervals
Approach: present an interval
With endpoints:
Estimate +- margin of error
I.e.
reflecting variability
How to choose ?
mX
m
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Confidence Intervals
Choice of “Confidence Interval radius”,
i.e. margin of error, :
Notes:
• No Absolute Range (i.e. including “everything”) is available
• From infinite tail of normal dist’n
• So need to specify desired accuracy
m
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Confidence Intervals
HW: 6.1
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Confidence IntervalsChoice of margin of error, :Approach:• Choose a Confidence Level• Often 0.95
(e.g. FDA likes this number for approving new drugs, and it is a common standard for publication in many fields)
• And take margin of error to include that part of sampling distribution
m
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Confidence Intervals
E.g. For confidence level 0.95, want
distribution
0.95 = Area
= margin of errorm
X
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Confidence Intervals
Computation: Recall NORMINV takes
areas (probs), and returns cutoffs
Issue: NORMINV works with lower areas
Note: lower tail
included
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Confidence Intervals
So adapt needed probs to lower areas….
When inner area = 0.95,
Right tail = 0.025
Shaded Area = 0.975
So need to compute:
nNORMINV
,,975.0
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Confidence Intervals
Need to compute:
Major problem: is unknown
• But should answer depend on ?
• “Accuracy” is only about spread
• Not centerpoint
• Need another view of the problem
nNORMINV
,,975.0
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Confidence Intervals
Approach to unknown :
Recenter, i.e. look at dist’n
Key concept:
Centered at 0
Now can calculate as:
nNORMINVm
,0,975.0
X
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Confidence Intervals
Computation of:
Smaller Problem: Don’t know
Approach 1: Estimate with
• Leads to complications
• Will study later
Approach 2: Sometimes know
nNORMINVm
,0,975.0
s
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Confidence Intervals
E.g. Crop researchers plant 15 plots
with a new variety of corn. The
yields, in bushels per acre are:
Assume that = 10 bushels / acre
138
139.1
113
132.5
140.7
109.7
118.9
134.8
109.6
127.3
115.6
130.4
130.2
111.7
105.5
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Confidence IntervalsE.g. Find:
a) The 90% Confidence Interval for the mean value , for this type of corn.
b) The 95% Confidence Interval.
c) The 99% Confidence Interval.
d) How do the CIs change as the confidence level increases?
Solution, part 1 of:http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg22.xls
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Confidence Intervals
An EXCEL shortcut:
CONFIDENCE
Careful: parameter is:
2 tailed outer area
So for level = 0.90, = 0.10
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Confidence Intervals
HW: 6.5, 6.9, 6.13, 6.15, 6.19
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Choice of Sample Size
Additional use of margin of error idea
Background: distributions
Small n Large n
X
n
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Choice of Sample Size
Could choose n to make = desired value
But S. D. is not very interpretable, so make “margin of error”, m = desired value
Then get: “ is within m units of ,
95% of the time”
n
X
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Choice of Sample Size
Given m, how do we find n?
Solve for n (the equation):
n
mn
XPmXP
95.0
nm
ZP
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Choice of Sample Size
Graphically, find m so that:
Area = 0.95 Area = 0.975
nm
nm
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Choice of Sample Size
Thus solve:
2
1,0,975.0
NORMINVm
n
1,0,975.0NORMINVn
m
1,0,975.0NORMINVm
n
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Choice of Sample Size
Numerical fine points:
• Change this for coverage prob. ≠ 0.95
• Round decimals upwards,
To be “sure of desired coverage”
2
1,0,975.0
NORMINVm
n
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Choice of Sample Size
EXCEL Implementation:
Class Example 22, Part 2:http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg22.xls
HW: 6.22 (1945), 6.23
2
1,0,975.0
NORMINVm
n
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Interpretation of Conf. Intervals
2 Equivalent Views:
Distribution Distribution
95%
pic 1 pic 2
m m m 0 m
X X
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Interpretation of Conf. Intervals
Mathematically:
pic 1 pic 2
no pic
"",.. bracketsmXmXICtheP
mXPmXmP 95.0
mXmXP
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Interpretation of Conf. Intervals
Frequentist View: If repeat the
experiment many times,
About 95% of the time, CI will contain
(and 5% of the time it won’t)
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Confidence Intervals
Nice Illustration:
Publisher’s Website
• Statistical Applets
• Confidence Intervals
Shows proper interpretation:
• If repeat drawing the sample
• Interval will cover truth 95% of time
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Interpretation of Conf. Intervals
Revisit Class Example 17http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg17.xls
Recall Class HW:
Estimate % of Male Students at UNC
C.I. View: Class Example 23http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg23.xls
Illustrates idea:
CI should cover 95% of time
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Interpretation of Conf. Intervals
Class Example 23:http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg23.xls
Q1: SD too small Too many cover
Q2: SD too big Too few cover
Q3: Big Bias Too few cover
Q4: Good sampling About right
Q5: Simulated Bi Shows “natural var’n”
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Interpretation of Conf. Intervals
HW: 6.27, 6.29, 6.31
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And now for somethingcompletely different….
A fun dance video:
http://ebaumsworld.com/2006/07/robotdance.html
Suggested by David Moltz
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Sec. 6.2 Tests of Significance
= Hypothesis Tests
Big Picture View:
Another way of handling random error
I.e. a different view point
Idea: Answer yes or no questions, under uncertainty
(e.g. from sampling or measurement error)
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Hypothesis Tests
Some Examples:
• Will Candidate A win the election?
• Does smoking cause cancer?
• Is Brand X better than Brand Y?
• Is a drug effective?
• Is a proposed new business strategy effective?
(marketing research focuses on this)
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Hypothesis Tests
E.g. A fast food chain currently brings in
profits of $20,000 per store, per day. A
new menu is proposed. Would it be
more profitable?
Test: Have 10 stores (randomly selected!)
try the new menu, let = average of
their daily profits.
X
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Fast Food Business ExampleSimplest View: for :
new menu looks better.
Otherwise looks worse.
Problem: New menu might be no better (or
even worse), but could have
by bad luck of sampling
(only sample of size 10)
000,20$X
000,20$X
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Fast Food Business Example
Problem: How to handle & quantify gray area in these decisions.
Note: Can never make a definite conclusion e.g. as in Mathematics,
Statistics is more about real life…
(E.g. even if or , that might be bad luck of sampling, although very unlikely)
0$X 000,000,1$X
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Hypothesis Testing
Note: Can never make a definite conclusion,
Instead measure strength of evidence.
Approach I: (note: different from text)
Choose among 3 Hypotheses:
H+: Strong evidence new menu is better
H0: Evidence is inconclusive
H-: Strong evidence new menu is worse
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Caution!!!
• Not following text right now
• This part of course can be slippery
• I am “breaking this down to basics”
• Easier to understand
(If you pay careful attention)
• Will “tie things together” later
• And return to textbook approach later
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Hypothesis Testing
Terminology:
H0 is called null hypothesis
Setup: H+, H0, H- are in terms of
parameters, i.e. population quantities
(recall population vs. sample)
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Fast Food Business Example
E.g. Let = true (over all stores) daily
profit from new menu.
H+: (new is better)
H0: (about the same)
H-: (new is worse)000,20$
000,20$
000,20$
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Fast Food Business Example
Base decision on best guess:
Will quantify strength of the evidence using
probability distribution of
E.g. Choose H+
Choose H0
Choose H-000,20$X
000,20$X
000,20$X
X
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Fast Food Business Example
How to draw line?
(There are many ways,
here is traditional approach)
Insist that H+ (or H-) show strong evidence
I.e. They get burden of proof
(Note: one way of solving
gray area problem)
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Fast Food Business Example
Assess strength of evidence by asking:
“How strange is observed value ,
assuming H0 is true?”
In particular, use tails of H0 distribution as
measure of strength of evidence
X
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Fast Food Business ExampleUse tails of H0 distribution as measure of
strength of evidence: distribution
under H0
observed value ofUse this probability to measure
strength of evidence
X
X
k20$
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Hypothesis Testing
Define the p-value, for either H+ or H-, as:
P{what was seen, or more conclusive | H0}
Note 1: small p-value strong evidence against H0, i.e. for H+ (or H-)
Note 2: p-value is also called observed significance level.
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Fast Food Business Example
Suppose observe: ,
based on
Note , but is this conclusive?
or could this be due to natural sampling variation?
(i.e. do we risk losing money from new menu?)
400,2$s000,21$X10n
000,20$X
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Fast Food Business Example
Assess evidence for H+ by:
H+ p-value = Area
10400,2
,000,20' NndistX
000,21$000,20$
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Fast Food Business Example
Computation in EXCEL:
Class Example 22, Part 1:http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg24.xls
P-value = 0.094.
“1 in 10”, “could be random variation”,
“not very strong evidence”