Stats Lecture 08 Hypothesis Testing

8/3/2019 Stats Lecture 08 Hypothesis Testing

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Tests of Hypothesis

For Means and Proportions

Quantitative Methods for Economics

Dr. Katherine Sauer

Metropolitan State College of Denver


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Chapter Overview:

I. Hypothesis Tests for Means

II. Type I and Type II ErrorsII. Hypothesis Tests for Proportions

IV. Hypothesis Test for Difference in Means

V. Hypothesis Test for Difference in Proportions


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I. Hypothesis Tests for Means

A null hypothesis is a claim about a population characteristic.

H0

Ex: A property developer makes a claim that the average annual

rental income per student is at most $5000.

H0: < 5000


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If the null hypothesis is rejected, then the alternative hypothesis

prevails.

The alternative hypothesis is the complement of the null.

H1 HA

Ex: If H0: < 5000, then H1: > 5000

The null hypothesis is usually a statement about the status quo

and its expression always contains the equality.


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Ex: A property developer claims the average rental income per

room in student housing is at most $5000 for the year. A sample of36 students is randomly selected. The mean annual rent is

calculated to be $5200 from this sample. The standard deviation for

all rents is $735.

1. State the null and alternative hypothesis.H0: < 5000

H1: > 5000

Because the alternative hypothesis only refers tovalues greater than 5000, well use a one-sided test

(upper tail).


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2. Decide at what level of probability the null hypothesis will be

rejected.

- the null hypothesis will be rejected if the p-value is less

than a given probability- usually 0.05 or 0.01

For this example, lets use = 0.05.

3. Sketch the curve and identify the critical region.

for = 0.05 Z0.05 = 1.6449

= 5000

Z = 1.6449

Reject

H0

Accept H0 Accept H0 if Z 1.6449


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x

H

x

xZ

0

nx

4. Calculate Z:

5.12236

735

nx

63.1

5.122

50005200

xZ

= 5000

Z = 1.6449

Reject

H0

Accept H0

Z = 1.63

Accept H0 because

1.63


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5. Calculate the p-value.

The p-value is the area under the curve and to the right of our

sample Z.

look up 1.63 in the table

p-value = 0.0516

= 5000

Z = 1.6449

Reject

H0

Accept H0

Z = 1.63

Close to 0.05, but not

close enough.


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room in student housing is at least $5000 for the year. A sample of36 students is randomly selected. The mean annual rent is

calculated to be $5200 from this sample. The standard deviation for

all rents is $735. Use = 0.05

1. State the null and alternative hypothesis.H0: > 5000

H1: < 5000

Because the alternative hypothesis only refers tovalues less than 5000, well use a one-sided test

(lower tail).


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for = 0.05 Z0.05 = -1.6449

= 5000

Z = -1.6449

Reject

H0

Accept H0 Accept H0 if Z >-1.6449

Reject H0 if Z < -1.6449

x

H

x

x

Z

0

nx

3. Calculate Z:

5.12236

735

nx

63.1

5.122

50005200

xZ


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= 5000

Z = -1.6449

Z = 1.63

Accept H0 because

1.63 > -1.6449.

Reject

H0

Accept H0


The p-value is the area under the curve and to the leftof our

sample Z.look up 1.63 in the table

0.0516

10.0516 = 0.9484


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Comparing the two scenarios:

H0: < 5000H1: > 5000

Accept the null hypothesis that the true mean is less than

or equal to $5000 with a p-value of 0.0516.

H0: > 5000

H1: < 5000

Accept the null hypothesis that the true mean is greater

than or equal to $5000 with a p-value of 0.9485.


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room in student housing is $5000 for the year. A sample of 36

students is randomly selected. The mean annual rent is calculated tobe $5200 from this sample. The standard deviation for all rents is

$735. Use = 0.05

1. State the null and alternative hypothesis.H0: = 5000

H1: 5000

Because the alternative hypothesis refers to all

values other than 5000, well use a two-sided test.


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for = 0.05 /2 = 0.025 Z0.025 = 1.96

= 5000

Z = -1.96

Reject

H0

Accept H0 Accept H0 if

-1.96 < Z < 1.96

Reject H0 ifZ < -1.96 or Z > 1.96

Z = 1.96

Reject

H0

x

Hx xZ

0

n

x 3. Calculate Z:

5.12236

735

nx

63.1

5.122

50005200

xZ


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= 5000

Z = -1.96

Reject

H0

Accept H0

Z = 1.96

Reject

H0

Z = 1.63

Accept H0 because-1.96 < 1.63 < 1.96


The p-value is the area under the curve and to the leftof -1.63

and the area under the curve to the rightof 1.63.

look up 1.63 in the table

0.0516

p-value = 0.0516 + 0.0516

p-value = 0.1032


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The Relationship between Confidence Intervals

and Hypothesis Testing

Same example: Lets calculate the 95% confidence interval.

xZx 2/

5200 + 1.96(122.5)5200 + 240.1

4959.9 to 5440.1

We are 95% sure that the population mean is between $4959.9

and $5440.1.

Since the claim is = 5000, our CI supports the claim.


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A two-sided Confidence Interval is a set of acceptable two-

sided null hypotheses at the same level of significance.(same argument for one-sided CIs and hypotheses)


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II. Type I and Type II errors

If a sample mean falls in the acceptance region, we say

there is no significant difference between the

sample mean and the hypothesized population mean.

If the sample mean falls in the rejection region, we say

there is a significant difference between the sample

mean and the hypothesized population mean.

- the difference is too great to attribute to chance


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If you reject the null when it is actually true, it is called a Type I

error.

is the probability of committing a Type I error.

If you reject the alternative hypothesis when it is actually true, it is

called a Type II error.

is the probability of committing a Type II error.


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Thepower of a testis the probability of accepting a true

alternative hypothesis.

The power of a test is also defined as the probability of rejectinga false null hypothesis.

The power of a test increases as increases.


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III. Hypothesis tests for Proportions

Ex: A Budget airline claims that 96% of its flights depart on time. A

researcher working for the company records departure informationfor 80 randomly selected flights and finds that 5 departed late. Test

the airlines claim at the 1% significance level.

1. State the null and alternative hypothesesp = 5/80 = 0.0625 departed late

= 0.96 depart on time so = 0.04 departed late

H0: = 0.04

H1: 0.04

(2-sided test)


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for = 0.01 /2 = 0.005 Z0.005 = 2.5758

=0.04

Z = -2.5758

Reject

H0


-2.5758 < Z < 2.5758

Reject H0 ifZ < -2.5758 or Z > 2.5758

Z = 2.5758

Reject

H0

3. Calculate Z:

0274.10219.0

04.00625.0

pZ

p

p

p

Z

nsp

)1(

0219.080

)04.01(04.0

ps


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=0.04

Z = -2.5758

Reject

H0

Accept H0

Z = 2.5758

Reject

H0

Z = 1.0274

Accept H0 because-2.5758


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Alternative:

| p | = | 0.040.0625| = 0.0225

0.04 + 0.0225 = 0.0625

0.040.0225 = 0.0175

p-value = pr( p > 0.0625) + pr( p < 0.0175)

0.0625 = x / 80 x = 5

0.0175 = x / 80 x = 1.4

If the true percentage of late departures is 4%, there is a 30%

chance of selecting 5 or more late departures or 1.4 or fewer

departures in 80.


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IV. Hypothesis Test for Difference in Means

Ex: A group of students and a group of young professionals are

asked about their yearly rent expenditures.

Is there a difference in the average rental expenditure betweenthe two groups? (5% significance level)

1. State the null and alternative hypotheses

H0: A = B sometimes written as A - B = 0H1: A B(2-sided test)


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for = 0.05 /2 = 0.025 Z0.025 = 1.96

A - B = 0

Z = -1.96

Reject

H0


-1.96 < Z < 1.96

Reject H0 ifZ < -1.96 or Z > 1.96

Z = 1.96

Reject

H0

3. Calculate Z:

2046.20089.127

)0()49205200(

BxAxZ

BxAx

BABA

BxAxs

xx

Z

)()(

B

B

A

ABxAxn

s

n

ss

22

0089.12745

)225(

36

)735(22

BxAxs


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=0.04

Z = -2.5758

Reject

H0

Accept H0

Z = 2.5758

Reject

H0

Z = 2.2046

Accept H0 because-2.57582.20) + pr(Z< -2.20)

= 0.0139 + 0.0139

= 0.0278

There is a 2.78% chance that 2 sample means will differ by

52004920 = 280 or more when the population means

are equal


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V. Hypothesis Test for Difference in Proportions

When testing whether or not two population proportions are

different, combine the two sample proportions into a pool.

pc = # items with given characteristic in both samples

sum of all items in both samples

BA

cccombinedpBpAnn

ppss11

)1(


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Ex: Test the hypothesis that the support for the Green Party is thesame in both areas. Use 10% significance.

1. State the null and alternative hypotheses

H0: A = B

H1: A B

Calculate pc 88+54 = 142 =0.3944

200+160 360


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for = 0.10 /2 = 0.05 Z0.005 = 1.6449

A - B = 0

Z = -1.6449

Reject

H0


-1.6449 < Z < 1.6449

Reject H0 ifZ < -1.6449 or Z > 1.6449

Z = 1.6449

Reject

H0

3. Calculate Z: pBpA

BABA

pBpA s

pp

Z

)()(

BA

ccpBpAnnpps11

)1(

0518.0160

1

200

1)3944.01(3944.0

pBpAs 9788.1

0518.0

)0()3375.044.0(

pBpAZ


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Z = -1.6449

Reject

H0

Accept H0

Z = 1.6449

Reject

H0

Z = 1.98

Reject H0 because1.98>1.6449

4. p-value

p-value = pr(Z>1.98) + pr(Z< -1.98)

= 0.0239 + 0.0239

= 0.0478

When the null hypothesis is rejected, the p-value is the level of

significance.

A - B = 0


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Concepts:

Type I, Type II errors

Power of a test

Skills:Perform Hypothesis Tests for means, proportions, differences in

means, differences in proportions

Documents

Stats Lecture 08 Hypothesis Testing