math 1710 class 22pi.math.cornell.edu/~web1710/slides/oct17_v1.pdf · math 1710 class 22 dr. allen...

Math 1710Class 22

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A SimpleHypothesisTest Question

Logic ofHypothesisTesting

HypothesisTestingVocabulary

Three ways toCarry Out anHT

HT VocabPart II

AlternativeHypotheses

Sample Sizefor a GivenMOE

Examples

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PowerExample

PowerProperties

Math 1710 Class 22

Dr. Allen Back

Oct. 17, 2016

Math 1710Class 22

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Smoke Detectors

CPSC ’96: 90% of American homes have at least one smokedetector. After a public safety campaign, a city observes that376 out of 400 randomly selected homes have a detector.

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Smoke Detectors

CPSC ’96: 90% of American homes have at least one smokedetector. After a public safety campaign, a city observes that376 out of 400 randomly selected homes have a detector.Is this strong evidence the local rate is greater than thenational rate?

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Smoke Detectors

CPSC ’96: 90% of American homes have at least one smokedetector. After a public safety campaign, a city observes that376 out of 400 randomly selected homes have a detector.Is this strong evidence the local rate is greater than thenational rate?

So p̂ =376

400= .94.

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Logic of Hypothesis Testing

Null Hypothesis H0 - retained unless disproven.

Sometimes Preferred Terminology: “fail to reject H0” ratherthan “retain H0” to emphasize that the null is neverestablished.

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Alternative Hypothesis HA- Only thing which possibly can beproven in the HT.

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Alternative Hypothesis HA- Only thing which possibly can beproven in the HT.Procedure:

1 . . .2 . . .3 Consider the sampling distribution of p̂ which

would hold if H0 were true.4 Fail to reject H0 if p̂ is reasonably consistent with

this sampling distribution. Otherwise reject H0.5 . . .

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to emphasize that the null is never established.

Procedure:

1 . . .2 . . .3 Consider the sampling distribution of p̂ which

would hold if H0 were true.4 Fail to reject H0 if p̂ is reasonably consistent with

this sampling distribution. Otherwise reject H0.5 . . .

Three ways to carry out step 4:

1 Calculate a Z-statistic and determine a p-value.2 Calculate a Z-statistic and compare with a critical

value z∗.3 Use a confidence interval.

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Hypothesis Testing Vocabulary

Null Hypothesis H

Alternative Hypothesis H(One Sided vs Two Sided)

SD(p̂) vs. SE (p̂)

Z-statistic

Tail Probability

P-value

Significance Level

Fail to reject Null H (retain H0) vs. reject Null H (equiv.accept Alt H )

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CPSC Example

Let p be the true proportion of smoke detectors homes in ourcity have.

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CPSC Example

Let p be the true proportion of smoke detectors homes in ourcity have.H0 : p = .9, HA : p > .9

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CPSC Example


SD(p̂) =

√.9 · .1400

= .015.

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CPSC Example


SD(p̂) =

√.9 · .1400

= .015.

z-statistic

z =.94− .9.015

= 2.67.

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

z-statistic is the z-score of p̂ on the samp dist centered at thehypothesized value of p

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

z-statistic is the z-score of p̂ on the samp dist centered at thehypothesized value p0 of p

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:Method 1: Tail Prob = P(p̂ > .94) = P(Z > 2.67) = .0038Since the test is 1-sided, P-value=tail probability=.0038.

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:Method 1: Tail Prob = P(p̂ > .94) = P(Z > 2.67) = .0038Since the test is 1-sided, P-value=tail probability=.0038.This is small, so conclusion is we reject H0.

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:Method 2: Say significance level is α = .05.

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:Method 2: Say significance level is α = .05.Since this is a 1-sided test with α = .05, the appropriatecritical value is z∗ = 1.645.

N(0,1)

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CPSC Example

z-statistic

z =.94− .9.015

= 2.67.

Two primary ways to continue:Method 2: Say significance level is α = .05.Since our z-statistic of 2.67 is more extreme than z∗ = 1.645(and supports Ha), we reject Ha at α = .05.

N(0,1)

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CPSC Example

Method 3: Using a CI. Still α = .05.

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CPSC Example

Method 3: Using a CI. Still α = .05.The picture

shows that a 90% CI has the same critical value as an α = .05HT.

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

A 90% CI is

.94± 1.645 · .0119 = .94± .0196 = (.9204, .9596)

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

A 90% CI is

.94± 1.645 · .0119 = .94± .0196 = (.9204, .9596)

Since all the values of p in our CI support Ha, the result of theHT is to reject the null.Had even one been consistent with H0, we would have failed toreject the null.

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

A 90% CI is

.94± 1.645 · .0119 = .94± .0196 = (.9204, .9596)

Because of the extra error associated with SE instead of SD,Method 3 is not arithmetically equivalent to Methods 1 and 2.Method 3 could reject the null while method 1 failed to rejector vice-versa.

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

A 90% CI is

.94± 1.645 · .0119 = .94± .0196 = (.9204, .9596)

Because of the extra error associated with SE instead of SD,Method 3 is not arithmetically equivalent to Methods 1 and 2.Method 3 could reject the null while method 1 failed to rejector vice-versa.This small discrepancy is not of practical importance. It couldbe corrected by a more elaborate CI formula.

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CPSC Example


SE (p̂) =

√.94 · .06

400= .0119

A 90% CI is

.94± 1.645 · .0119 = .94± .0196 = (.9204, .9596)

Because of the extra error associated with SE instead of SD,Method 3 is not arithmetically equivalent to Methods 1 and 2.Method 3 could reject the null while method 1 failed to rejector vice-versa.This small discrepancy is not of practical importance. It couldbe corrected by a more elaborate CI formula.The general principle that many HT’s can be done via a CI isan important fact.

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Hypothesis Testing Vocabulary

Null Hypothesis H

Alternative Hypothesis H(One Sided vs Two Sided)

SD(p̂) vs. SE (p̂)

Z-statistic

Tail Probability

P-value

Significance Level

Fail to reject Null H vs. Accept Alt H (equiv. reject NullH )

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HT Vocabulary Details Part II

z-statistic - the z-score of p̂ in relation to the samplingdistribution of p̂ assuming H0 is true.

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z =p̂ − p0

SD(p̂)

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z =p̂ − p0

SD(p̂)

“. . . assuming H0 is true” is why we use SD(p̂) rather thanSE (p̂) in calculating our z-statistic.

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Tail Probability - The probability of a p̂ as extreme or more soand on the same side of p0 (assuming p̂ supports thealternative hypothesis) than the p̂ we actually observed.This is again assuming H0 is true.

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Tail Probability - The probability of a p̂ as extreme or more soand on the same side of p0 (assuming p̂ supports thealternative hypothesis) than the p̂ we actually observed.This is again assuming H0 is true.

(Tail probability shaded in gray.)

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P-value: The probability of a z-statistic as extreme as the givenone or more so (and in support of the alternative hypothesis)under the assumption that the null hypothesis is true.

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P-value: The probability of a z-statistic as extreme as the givenone or more so (and in support of the alternative hypothesis)under the assumption that the null hypothesis is true.So P-value =

(single) tail probability in 1-sided test casetwice the (single) tail probability in 2-sided test case.

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Ha : p > p0 P-value=tail probability

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Ha : p 6= p0 P-value=2(tail probability)

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P-value: The probability of a z-statistic as extreme as the givenone or more so (and in support of the alternative hypothesis)under the assumption that the null hypothesis is true.Significance Level α: The cutoff between ”small” and ”big”P-values.

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P-value: The probability of a z-statistic as extreme as the givenone or more so (and in support of the alternative hypothesis)under the assumption that the null hypothesis is true.Small P-value means reject the null.Big P-value means retain the null.

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Three Kinds of H ′as

Ha for a prop. HT can be any of the three possibilities

Two-Sided p 6= p0

One-Sided p > p0

One-Sided p < p0

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Two-Sided p 6= p0

One-Sided p > p0

One-Sided p < p0

One chooses among these based on the question being studied.

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Two-Sided p 6= p0

One-Sided p > p0

One-Sided p < p0


A question like “Is there strong evidence that p has changed. . . ” would point to 2-sided.

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Two-Sided p 6= p0

One-Sided p > p0

One-Sided p < p0


A question like “Is there strong evidence that p has increased. . . ” would point to 1-sided.

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Two-Sided p 6= p0

One-Sided p > p0

One-Sided p < p0


The value of p̂ never plays a role in formulating hypotheses!

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N(0,1) and the critical value z∗

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N(0,1)

Ha : p 6= p0 (2-sided), α = .10

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N(0,1)

Ha : p > p0 (1-sided), α = .05

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N(0,1)

Ha : p < p0 (1-sided), α = .05

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Sample Size for a Given MOE

We wish to produce a 95% CI with a margin of error of .1%.What sample size should we use?

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MOE = z∗√

p̂q̂

n

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MOE = z∗√

p̂q̂

n

Margin of error picture.

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MOE = z∗√

p̂q̂

n

Algebra gives

n =

(z∗

MOE

)2

p̂q̂

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We wish to produce a 95% CI with a margin of error of .1%.What sample size should we use?Algebra gives

n =

(z∗

MOE

)2

p̂q̂

Here we have

n =

(1.96

.001

)2

p̂q̂ = 19602p̂q̂.

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n =

(z∗

MOE

)2

p̂q̂

Here we have

n =

(1.96

.001

)2

p̂q̂ = 19602p̂q̂.

The challenge is we don’t know p̂ until we conduct the sample.

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n =

(z∗

MOE

)2

p̂q̂

Here we have

n =

(1.96

.001

)2

p̂q̂ = 19602p̂q̂.

Three Approaches:

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n =

(z∗

MOE

)2

p̂q̂

Here we have

n =

(1.96

.001

)2

p̂q̂ = 19602p̂q̂.

Three Approaches:1. Conservative: Just use p̂ = .5 since that gives the biggestpossible value of p̂q̂.

(Think about the graph of y = x(1− x).)

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We wish to produce a 95% CI with a margin of error of .1%.What sample size should we use?1. Conservative: Just use p̂ = .5 since that gives the biggestpossible value of p̂q̂.

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We wish to produce a 95% CI with a margin of error of .1%.What sample size should we use?Here we have

n =

(1.96

.001

)2

p̂q̂ = 19602p̂q̂.

Three Approaches:1. Conservative: Just use p̂ = .5 since that gives the biggestpossible value of p̂q̂.Here we get

n = 19602(.5)(.5) = 9802 = 960, 400

which explains why surveys DO NOT seek an MOE of .1%.

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Three Approaches:1. Conservative: Just use p̂ = .5 since that gives the biggestpossible value of p̂q̂.Let’s shift to producing a 95% CI with a margin of error of 2%.What sample size should we use?

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.

The conservative estimate here is n = 9604 · .25 = 2401.

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Three Approaches:Let’s shift to producing a 95% CI with a margin of error of 2%.What sample size should we use?

n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.

2. Use an estimate of p̂:

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.

2. Use an estimate of p̂:e.g. if we expect p̂ ∼ .2, we might use this value leading to anestimate n = 9604 · .2 · .8 = 1537. But no guarantees.

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.

3. If you expect p̂ to be in a range, use the most demanding p̂within that range.As long as your expectation is met, this is guaranteed to work.

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.


For example if we expect .05 ≤ p̂ ≤ .2, use p̂ = .2 againleading to 1537.

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n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.


How about if we expect .2 ≤ p̂ ≤ .7?

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Let’s shift to producing a 95% CI with a margin of error of 2%.What sample size should we use?

n =

(1.96

.02

)2

p̂q̂ = 982p̂q̂ = 9604p̂q̂.


How about if we expect .2 ≤ p̂ ≤ .7?The most demanding is p̂ = .5, so we use that and again arriveat 2401.

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Note the slow growth: A 100 fold increase in sample sizereduces the MOE by only a factor of 10.

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Smoking 2nd Edition Ch. 20 # 15

National data in the 1960s showed that about 44% of the adultpopulation had never smoked cigarettes. In 1995 a nationalhealth survey interviewed a random sample of 881 adults andfound that 52% had never been smokers.

(a) Create a 95% CI for the proportion of adults (in 1995)who had never been smokers.

(b) Does this provide evidence of a change in behavioramong Americans? Using your CI, test an appropriatehypothesis and state your conclusion.

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Notation: Let p denote the proportion of Americans in 1995who had never smoked.

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Hypotheses:

H0: p = .44

Ha: p 6= .44

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Random Sampling: Stated.

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10% Condition: Much less than the national population.

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Success/Failure:

881 · .52 ≥ 10

881 · .48 ≥ 10

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Women Executives Ch. 20 #25

A company is criticized because only 13 of 43 people inexecutive-level positions are women. The company explainsthat although this proportion is lower than it might wish, it’snot surprising given that only 40% of all its employees arewomen. What do you think? Test an appropriate hypothesisand state your conclusion.

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Notation: Let p denote the proportion of executives (incompanies like this one, perhaps) who are women.

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Hypotheses: If potentially proving the company is wrong:

H0: p = .4 (or p ≥ .4)

Ha: p < .4

If potentially proving the company is right:

H0: p = .4 (or p ≤ .4)

Ha: p > .4

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Random Sampling: Hopefully representative of a much largerpopulation.

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10% Condition: Depends on definition of the population.Hopefully much less than 10% of population.

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Success/Failure:

43

(13

43

)≥ 10

43

(30

43

)≥ 10

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Dropouts Ch. 20 #27

Some people are concerned that new tougher standards andhigh-stakes tests adopted in many states have driven up thehigh school dropout rate. The National Center for EducationStatistics reported that the high school dropout rate for theyear 2004 was 10.3%. One school district whose dropout ratehas always been very close to the national average reports that210 of their 1782 high school students dropped out last year. Isthis evidence that their dropout rate may be increasing?Explain.

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Dropouts Ch. 20 #27

Notation: Let p denote the proportion of students in districtslike this one who drop out.

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Dropouts Ch. 20 #27

Hypotheses:

H0: p = 10.3 (or p ≤ 10.3)

Ha: p > 10.3

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Dropouts Ch. 20 #27

Random Sampling: Hopefully representative of a much largerpopulation.

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Dropouts Ch. 20 #27

10% Condition: Depends on definition of the population.Hopefully much less than 10% of population.Certainly much less than the national population.

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Dropouts Ch. 20 #27

Success/Failure:

1782

(210

1782

)≥ 10

1782

(1572

1782

)≥ 10

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PowerProperties

Lost Luggage Ch. 20 #29

An airline’s public relations department says that the airlinerarely loses passengers’ luggage. It further claims that on thoseoccasions when luggage is lost, 90% is recovered and deliveredto its owner within 24 hours. A consumer group that surveyeda large number of air travelers found that only 103 of 122people who lost luggage on that airline were reunited with themissing items by the next day. Does this cast doubt on theairline’s claim? Explain.

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Notation: Let p denote the proportion of lost luggage that isreturned within 24 hours.

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Hypotheses:

H0: p = .9 (or p ≥ .9)

Ha: p < .9

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Random Sampling: Hopefully.

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10% Condition: Depends on definition of the population.Hopefully much less than 10% of population.Certainly much less than total volume of luggage.

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Success/Failure:

122

(103

122

)≥ 10

122

(19

122

)≥ 10

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Power, Type I/II Errors, α, and β

Given H0 : p = p0, there are two ways an HT can report aninaccurate result:

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Given H0 : p = p0, there are two ways an HT can report aninaccurate result:

H0 true H0 false

Retain H0 Good Type II Errorprobability = βdepends on value of p

Reject H0 Type I Error Goodprobability = α probability = 1-β

= power

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Given H0 : p = p0, there are two ways an HT can report aninaccurate result:Type I Error Examples:

a: False Positive in a diagnosis; i.e. deciding aperson is sick when they really are not. (H0 : Theperson is well.)

b: Convicting an innocent person. (H0 : The personis innocent.)

c: Producer Risk; the chance that a good goodshipment erroneously fails a a test for quality.(H0 : A product meets a specification.)

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Given H0 : p = p0, there are two ways an HT can report aninaccurate result:Type I Error Examples:

a: False Positive in a diagnosis; i.e. deciding aperson is sick when they really are not. (H0 : Theperson is well.)

b: Convicting an innocent person. (H0 : The personis innocent.)

c: Producer Risk; the chance that a good goodshipment erroneously fails a a test for quality.(H0 : A product meets a specification.)

Type II Error Examples:

a: False Negative; missing a sick person.b: Letting a guilty person go free.c: Consumer Risk; the chance that a bad shipment

erroneously passes a a test for quality.

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40% of employees are women.Woman under-represented as executives?What would it take in p̂ for the company to prove that womenare as well represented among executives?

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H0 : p = .4

Ha : p > .4

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H0 : p = .4

Ha : p > .4

One could also do a HT to see if a given p̂ demonstrateswomen are under represented. Then Ha would be p < .4.

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H0 : p = .4

Ha : p > .4

n = 420,SD(p̂) = .0239.

z =p̂ − .4.0239

z > z∗ means p̂ > .0239z∗ + .4.α = .05⇒ z∗ = 1.645; rejection means p̂ > .439.α = .01⇒ z∗ = 2.326; rejection means p̂ > .456.

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How a p̂ will be dealt with in the HT

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How does the power depend on the effect size?

i.e. on the actual value of p

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Sampling Dist of p̂ when effect size is large.

Power nearly 1

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Sampling Dist of p̂ when effect size is small.

Power nearly α

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Sampling Dist of p̂ when effect size is middle size.

Power in middle between 0 and 1; you could calulate it, but wewon’t ask you to.

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How does the power depend on the effect size?α = .05 case:

p β power

.4001 .95 .05

.41 .89 .11

.45 .33 .67

.5 .01 .99

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How does the power depend on the effect size?α = .01 case:

p β power

.4001 .99 .01

.41 .97 .03

.45 .59 .41

.5 .04 .96

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Power vs. alternative value of p in 2-sided case.

A Power Curve

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Prob of type I error = α .

(H0 1-sides, α = .05 below)

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Power = 1 - β always.

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Power depends on effect size. (i.e actual alternative value forp.)

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Power is approx. α when actual p is near p0 but not exactly p0.

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Power goes to 1 as effect size grows, assuming in the 1 sidedcase that the alternative value supports Ha.

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Power increases as sample size increases.

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Increasing α decreases β . (Easier to reject H0.)

math 1710 class 22pi.math.cornell.edu/~web1710/slides/oct17_v1.pdf · math 1710 class 22 dr. allen...

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