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Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Procedure for Hypothesis Testing 1. Establish the null hypothesis, H 0 . 2. Establish the alternate hypothesis: H 1 . 3. Use the level of significance and the alternate hypothesis to determine the critical region. 4. Find the critical values that form the boundaries of the critical region(s). 5. Use the sample evidence to draw a conclusion regarding whether or not to reject the null hypothesis.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Procedure for Hypothesis Testing 1. Establish the null hypothesis, H 0. 2.Establish

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Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1

Procedure for Hypothesis Testing

1. Establish the null hypothesis, H0.

2. Establish the alternate hypothesis: H1.

3. Use the level of significance and the alternate hypothesis to determine the critical region.

4. Find the critical values that form the boundaries of the critical region(s).

5. Use the sample evidence to draw a conclusion regarding whether or not to reject the null hypothesis.

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Null Hypothesis

Claim about or historical value of

H0: = k

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H0: = k

If you believe is less than the value stated in H0,

use a left-tailed test.

H1: < k

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H0: = k

If you believe is more than the value stated in H0,

use a right-tailed test.

H1: > k

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H0: = k

If you believe is different from the value stated in H0,

use a two-tailed test.

H1: k

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Hypothesis Testing About a

Population Mean when Sample Evidence Comes From

a Large Sample

Apply the Central Limit Theorem.

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Central Limit Theorem Indicates:

Central Limit Theorem Indicates:

Since we are working with assumptions concerning a population mean for a large sample, we can assume:

1. The distribution of sample means is (approximately) normal.

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Since we are working with assumptions concerning a population mean for a large sample, we can assume:

2. The mean of the sampling distribution is the same as the mean of the original

distribution.

Central Limit Theorem Indicates:

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Since we are working with assumptions concerning a population mean for a large sample, we can assume:

3. The standard deviation of the sampling distribution = the original standard deviation divided by the square root of the the sample size.

Assumptions:

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Use of the Level of SignificanceUse of the Level of Significance

For a one-tailed test, is the area in the tail (the rejection area).

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Use of the Level of SignificanceUse of the Level of Significance

For a two-tailed test, is the total area in the two tails.

Each tail = /2.

/2 /2

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Critical z Values for Two-Tailed Test: = 0.05

– 1.96 0 1.96

If test statistic is at or near the

claimed mean, we

do not reject the Null

Hypothesis

The critical regions: z < – 1.96 with z > 1.96.

H0: = k

H1: k

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Critical z Values for Two-Tailed Test: = 0.01

– 2.58 0 2.58

If test statistic is at or near the

claimed mean, we

do not reject the Null

Hypothesis

The critical regions: z < – 2.58 with z > 2.58.

H0: = k

H1: k

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Critical z Value for Right-Tailed Test: = 0.05

Critical z Value for Right-Tailed Test: = 0.05

0 1.645

If test statistic is at, near, or below the

claimed mean, we do not reject the Null

Hypothesis

The critical region: z > 1.645.

H0: = k

H1: > k

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Critical z Value for Right-Tailed Test: = 0.01

Critical z Value for Right-Tailed Test: = 0.01

0 2.33

If test statistic is at, near, or below the

claimed mean, we do not reject the Null

Hypothesis

The critical region: z > 2.33.

H0: = k

H1: > k

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Critical z Value for Left-Tailed Test: = 0.05

Critical z Value for Left-Tailed Test: = 0.05

-1.645 0

If test statistic is at, near, or above the

claimed mean, we do not reject the Null

Hypothesis

The critical region: z < – 1.645

H0: = k

H1: < k

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Critical z Value for Left-Tailed Test: = 0.01

Critical z Value for Left-Tailed Test: = 0.01

-2.33 0

If test statistic is at, near, or above the

claimed mean, we do not reject the Null

Hypothesis

The critical region: z < – 2.33

H0: = k

H1: < k

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Hypothesis Test Example

Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of = 0.05.

A random sample of 49 students has a mean age of 26 years with a standard deviation of 2.3 years.

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two

Test H0: = 28Against H1: 28

Perform a ________-tailed test.

Hypothesis Test Example

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two

Using = 0.05 Critical z value(s) = _________

Test H0: = 28Against H1: 28

Perform a ________-tailed test.

Hypothesis Test Example

±1.96

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Critical z Values for Two-Tailed Test: = 0.05

– 1.96 0 1.96

If test statistic is at or near the

claimed mean, we

do not reject the Null

Hypothesis

The critical regions: z < – 1.96 with z > 1.96.

H0: = 28

H1: 28

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Since z < – 1.96, we _________ the null hypothesis.

73.2

2826

n

xz

09.632857.

2

reject

Sample Results

:z statistictest the Calculate

.3.2,26 sx

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Hypothesis Test Example

Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of = 0.05.

A random sample of 49 students has a mean age of 27.5 years with a standard deviation of 2.3 years.

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Using = 0.05 Critical z values = 1.96

Test H0: = 28Against H1: 28

So, perform a two-tailed test.

Hypothesis Test ExampleHypothesis Test Example

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Since the test statistic is neither < – 1.96 nor > 1.96 , we _______________ the null

hypothesis.

73.2

285.27

n

xz

52.132857.

5.

do not reject

Sample ResultsSample Results

:z statistic,test the Calculate

2.3.s 27.5,x

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Hypothesis Test Example

The manufacturer of light bulbs claims that they will burn for 1000 hours. I will test a sample of the bulbs before deciding whether to keep them. The bulbs will be returned to the manufacturer only if my sample indicates that they will burn less than 1000 hours.

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The manufacturer of light bulbs claims that they will burn for 1000 hours. ...The bulbs will be returned ... if my sample indicates that they will burn less than 1000 hours.

H0: = 1000H1: < 1000

Hypothesis Test ExampleHypothesis Test Example

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left

Using = 0.01 (So, critical z value = _______)

Test H0: = 1000Against H1: < 1000

Perform a ____-tailed test.)

Hypothesis Test ExampleHypothesis Test Example

–2.33

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Critical z Value for Left-Tailed Test: = 0.01

Critical z Value for Left-Tailed Test: = 0.01

-2.33 0

If test statistic is at, near, or above the

claimed mean, we do not reject the Null

Hypothesis

The critical region: z < – 2.33

H0: = 1000

H1: < 1000

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Since the test statistic is not < – 2.33 we _____________ the null hypothesis.

64.3

1000999

n

xz

76.1567.

1

do not reject

Sample ResultsSample Results

:z Calculate

hours. 3.4 s and hours 999x

shows bulbs 36 of sampleour Suppose

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Comparison of Critical z Values for Left-Tailed Tests:

= 0.01 and = 0.05

Comparison of Critical z Values for Left-Tailed Tests:

= 0.01 and = 0.05

– 2.33 0 – 1.645 0

.01 .05

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In our last hypothesis test example, we calculated z = – 1.76.

Since we were using = 0.01, the boundary of the critical region was

– 2.33.

Our conclusion was not to reject the null hypothesis.

Had we been using = 0.05, our conclusion would have been to reject H0.

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Comparison of Critical z Values for Left-Tailed Tests:

= 0.01 and = 0.05

Comparison of Critical z Values for Left-Tailed Tests:

= 0.01 and = 0.05

– 2.33 0 – 1.645 0

=.01 = .05

z = – 1.76Do not reject H0.

z = – 1.76Reject H0.

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Statistical Significance

• If we reject H0, we say that the data collected in the hypothesis testing process are statistically significant.

• If we do not reject H0, we say that the data collected in the hypothesis testing process are not statistically significant.