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Hypothesis Testing

Chapter 9BA 201

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Hypothesis Testing

The null hypothesis, denoted by H0 , is a tentative assumption about a population parameter. The alternative hypothesis, denoted by Ha, is the opposite of what is stated in the null hypothesis.

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One-tailed(lower-tail)

One-tailed(upper-tail)

Two-tailed

0 0: H

0: aH 0 0: H

0: aH 0 0: H

0: aH

Summary of Forms for Null and Alternative Hypotheses about a

Population Mean Three Scenarios:

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Type I and Type II Errors

CorrectDecision

Type II Error

CorrectDecisionType I Error

Reject H0

(Conclude m > 12)

Accept H0

(Conclude m < 12)

H0 True(m < 12)

H0 False(m > 12)Conclusion

Population Condition

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Two Scenarios

Known z Score Unknown t Value

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HYPOTHESIS TESTING KNOWN

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p-Value Approach

Reject H0 if the p-value < .

The p-value is the probability, computed using the test statistic, that measures the support (or lack of support) provided by the sample for the null hypothesis.

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Rejection Rule: p -Value Approach

H0: Reject H0 if z > z

Reject H0 if z < -z

Reject H0 if z < -z or z > z

H0:

H0:

Tests About a Population Mean:s Known

Rejection Rule: Critical Value Approach

Reject H0 if p–value < a

Ha:

Ha: >

Ha:

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00

zz

-za = -1.28 -za = -1.28

a = 0.10a = 0.10

p-value 072p-value 072

z =-1.46 z =-1.46

Lower-Tailed Test About a Population Mean

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00

zz

za = 1.75 za = 1.75

a = 0.04a = 0.04

p-Value 011p-Value 011

z =2.29 z =2.29

Upper-Tailed Test About a Population Mean

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Two-Tailed Tests About a Population Mean

00zz

a/2 = .015

-za/2 = -2.17

a/2 = .015

za/2 = 2.17z = 2.74

1/2p -value= .0031

z = -2.74

1/2p -value= .0031

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z Statistics for s Known

z xn

0

/Where

0 is the mean under H0

is the population standard deviation

n is the sample size

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Metro EMS

The EMS director wants to perform a hypothesistest, with a .05 level of significance, to determinewhether the service goal of 12 minutes or less is being achieved.

The response times for a random sample of 40medical emergencies were tabulated. The samplemean is 13.25 minutes. The population standarddeviation is believed to be 3.2 minutes.

One-Tailed Tests About a Population Mean:s Known

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1. Develop the hypotheses.

2. Specify the level of significance. a = 0.05

H0: Ha:

p -Value and Critical Value Approaches

One-Tailed Tests About a Population Mean:s Known

3. Compute the value of the test statistic.

13.25 12 2.47

/ 3.2/ 40x

zn

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5. Determine whether to reject H0.

p –Value Approach

One-Tailed Tests About a Population Mean:s Known

4. Compute the p –value.

For z = 2.47, cumulative probability = 0.9932.

p–value = 1 - 0.9932 = 0.0068

Because p–value = 0.0068 < a = .05, we reject H0.

There is sufficient statistical evidence

to infer that Metro EMS is not meeting

the response goal of 12 minutes.

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p –Value Approach p –Value Approach

0

z

za =1.645

a = .05

p-value

z =2.47

One-Tailed Tests About a Population Mean:s Known

Samplingdistribution

of

Samplingdistribution

of z xn

0

/

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5. Determine whether to reject H0.

There is sufficient statistical evidence

to infer that Metro EMS is not meeting

the response goal of 12 minutes.

Because 2.47 > 1.645, we reject H0.

Critical Value Approach

One-Tailed Tests About a Population Mean:s Known

For a = 0.05, z0.05 = 1.645

4. Determine the critical value and rejection rule.

Reject H0 if z > 1.645

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p-Value Approach toTwo-Tailed Hypothesis Testing

Reject H0 if the p-value < .

Compute the p-value using the following three steps:

3. Double the tail area obtained in step 2 to obtain the p–value.

2. If z is in the upper tail (z > 0), find the area under the standard normal curve to the right of z.

If z is in the lower tail (z < 0), find the area under the standard normal curve to the left of z.

1. Compute the value of the test statistic z.

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Glow Toothpaste

Quality assurance procedures call for thecontinuation of the filling process if the sampleresults are consistent with the assumption that themean filling weight for the population of toothpastetubes is 6 oz.; otherwise the process will be adjusted.

The production line for Glow toothpaste isdesigned to fill tubes with a mean weight of 6 oz.Periodically, a sample of 30 tubes will be selected inorder to check the filling process.

Two-Tailed Tests About a Population Mean:s Known

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Perform a hypothesis test, at the 0.03 level ofsignificance, to help determine whether the fillingprocess should continue operating or be stopped andcorrected.

Assume that a sample of 30 toothpaste tubesprovides a sample mean of 6.1 oz. The populationstandard deviation is believed to be 0.2 oz.

Two-Tailed Tests About a Population Mean:s Known

Glow Toothpaste

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1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

a = .03

p –Value and Critical Value Approaches

H0: = 6Ha:

6

Two-Tailed Tests About a Population Mean:s Known

0 6.1 6 2.74

/ .2/ 30x

zn

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Two-Tailed Tests About a Population Mean:s Known

5. Determine whether to reject H0.

p –Value Approach

4. Compute the p –value.

For z = 2.74, cumulative probability = 0.9969

p–value = 2(1 - 0.9969) = 0.0062

Because p–value = 0.0062 < a = 0.03, we reject H0.

There is sufficient statistical evidence toinfer that the alternative hypothesis is

true (i.e. the mean filling weight is not 6

ounces).

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Two-Tailed Tests About a Population Mean

00zz

a/2 = .015

-za/2 = -2.17

a/2 = .015

za/2 = 2.17z = 2.74

1/2p -value= .0031

z = -2.74

1/2p -value= .0031

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Critical Value Approach

Two-Tailed Tests About a Population Mean:s Known

5. Determine whether to reject H0.

There is sufficient statistical evidence toinfer that the alternative hypothesis is

true (i.e. the mean filling weight is not 6

ounces).

Because 2.74 > 2.17, we reject H0.

For a/2 = .03/2 = .015, z.015 = 2.17

4. Determine the critical value and rejection rule.

Reject H0 if z < -2.17 or z > 2.17

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0

Do Not Reject H0

z

Reject H0

-2.17 2.17

Reject H0

Critical Value Approach

Two-Tailed Tests About a Population Mean:s Known

z = 2.74

1/2p -value= .0031

z = -2.74

1/2p -value= .0031

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PRACTICE KNOWN

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9.16

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9.18

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HYPOTHESIS TESTING UNKNOWN

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Tests About a Population Mean:s Unknown

txs n

0/

This test statistic has a t distributionwith n - 1 degrees of freedom.

Where

0 is the mean under H0

s is the sample standard deviation

n is the sample size

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Rejection Rule: p -Value Approach

H0: Reject H0 if t > t

Reject H0 if t < -t

Reject H0 if t < - t or t > t

H0:

H0:

Tests About a Population Mean:s Unknown

Rejection Rule: Critical Value Approach

Reject H0 if p –value < a

Ha:

Ha: >

Ha:

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p -Values and the t Distribution

The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact p-value for a hypothesis test.

However, we can still use the t distribution table to identify a range for the p-value.

An advantage of computer software packages is that the computer output will provide the p-value for the t distribution.

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A State Highway Patrol periodically samplesvehicle speeds at various locations on a particularroadway. The sample of vehicle speeds is used totest the hypothesis H0: m < 65.

Example: Highway Patrol

One-Tailed Test About a Population Mean: s Unknown

The locations where H0 is rejected are deemed thebest locations for radar traps. At Location F, asample of 64 vehicles shows a mean speed of 66.2mph with a standard deviation of 4.2 mph. Use a= .05 to test the hypothesis.

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One-Tailed Test About a Population Mean:s Unknown

1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

a = .05

p –Value and Critical Value Approaches

H0: < 65

Ha: m > 65

0 66.2 65

2.286/ 4.2/ 64

xt

s n

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One-Tailed Test About a Population Mean:s Unknown

p –Value Approach

5. Determine whether to reject H0.

4. Compute the p –value.

For t = 2.286, the p–value must be less than .025(for t = 1.998) and greater than .01 (for t = 2.387).

.01 < p–value < .025

Because p–value < a = .05, we reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph.

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Critical Value Approach

5. Determine whether to reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap.

Because 2.286 > 1.669, we reject H0.

One-Tailed Test About a Population Mean:s Unknown

For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669

4. Determine the critical value and rejection rule.

Reject H0 if t > 1.669

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00

00 ta =1.669 ta =1.669

Reject H0Reject H0

Do Not Reject H0Do Not Reject H0

t

One-Tailed Test About a Population Mean:s Unknown

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PRACTICE UNKNOWN

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9.28

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9.30

(C.L. = 95%)

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