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Chapters 20, 21 Testing Hypotheses about Proportions

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Chapters 20, 21 Testing Hypotheses about Proportions Part II: Significance Levels, Type I and Type II Errors, Power. 1. Alpha Levels: a Threshold for the P-value. Sometimes we need to make a firm decision about whether or not to reject the null hypothesis. - PowerPoint PPT Presentation
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Chapters 20, 21 Testing Hypotheses about Proportions Part II: Significance Levels, Type I and Type II Errors, Power 1
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Page 1: Chapters 20, 21 Testing Hypotheses about Proportions

Chapters 20, 21Testing Hypothesesabout Proportions

Part II: Significance Levels, Type I and Type II Errors,

Power

1

Page 2: Chapters 20, 21 Testing Hypotheses about Proportions

- 2

Alpha Levels: a Threshold for the P-value. Sometimes we need to make a firm decision about

whether or not to reject the null hypothesis. When the P-value is small, it tells us that our data

are rare if the null hypothesis is true. How rare is “rare”?

Page 3: Chapters 20, 21 Testing Hypotheses about Proportions

Alpha Levels (cont.) We can define “rare event” arbitrarily by setting a

threshold for our P-value. ◦ If our P-value falls below that threshold, we’ll reject H0.

We call such results statistically significant.◦ The threshold is called an alpha level, denoted by α.

Page 4: Chapters 20, 21 Testing Hypotheses about Proportions

Alpha Levels (cont.)

Common alpha levels are 0.10, 0.05, and 0.01. ◦ You have the option—almost the obligation—to consider

your alpha level carefully and choose an appropriate one for the situation.

The alpha level is also called the significance level. ◦ When we reject the null hypothesis, we say that the test is

“significant at that level.” Rejection Region (RR): values of the test statistic z

that lead to rejection of the null hypothesis H0.

Page 5: Chapters 20, 21 Testing Hypotheses about Proportions

5

a Levels andRejection Regions 1-Tail

If HA: p > p0 and =.10then RR={z: z > 1.28}

If HA: p > p0 and =.05 then RR={z: z > 1.645}

If HA: p > p0 and =.01then RR={z: z > 2.33}

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 0 1.6452.33 4

Chart Title

Z

RR: z > 1.645

area = =.05

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 0 1.281.6452.33 4

Chart Title

Z

RR: z > 1.28area = =.10

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 0 1.6452.33 4

Chart Title

Z

RR: z > 2.33area = =.01

Rej Region

.10 Z > 1.28

.05 Z > 1.645

.01 Z > 2.33

0 0

0

0 0

:

ˆ

(1 )

H p p

p pz

p pn

Page 6: Chapters 20, 21 Testing Hypotheses about Proportions

Medication side effects (hypothesis test for p)Arthritis is a painful, chronic inflammation of the joints.

An experiment on the side effects of the pain reliever ibuprofen

examined arthritis patients to find the proportion of patients who suffer

side effects. If more than 3% of users suffer side effects, the FDA will

put a stronger warning label on packages of ibuprofenSerious side effects (seek medical attention immediately):

Allergic reaction (difficulty breathing, swelling, or hives),Muscle cramps, numbness, or tingling,Ulcers (open sores) in the mouth,Rapid weight gain (fluid retention),Seizures,Black, bloody, or tarry stools,Blood in your urine or vomit,Decreased hearing or ringing in the ears,Jaundice (yellowing of the skin or eyes), orAbdominal cramping, indigestion, or heartburn,

Less serious side effects (discuss with your doctor):Dizziness or headache,Nausea, gaseousness, diarrhea, or constipation,Depression,Fatigue or weakness,Dry mouth, orIrregular menstrual periods

What are some side effects of ibuprofen?

Page 7: Chapters 20, 21 Testing Hypotheses about Proportions

Test statistic:

H0 : p =.03 HA : p > .03 where p is the proportion of

Ibuprofen users who suffer side effects. Let = .05

0ˆ .0523 .032.75

ˆ( ) .0081

p pzSD p

23ˆ 0.0523

440p

Conclusion: since the test statistic value 2.75 is

in the RR, reject H0 : p =.03;

there is sufficient evidence to conclude that the

proportion of ibuprofen users who suffer side

effects is greater than .03

Note that ( 2.75) .0030

is less than =.05

P value P z

440 subjects with chronic arthritis were given ibuprofen for pain relief; 23 subjects suffered from adverse side effects.

(.03)(.97)ˆ( ) .0081

440SD p

Rejection Region: z > 1.645

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 0 1.6452.33 4

Chart Title

Z

RR: z > 1.645

area = =.05

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 -2.58-2.33-1.645 0 1.6452.332.582.75 4

Chart Title

Z

P-value=P(z>2.75) = . 003

Page 8: Chapters 20, 21 Testing Hypotheses about Proportions

8

a Levels andRejection Regions 1-Tail

If HA: p < p0 and =.10then RR={z: z < -1.28}

If HA: p < p0 and =.05 then RR={z: z < -1.645}

If HA: p < p0 and =.01then RR={z: z < -2.33}

Rej Region

.10 Z < -1.28

.05 Z < -1.645

.01 Z < -2.33

0 0

0

0 0

:

ˆ

(1 )

H p p

p pz

p pn

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -1.28 0 1.6452.33 4

Chart Title

Z

RR: z < -1.28area = =.10

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 -1.645 0 1.6452.33 4

Chart Title

Z

RR: z < -1.645

area = =.05

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 -2.33-1.645 0 1.6452.33 4

Chart Title

Z

RR: z < -2.33area = =.01

Page 9: Chapters 20, 21 Testing Hypotheses about Proportions

Rejection region for a 2-tail test of p with α = 0.05

A 2-tailed test means that area α/2 is in each tail, thus:

-A middle area of 1 − α = .95, and tail areas of α /2 = 0.025.

RR={z < -1.96, z > 1.96}

From Z Table

0.0250.025

0 0 0

0

0 0

: , :

ˆ

(1 )

AH p p H p p

p pz

p pn

Page 10: Chapters 20, 21 Testing Hypotheses about Proportions

10

a Levels and Rejection Regions 2-Tail

If HA: p p0 and =.10, thenRR={z: z < -1.645, z>1.645}

If HA: p p0 and =.05, then RR={z: z < -1.96, z > 1.96}

If HA: p p0 and =.01, thenRR={z: z < -2.58, z > 2.58}

Rejection Region

.10 Z < -1.645, Z > 1.645

.05 Z < -1.96, Z > 1.96

.01 Z < -2.58, Z > 2.58

0 0

0

0 0

:

ˆ

(1 )

H p p

p pz

p pn

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -2.33-1.645 0 1.6452.33 4

Chart Title

Z

RR: z < -1.645, z > 1.645

area = =.05 + .05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -2.33-1.96-1.645 0 1.6451.962.33 4

Chart Title

Z

RR: z < -1.96, z > 1.96area = =.025 + .025

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -2.58-2.33-1.645 0 1.6452.332.58 4

Chart Title

Z

RR: z < -2.58, z > 2.58area = =.005 + .005

Page 11: Chapters 20, 21 Testing Hypotheses about Proportions

Chap 9-11

Example: one-proportion z test

A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Perform a 2-sided hypothesis test to evaluate the company’s claim. Use = .05

Check:

n p = (500)(.08) = 40

n(1-p) = (500)(.92) = 460

Page 12: Chapters 20, 21 Testing Hypotheses about Proportions

SolutionH0: p = .08

HA: p ¹ .08

= .05Test Statistic:

z0

.0068.0068

2.47

Conclusion: since the test statistic valuez = -2.47 is in the rejection region, we reject the company’s claim of 8% response rate.

0

0 0

ˆ .05 .08z 2.47

(1 ) .08(1 .08)500

p p

p pn

-2.47

25ˆ500, 25, .05

500n x p

Note that the :2 ( | 2.47 |)2(1 .9932) .0136is less than =.05

P valueP z

00.05

0.10.15

0.20.25

0.30.35

0.40.45

-4 -2.33-1.96-1.645 0 1.6451.962.33 4

Chart Title

Z

RR: z < -1.96, z > 1.96

area = =.025 + .025

Rejection region:

1.96, 1.96z z

Page 13: Chapters 20, 21 Testing Hypotheses about Proportions

Levels and Rejection Regions HA: p > p0 HA: p < p0

HA: p ≠ p0

Rejection Region

.01 Z > 2.33

.05 Z > 1.645

.10 Z > 1.28

Rejection Region

.01 Z < -2.33

.05 Z < -1.645

.10 Z < -1.28

Rejection Region

.01 Z < -2.58, Z > 2.58

.02 Z < -2.33, Z > 2.33

.05 Z < -1.96, Z > 1.96

.10 Z < -1.645, Z > 1.645

Page 14: Chapters 20, 21 Testing Hypotheses about Proportions

Alpha Levels (cont.) What can you say if the P-value does not fall below α? ◦ You should say that “The data have failed to provide

sufficient evidence to reject the null hypothesis.” ◦ Don’t say that you “accept the null hypothesis.”

Page 15: Chapters 20, 21 Testing Hypotheses about Proportions

Alpha Levels (cont.) Recall that, in a jury trial, if we do not find the

defendant guilty, we say the defendant is “not guilty”—we don’t say that the defendant is “innocent.”

Page 16: Chapters 20, 21 Testing Hypotheses about Proportions

Alpha Levels (cont.)

The P-value gives the reader far more information than just stating that you reject or fail to reject the null.

In fact, by providing a P-value to the reader, you allow that person to make his or her own decisions about the test. ◦ What you consider to be statistically significant might not

be the same as what someone else considers statistically significant.

◦ There is more than one alpha level that can be used, but each test will give only one P-value.

Page 17: Chapters 20, 21 Testing Hypotheses about Proportions

Confidence Intervals and Hypothesis Tests

Because confidence intervals are two-sided, they correspond to two-sided (two-tailed) hypothesis tests.

In general, a confidence interval with a confidence level of C% corresponds to a two-sided hypothesis test with an α-level of 100 – C%. For example:◦ If a 2-sided hypothesis test at level .05 rejects H0 , then the

null hypothesized value of p will not be in a 95% confidence interval calculated from the same data.

◦ A 95% confidence interval shows the values of null hypothesis values p for which a 2-sided hypothesis test at level .05 will NOT reject the null hypothesis.

Page 18: Chapters 20, 21 Testing Hypotheses about Proportions

Chap 9-18

Confidence Intervals and Hypothesis Tests: Example

A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Calculate a 95% confidence interval to estimate the company’s response rate.

Check:

n p = (500)(.08) = 40

n(1-p) = (500)(.92) = 460

Page 19: Chapters 20, 21 Testing Hypotheses about Proportions

Solution

Recall that we rejected H0: p=.08 for the hypothesis

test H0: p = .08 HA: p ¹ .08 with = .05

25ˆ500, 25, .05

500n x p

95% confidenceinterval:

ˆ ˆ(1 ) .05(.95)ˆ 1.96 .05 1.96

500.05 .019 (.031,.069)

p pp

n

The 95% confidence interval (.031, .069) gives the values of p0 for which a 2-tailed hypothesis testH0: p = p0 , HA: p ¹ p0 at =.05 will NOT reject H0: p = p0

Page 20: Chapters 20, 21 Testing Hypotheses about Proportions

Making Errors Here’s some shocking news for you: nobody’s

perfect. Even with lots of evidence we can still make the wrong decision.

When we perform a hypothesis test, we can make mistakes in two ways:

I. The null hypothesis is true, but we mistakenly reject it. (Type I error)

II. The null hypothesis is false, but we fail to reject it. (Type II error)

Page 21: Chapters 20, 21 Testing Hypotheses about Proportions

Making Errors (cont.) Which type of error is more serious depends on the

situation at hand. In other words, the gravity of the error is context dependent.

Here’s an illustration of the four situations in a hypothesis test:

Page 22: Chapters 20, 21 Testing Hypotheses about Proportions

Making Errors (cont.) How often will a Type I error occur?

◦ Since a Type I error is rejecting a true null hypothesis, the probability of a Type I error is our α level.

When H0 is false and we reject it, we have done the right thing.◦ A test’s ability to detect a false hypothesis is called the

power of the test.

Page 23: Chapters 20, 21 Testing Hypotheses about Proportions

Making Errors (cont.) When H0 is false and we fail to reject it, we have

made a Type II error.◦ We assign the letter β to the probability of this mistake.◦ It’s harder to assess the value of β because we don’t know

what the value of the parameter really is.◦ There is no single value for β--we can think of a whole

collection of β’s, one for each incorrect parameter value.

Page 24: Chapters 20, 21 Testing Hypotheses about Proportions

Making Errors (cont.) One way to focus our attention on a particular β is

to think about the effect size. ◦ Ask “How big a difference would matter?”

We could reduce β for all alternative parameter values by increasing α.◦ This would reduce β but increase the chance of a Type I

error.◦ This tension between Type I and Type II errors is

inevitable. The only way to reduce both types of errors is to

collect more data. Otherwise, we just wind up trading off one kind of error against the other.

Page 25: Chapters 20, 21 Testing Hypotheses about Proportions

Example The proportion of NCSU undergraduates with student

loans historically has been approximately 35%. The director of financial aid thinks this percentage has increased recently because of tuition increases. A random sample of 225 students results in 81 that have student loans.

Perform a hypothesis test to determine if the percentage of NCSU undergraduates with student loans has increased. Use =.05

Page 26: Chapters 20, 21 Testing Hypotheses about Proportions

Solution0 : .35, .05

: .35, where is the proportion of NCSU undergrads

with student loans.81 .36 .35ˆ .36; : 1.645; : .314

225 .35(.65)225

A

H p

H p p

p z z

test statisticRR

Conclusion: since the value of the test statistic is not in the rejection region, we DO NOT rejectH0: p=.35. There is insufficient evidence to conclude that the proportion of NCSU undergrads with student loans has increased.

What type of error might we be making? Type II

What is the probability we are making a type I error? 0

Page 27: Chapters 20, 21 Testing Hypotheses about Proportions

Solution (cont.) What is the probability of a type II error when the true

value of p is .37?

.35(.65)

225ˆ( .402).

Rejection region: 1.645

ˆ .35ˆRejection region in terms of : ( 1.645) 1.645.35(.65)

225

ˆ .35 1.645 P p

z

pp P z P

P p

ˆ .37 .402 .37ˆ(.37) ( .402 when .37) ( .994) 0.84

.37(.63) .37(.63)225 225

pP p p P P z

Page 28: Chapters 20, 21 Testing Hypotheses about Proportions

Power The power of a test is the probability that it

correctly rejects a false null hypothesis. When the power is high, we can be confident that

we’ve looked hard enough at the situation. The power of a test is 1 – β ; because β is the

probability that a test fails to reject a false null hypothesis and power is the probability that it does reject.

Page 29: Chapters 20, 21 Testing Hypotheses about Proportions

Power (cont.) Whenever a study fails to reject its null hypothesis,

the test’s power comes into question. When we calculate power, we imagine that the null

hypothesis is false. The value of the power depends on how far the truth

lies from the null hypothesis value.◦ The distance between the null hypothesis value, p0, and the

truth, p, is called the effect size.◦ Power depends directly on effect size.

Page 30: Chapters 20, 21 Testing Hypotheses about Proportions

A Picture Worth Words The larger the effect size, the easier it should be to

see it. Obtaining a larger sample size decreases the

probability of a Type II error, so it increases the power.

It also makes sense that the more we’re willing to accept a Type I error, the less likely we will be to make a Type II error.

1

P(z 3.09)

Page 31: Chapters 20, 21 Testing Hypotheses about Proportions

A Picture is Worth Words (cont.)

This diagram shows the relationship between these concepts:

1

P(z 3.09)

Page 32: Chapters 20, 21 Testing Hypotheses about Proportions

Reducing Both Type I and Type II Error The previous figure seems to show that if we reduce

Type I error, we must automatically increase Type II error.

But, we can reduce both types of error by making both curves narrower.

How do we make the curves narrower? Increase the sample size.

Page 33: Chapters 20, 21 Testing Hypotheses about Proportions

Reducing Both Type I and Type II Error (cont.) This figure has means that are just as far apart as in

the previous figure, but the sample sizes are larger, the standard deviations are smaller, and the error rates are reduced:

Page 34: Chapters 20, 21 Testing Hypotheses about Proportions

Slide 21 -

34

Reducing Both Type Iand Type II Error (cont.)

Original comparison of errors:

Comparison of errors with a larger sample size:

Page 35: Chapters 20, 21 Testing Hypotheses about Proportions

Sample size n for which -level test also satisfies (p1)= 1-tailed test:

2-tailed test (an approximate solution):

where

0 0 1 1

1 0

2(1 ) (1 )z p p z p p

p pn

/2 0 0 1 1

1 0

2(1 ) (1 )z p p z p p

p pn

( )P z z


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