Stat 155, Section 2, Last Time Binomial Distribution –Normal Approximation –Continuity Correction –Proportions (different scale from “counts”) Distribution

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

Reading In Textbook

Approximate Reading for Today’s Material:

Pages 382-396, 400-416

Approximate Reading for Next Class:

Pages 425-428, 431-439

Chapter 6: Statistical Inference

Main Idea:

Form conclusions by

quantifying uncertainty

(will study several approaches,

first is…)

Section 6.1: Confidence Intervals

Background:

The sample mean, , is an “estimate”

of the population mean,

How accurate?

(there is “variability”, how

much?)

X

Confidence Intervals

Recall the Sampling Distribution:

(maybe an approximation)

nNX

,~


Thus understand error as:

How to explain to untrained consumers?

(who don’t know randomness,

distributions, normal curves)

ndistX 'n


Approach: present an interval

With endpoints:

Estimate +- margin of error

I.e.

reflecting variability

How to choose ?

mX

m


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


HW: 6.1

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


E.g. For confidence level 0.95, want

distribution

0.95 = Area

= margin of errorm

X


Computation: Recall NORMINV takes

areas (probs), and returns cutoffs

Issue: NORMINV works with lower areas

Note: lower tail

included


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


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


Approach to unknown :

Recenter, i.e. look at dist’n

Key concept:

Centered at 0

Now can calculate as:

nNORMINVm

,0,975.0

X


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


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

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

http://stat-or.unc.edu/webspace/postscript/marron/Teaching/stor155-2007/Stor155Eg22.xls


An EXCEL shortcut:

CONFIDENCE

Careful: parameter is:

2 tailed outer area

So for level = 0.90, = 0.10


HW: 6.5, 6.9, 6.13, 6.15, 6.19

Choice of Sample Size

Additional use of margin of error idea

Background: distributions

Small n Large n

X

n


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


Given m, how do we find n?

Solve for n (the equation):

n

mn

XPmXP

95.0

nm

ZP


Graphically, find m so that:

Area = 0.95 Area = 0.975

nm

nm


Thus solve:

2

1,0,975.0

NORMINVm

n

1,0,975.0NORMINVn

m

1,0,975.0NORMINVm

n


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


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


Interpretation of Conf. Intervals

2 Equivalent Views:

Distribution Distribution

95%

pic 1 pic 2

m m m 0 m

X X


Mathematically:

pic 1 pic 2

no pic

"",.. bracketsmXmXICtheP

mXPmXmP 95.0

mXmXP


Frequentist View: If repeat the

experiment many times,

About 95% of the time, CI will contain

(and 5% of the time it won’t)


Nice Illustration:

Publisher’s Website

• Statistical Applets

• Confidence Intervals

Shows proper interpretation:

• If repeat drawing the sample

• Interval will cover truth 95% of time

http://bcs.whfreeman.com/ips5e/


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




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”



HW: 6.27, 6.29, 6.31

And now for somethingcompletely different….

A fun dance video:

http://ebaumsworld.com/2006/07/robotdance.html

Suggested by David Moltz

http://ebaumsworld.com/2006/07/robotdance.html

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)

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)

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

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

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

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

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

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)


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$


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


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)


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

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$

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.


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


Assess evidence for H+ by:

H+ p-value = Area

10400,2

,000,20' NndistX

000,21$000,20$


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”


Documents

Stat 155, Section 2, Last Time Binomial Distribution –Normal Approximation –Continuity Correction –Proportions (different scale from “counts”) Distribution