3-1 HYPOTHESIS TESTING. 3-2 INTRODUCTION One of the ways of estimating the true value of population parameters is to test the validity of the claim(assertion

3-1

HYPOTHESIS TESTING

3-2

INTRODUCTION

One of the ways of estimating the true value of population parameters is to test the validity of the claim(assertion or statement) made about this value using simple statistics

• e.g. a pharmaceutical company claims the efficacy of a medicine against a disease that 95 % of all persons suffering from the disease get cured

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INTRODUCTION

To test such claims sample data is collected and analyzed. On the basis of sample findings the hypothesized value of population parameter is either accepted or rejected

• The process that enables a decision maker to test the validity (or significance ) of the claim by analyzing the difference between the value of sample statistic and the corresponding hypothesized population parameter value is called hypothesis testing

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Hypothesis

• Hypotheses are tentative, intelligent guesses as to the solution of the problem.

A hypothesis is an assumption about the population parameter.

– A parameter is a characteristic of the population, like its mean or variance.

Goal:Make statement(s) regarding unknown population parameter values based on sample data

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Types of Hypotheses• Null

– that no statistically significant difference exists between the parameter and the statistic being compared

• Alternative– logical opposite of the null hypothesis– that a statistically significant difference does exist

between the parameter and the statistic being compared.

H0 always contains =

HA matches the question asked; it is what we’re trying to prove.

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Various types of H0 and HA Case

Null Hypothesis Alternate Hypothesis

H 0 H A

1. µ = µ0 µ ≠ µ0

2 . µ ≤ µ0 µ > µ0

3 . µ ≥ µ0 µ < µ0

µ0 is Hypothesised mean

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Always the null hypothesis is tested, i.e., we want to either accept or reject the null hypothesis because we have information only for the null hypothesis.

Ho is taken as true and tested for its possible rejectionEither H0 or HA must be true, making the other false.

H0 and H1 are: Mutually exclusive

– Only one can be true.Exhaustive

– Together they cover all possibilities, so one or the other must be true.

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FORMATS OF HYPOTHESIS• Directional Hypothesis.

– A hypothesis is a statement to be tested about the true value of population parameter

– A hypothesis whether there exists any significant difference between two or more populations with respect to any of their common parameter can also be tested. In this case the hypothesis can be stated in the form if- then statement .Consider

• If inflation rate has increased then wholesale index will also increase

• If employees are healthy then they will take less sick leave

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FORMATS OF HYPOTHESIS

• Directional Hypothesis.– If terms such as positive, negative , more than etc are used

to make a statement , such a hypothesis is called directional hypothesis as it indicates the direction of the relationship

• Non Directional Hypothesis.– It indicates a relationship but offers no direction of

relationship ( or difference)e.g.• There is a relationship between age and job satisfaction• There is difference between average pulse rate of men

and women

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• Hypotheses about other parameters such as population proportions and and population variances are also possible. For example

H0: p 40% H1: p < 40%

H0: s2 50 H1: s2 > 50

Hypothesis about other Parameters

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RATIONALE FOR HYPOTHESIS TESTING• Since statistics are random values , therefore their

sampling distributions show the tendency of variation

• Consequently we do not expect the sample statistic value to be equal to hypothesized value

• If value of sample statistic differ significantly from the hypothesized parameter value then the question arises whether the hypothesized parameter value is correct or not

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RATIONALE FOR HYPOTHESIS TESTING

The probability level at which the decision maker concludes that observed difference between the value of the test statistic and hypothesized value cannot be due to chance is called the level of significance

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PROCEDURE FOR HYPOTHESIS TESTING

STEP 1 –STATE THE NULL HYPOTHESIS (Ho) AND ALTERNATIVE HYPOTHESIS(H1)

• Null hypothesis Ho represents the claim or statement made about the value or range of value of the population parameter. The letter H stands for hypothesis and zero implies no difference between sample statistic and parameter value

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• Null hypothesis is always expressed in the form of mathematical statement which includes (≤, ≥,=)making a claim regarding the specific value of the population parameter

• Ho: μ (≤, ≥,=) μo where μ is population mean and μo represents a hypothesized value of μ

• Alternate hypothesis H1 is the counter claim made against the value of particular population parameter. That is alternate hypothesis must be true when null hypothesis is found to be false.

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In other words alternative hypothesis states that specific population parameter is not equal to the value stated in the null hypothesis

H1:μ ≠ μo

Consequently

H1 :μ<μo or μ>μo

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ERRORS IN HYPOTHESIS TESTING

• Ideally the hypothesis testing procedure should lead to acceptance of the null hypothesis Ho when it is true and the rejection of Ho when it is not true

• However the correct decision is not always possible

• Since the decision to reject or accept a hypothesis is based on sample date there is a possibility of an incorrect decision or error

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ERRORS IN HYPOTHESIS TESTINGThere are two types of error

STATE OF NATURE

Sample decision

Ho is true Ho is false

Accept Ho Correct Decision=1-

Type II error. P(Type II)=

Reject Ho Type I error.p(Type I)=

Correct Decision

Power=1-

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TYPE 1 ERROR

• This is the probability of rejecting the null hypothesis when it is true and some alternative hypothesis is wrong

• The probability of making a Type I error is denoted by the symbol α

• It is represented by the area under the sampling distribution curve over the region of rejection

• The probability of making a Type I error is referred to as level of significance

• The complement 1-α of the probability of Type I error measure the probability level of not rejecting a true null hypothesis. It is also referred to as confidence level

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TYPE II ERROR

• This is the probability of accepting the null hypothesis when it is false and some alternative hypothesis is true

• The probability of Type II error varies with the actual values of the population parameter being tested when null hypothesis Ho is false

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The power of a statistical hypothesis test is the probability of rejecting the null hypothesis when the null hypothesis is false.

Power = (1 - )

The power of a statistical hypothesis test is the probability of rejecting the null hypothesis when the null hypothesis is false.

Power = (1 - )

The Power of a Test

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The probability of a type II error, and the power of a test, depends on the actual value of the unknown population parameter. The relationship between the population mean and the power of the test is called the power function.

The probability of a type II error, and the power of a test, depends on the actual value of the unknown population parameter. The relationship between the population mean and the power of the test is called the power function.

7069686766656463626160

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Power of a One-Tailed Test: =60, =0.05

P

ower

Value of m b Power = (1 - b)

61 0.8739 0.1261 62 0.7405 0.2695 63 0.5577 0.4423 64 0.3613 0.6387 65 0.1963 0.8037 66 0.0877 0.9123 67 0.0318 0.9682 68 0.0092 0.9908 69 0.0021 0.9972

Value of m b Power = (1 - b)

61 0.8739 0.1261 62 0.7405 0.2695 63 0.5577 0.4423 64 0.3613 0.6387 65 0.1963 0.8037 66 0.0877 0.9123 67 0.0318 0.9682 68 0.0092 0.9908 69 0.0021 0.9972

The Power Function

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The power depends on the distance between the value of the parameter under the null hypothesis and the true value of the parameter in question: the greater this distance, the greater the power.

The power depends on the population standard deviation: the smaller the population standard deviation, the greater the power.

The power depends on the sample size used: the larger the sample, the greater the power.

The power depends on the level of significance of the test: the smaller the level of significance,, the smaller the power.

Factors Affecting the Power Function

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To Note:

1. Would like a and b to be as small as possible.

2. a and b are inversely related.

3. Usually set a (and don’t worry too much about b.)

4. Most common values for a and b are 0.01 and 0.05.

23

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STEP 2-STATE THE LEVEL OF SIGNIFICANCE

• Select an appropriate significance level

(Type I error), it is denoted by α. It is the risk a decision maker takes of rejecting the null hypothesis when it is true

• Traditionally α = 0.05 for consumer research projects, α = 0.01 for quality assurance and α=0.10 for political polling

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Significance level gives the confidence with which a researcher rejects or retain the null hypothesis.

If we take 5% level of significance then there are about 5 chances out of 100 that we would reject the Null hypothesis when it should be accepted or

that we are 95% confident about the right decision

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STEP 3- ESTABLISH CRITICAL OR REJECTION REGION

• The area under the sampling distribution curve of test statistic is divided into two regions. These regions are called acceptance region and the rejection region ( critical region)

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Acceptance region and critical regions for two tailed test in hypothesis testing

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One-sided hypothesis where the critical region lies

in the right tail

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One-sided hypothesis where the critical region lies in the left tail

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Given the same level of significance the two tailed test is more conservative, i.e., it is more rigorous than the one-tailed test because the rejection point is farther out in the tail. It is more difficult to reject H0 with a two-tailed test than with a one-tailed test.

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The tails of a statistical test are determined by the need for an action. If action is to be taken if a parameter is greater than some value a, then the alternative hypothesis is that the parameter is greater than a, and the test is a right-tailed test. H0: 50

H1: 50

The tails of a statistical test are determined by the need for an action. If action is to be taken if a parameter is greater than some value a, then the alternative hypothesis is that the parameter is greater than a, and the test is a right-tailed test. H0: 50

H1: 50

If action is to be taken if a parameter is less than some value a, then the alternative hypothesis is that the parameter is less than a, and the test is a left-tailed test. H0: 50

H1: 50

If action is to be taken if a parameter is less than some value a, then the alternative hypothesis is that the parameter is less than a, and the test is a left-tailed test. H0: 50

H1: 50

If action is to be taken if a parameter is either greater than or less than some value a, then the alternative hypothesis is that the parameter is not equal to a, and the test is a two-tailed test. H0: 50

H1: 50

If action is to be taken if a parameter is either greater than or less than some value a, then the alternative hypothesis is that the parameter is not equal to a, and the test is a two-tailed test. H0: 50

H1: 50

1-Tailed and 2-Tailed Tests

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STEP 4 –SELECT THE TEST OF SIGNIFICANCE OR TEST STATISTIC

• The tests of significance or test statistic are divided into two categories: parametric and non parametric

• Parametric tests are more powerful because their data is derived from interval and ratio measurements

• Non parametric tests are used to test hypothesis with nominal and ordinal data

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Univariate Techniques

Parametric Test

One Sample•t test•Z test

Two or More SamplesInde

pendent

•t test•Z

test•On

e way ANOVA

Related•Paired t test

Nonparametric Test

One Sample•Frequency•Chi-square

•K-S•Runs

•Binomial

Two or More Samples

Independent•Chi

-Square

•Median

Related•Chi

-Square•Sig

n

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How to Select a Test

• Which does the test involve?– one sample, – two samples– k samples

• If two or k samples,are the individual cases independent or related?

• Is the measurement scale nominal, ordinal, interval, or ratio?

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• Cases in which the test statistic is Z

s is known and the population is normal.s is known and the sample size is at least 30. (The population need not be normal)

• Cases in which the test statistic is Z

s is known and the population is normal.s is known and the sample size is at least 30. (The population need not be normal)

Testing Population Means

n

xz

isZgcalculatinforformulaThe

s

:

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• Cases in which the test statistic is t

s is unknown but the sample standard deviation is known and the population is normal.

• Cases in which the test statistic is t

s is unknown but the sample standard deviation is known and the population is normal.

Testing Population Means

ns

xt

istgcalculatinforformulaThe

:

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STEP 5 – FORMULATE A DECISION RULE TO ACCEPT NULL HYPOTHESIS

• Compare the calculated value of test statistic with the critical value ( also called standard value of test statistic). The decision rule for null hypothesis are as follows – Accept Ho if the test statistic value falls within the area of

acceptance– Reject otherwise

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The p-value is the probability of obtaining a value of the test statistic as extreme as, or more extreme than, the actual value obtained, when the null hypothesis is true.

The p-value is the smallest level of significance, , at which the null hypothesis may be rejected using the obtained value of the test statistic.

Policy: When the p-value is less than a , reject H0.

The p-value is the probability of obtaining a value of the test statistic as extreme as, or more extreme than, the actual value obtained, when the null hypothesis is true.

The p-value is the smallest level of significance, , at which the null hypothesis may be rejected using the obtained value of the test statistic.

Policy: When the p-value is less than a , reject H0.

PROCEDURE FOR HYPOTHESIS TESTINGThe p-Value approach

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3-40

EXAMPLE 1

An automatic bottling machine fills cola into two liter (2000 cc) bottles. A consumer advocate wants to test the null hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of 40 bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The sample mean was 1999.6 cc. The population standard deviation is known from past experience to be 1.30 cc.

Test the null hypothesis at the 5% significance level.

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H0: 2000H1: 2000n = 40For = 0.05, the critical valueof z is -1.645

The test statistic is:

Do not reject H0 if: [z -1.645]Reject H0 if: z ]

zxs

n

0

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0

HReject 1.95 =

= 0

1.3 =

1999.6 = x

40 =n

401.3

2000-1999.6

n

xz s

s

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Do not reject H0 if: [p-value 0.05]Reject H0 if: p-value 0.05]



Do not reject H0 if: [p-value 0.05]Reject H0 if: p-value 0.05]

zxs

n

0

0.05 0.0256 since 0

HReject 0.0256

0.4744-0.5000

1.95)- P(Z value-

1.95 =

= 0

401.3

2000-1999.6

p

n

xz s

Example : p-value approach

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As part of a survey to determine the extent of required in-cabin storage capacity, a researcher needs to test the null hypothesis that the average weight of carry-on baggage per person is 0 = 12 pounds, versus the alternative hypothesis that the average weight is not 12 pounds. The analyst wants to test the null hypothesis at = 0.05.

As part of a survey to determine the extent of required in-cabin storage capacity, a researcher needs to test the null hypothesis that the average weight of carry-on baggage per person is 0 = 12 pounds, versus the alternative hypothesis that the average weight is not 12 pounds. The analyst wants to test the null hypothesis at = 0.05.

H0: = 12H1: 12

For = 0.05, critical values of z are ±1.96


Do not reject H0 if: [-1.96 z 1.96]

Reject H0 if: [z <-1.96] or z 1.96]

H0: = 12H1: 12




Reject H0 if: [z <-1.96] or z 1.96]

zxsn

0

Lower Rejection Region

Upper Rejection Region

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.025 .025

.95

Nonrejection Region

z 1.96-1.96

The Standard Normal Distribution

Examples :2

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n = 144

x = 14.6

s = 7.8

=14.6-12

7.8

144

= 2.6

0.65

zxs

n

0

4

Since the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average amount of carry-on baggage is more than 12 pounds.

Since the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average amount of carry-on baggage is more than 12 pounds.



0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.025 .025

.95

Nonrejection Region

z 1.96-1.96


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An insurance company believes that, over the last few years, the average liability insurance per board seat in companies defined as “small companies” has been $2000. Using = 0.01, test this hypothesis using Growth Resources, Inc. survey data.

An insurance company believes that, over the last few years, the average liability insurance per board seat in companies defined as “small companies” has been $2000. Using = 0.01, test this hypothesis using Growth Resources, Inc. survey data.

H0: = 2000H1: 2000 For = 0.01, critical values of z are ±2.576



Reject H0 if: [z <-2.576] or z 2.576]

H0: = 2000H1: 2000 For = 0.01, critical values of z are ±2.576



Reject H0 if: [z <-2.576] or z 2.576]

zxs

n

0

n = 100

x = 2700

s = 947

=2700 - 2000

947

100

= 700

94.7 Reject H

0

zx

s

n

0

7 39.

Examples :3

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Since the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average insurance liability per board seat in “small companies” is more than $2000.

Since the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average insurance liability per board seat in “small companies” is more than $2000.



0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.005 .005

.99

Nonrejection Region

z 2.576-2.576


Examples 3 : Continued

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The average time it takes a computer to perform a certain task is believed to be 3.24 seconds. It was decided to test the statistical hypothesis that the average performance time of the task using the new algorithm is the same, against the alternative that the average performance time is no longer the same, at the 0.05 level of significance.

The average time it takes a computer to perform a certain task is believed to be 3.24 seconds. It was decided to test the statistical hypothesis that the average performance time of the task using the new algorithm is the same, against the alternative that the average performance time is no longer the same, at the 0.05 level of significance.H0: = 3.24H1: 3.24




Reject H0 if: [z < -1.96] or z 1.96]

H0: = 3.24H1: 3.24




Reject H0 if: [z < -1.96] or z 1.96]

zxs

n

0

n = 200

x = 3.48

s = 2.8

=

200

= Do not reject H0

3.48- 3.242.8

0.240.20

zx

s

n

0

1 21.

Examples :4

3-49

Since the test statistic falls in the nonrejection region, H0 is not rejected, and we may conclude that the average performance time has not changed from 3.24 seconds.

Since the test statistic falls in the nonrejection region, H0 is not rejected, and we may conclude that the average performance time has not changed from 3.24 seconds.



0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.025 .025

.95

Nonrejection Region

z 1.96-1.96



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According to the Japanese National Land Agency, average land prices in central Tokyo soared 49% in the first six months of 1995. An international real estate investment company wants to test this claim against the alternative that the average price did not rise by 49%, at a 0.01 level of significance.

According to the Japanese National Land Agency, average land prices in central Tokyo soared 49% in the first six months of 1995. An international real estate investment company wants to test this claim against the alternative that the average price did not rise by 49%, at a 0.01 level of significance.

H0: = 49H1: 49n = 18For = 0.01 and (18-1) = 17 df , critical values of t are ±2.898


Do not reject H0 if: [-2.898 t 2.898]

Reject H0 if: [t < -2.898] or t 2.898]

H0: = 49H1: 49n = 18For = 0.01 and (18-1) = 17 df , critical values of t are ±2.898


Do not reject H0 if: [-2.898 t 2.898]

Reject H0 if: [t < -2.898] or t 2.898]

0

HReject 33.3 =

= 0

14 = s

38 = x

18 =n

3.3

11-

1814

49-38

n

s

xt

txs

n

0

Examples :5

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Since the test statistic falls in the rejection region, H0 is rejected, and we may conclude that the average price has not risen by 49%. Since the test statistic is in the lower rejection region, we may conclude that the average price has risen by less than 49%.

Since the test statistic falls in the rejection region, H0 is rejected, and we may conclude that the average price has not risen by 49%. Since the test statistic is in the lower rejection region, we may conclude that the average price has risen by less than 49%.



0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

.005 .005

.99

Nonrejection Region

t 2.898-2.898

The t Distribution

Examples 5: Continued

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An investment analyst for Goldman Sachs and Company wanted to test the hypothesis made by British securities experts that 70% of all foreign investors in the British market were American. The analyst gathered a random sample of 210 accounts of foreign investors in London and found that 130 were owned by U.S. citizens. At the = 0.05 level of significance, is there evidence to reject the claim of the British securities experts?

An investment analyst for Goldman Sachs and Company wanted to test the hypothesis made by British securities experts that 70% of all foreign investors in the British market were American. The analyst gathered a random sample of 210 accounts of foreign investors in London and found that 130 were owned by U.S. citizens. At the = 0.05 level of significance, is there evidence to reject the claim of the British securities experts?

H0: p = 0.70H1: p 0.70n = 210For = 0.05 critical values of z are ±1.96The test statistic is:

Do not reject H0 if: [-1.96 z 1.96]Reject H0 if: [z < -1.96] or z 1.96]

H0: p = 0.70H1: p 0.70n = 210For = 0.05 critical values of z are ±1.96The test statistic is:

Do not reject H0 if: [-1.96 z 1.96]Reject H0 if: [z < -1.96] or z 1.96]

n = 210

p =130

210

=

p - p0

p0

=

= Reject H0

0.619 - 0.70(0.70)(0.30)

-0.0810.0316

.

.

0 619

0

2 5614

210

zq

n

zp p

p qn

0

0 0

Examples :6

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A floodlight is said to last an average of 65 hours. A competitor believes that the average life of the floodlight is less than that stated by the manufacturer and sets out to prove that the manufacturer’s claim is false. A random sample of 21 floodlight elements is chosen and shows that the sample average is 62.5 hours and the sample standard deviation is 3. Using =0.01, determine whether there is evidence to conclude that the manufacturer’s claim is false.

A floodlight is said to last an average of 65 hours. A competitor believes that the average life of the floodlight is less than that stated by the manufacturer and sets out to prove that the manufacturer’s claim is false. A random sample of 21 floodlight elements is chosen and shows that the sample average is 62.5 hours and the sample standard deviation is 3. Using =0.01, determine whether there is evidence to conclude that the manufacturer’s claim is false.H0: 65H1: 65n = 21For = 0.01 an (21-1) = 20 df, the critical value -2.528


Do not reject H0 if: [t -2.528]Reject H0 if: z 2.52]

H0: 65H1: 65n = 21For = 0.01 an (21-1) = 20 df, the critical value -2.528


Do not reject H0 if: [t -2.528]Reject H0 if: z 2.52]

Examples :7

3-54

Since the test statistic falls in the rejection region, H0 is rejected, and we

may conclude that the manufacturer’s claim is false, that the average floodlight life is less than 65 hours.

Since the test statistic falls in the rejection region, H0 is rejected, and we

may conclude that the manufacturer’s claim is false, that the average floodlight life is less than 65 hours.

0.95

-2.52850-5

0 .4

0 .3

0 .2

0 .1

0 .0

t

f(t)

NonrejectionRegion

RejectionRegion

Critical Point for a Left-Tailed Test

-3.82


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“After looking at 1349 hotels nationwide, we’ve found 13 that meet our standards.” This statement by the Small Luxury Hotels Association implies that the proportion of all hotels in the United States that meet the association’s standards is 13/1349=0.0096. The management of a hotel that was denied acceptance to the association wanted to prove that the standards are not as stringent as claimed and that, in fact, the proportion of all hotels in the United States that would qualify is higher than 0.0096. The management hired an independent research agency, which visited a random sample of 600 hotels nationwide and found that 7 of them satisfied the exact standards set by the association. Is there evidence to conclude that the population proportion of all hotels in the country satisfying the standards set by the Small Luxury hotels Association is greater than 0.0096?

“After looking at 1349 hotels nationwide, we’ve found 13 that meet our standards.” This statement by the Small Luxury Hotels Association implies that the proportion of all hotels in the United States that meet the association’s standards is 13/1349=0.0096. The management of a hotel that was denied acceptance to the association wanted to prove that the standards are not as stringent as claimed and that, in fact, the proportion of all hotels in the United States that would qualify is higher than 0.0096. The management hired an independent research agency, which visited a random sample of 600 hotels nationwide and found that 7 of them satisfied the exact standards set by the association. Is there evidence to conclude that the population proportion of all hotels in the country satisfying the standards set by the Small Luxury hotels Association is greater than 0.0096?

H0: p 0.0096H1: p 0.0096n = 600

For = 0.10 the critical value 1.282


Do not reject H0 if: [z 1.282]Reject H0 if: z .22]

H0: p 0.0096H1: p 0.0096n = 600

For = 0.10 the critical value 1.282


Do not reject H0 if: [z 1.282]Reject H0 if: z .22]

Examples :8

3-56

Since the test statistic falls in the nonrejection region, H0 is not rejected, and we may not conclude that proportion of all hotels in the country that meet the association’s standards is greater than 0.0096.

Since the test statistic falls in the nonrejection region, H0 is not rejected, and we may not conclude that proportion of all hotels in the country that meet the association’s standards is greater than 0.0096.

0.90

1.28250-5

0.4

0.3

0.2

0.1

0.0

z

f(z)

NonrejectionRegion

RejectionRegion

Critical Point for a Right-Tailed Test

0.519


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EXAMPLE 9

A packaging device is set to fill detergent powder packets with a mean weight of 3 kg, with a standard deviation of 0.21 kg. The weight of packets can be assumed to be normally distributed.

The weight of packets is known to drift upwards over a period of time due to machine fault which is no tolerable. A sample of 100 packets is taken and weighed. This sample has a mean weight of 5.03 kg. can we conclude that the mean weight produced by the machine has increased? Use a 5 % level of significance

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EXAMPLE 9: Continued

• Let null hypothesis Ho that mean weight has increased

• Ho : μ ≥ 5 and H1 ; μ < 5

• Given n=100,x=5.3 kg, σ=0.21, α= 5%

• Z= (x- μ) / σ√n = 5.03-5 /0.21/√100=1.428

• Since calculated value z(cal) = 1.428 is less than its critical value z(α) =1.645 at α = 0.05, the null hypothesis Ho is accepted

• Hence we conclude that mean weight is likely to be more than 5 kg

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HYPOTHESIS TESTING FOR DIFFERENCE BETWEEN TWO POPULATION MEANS

Idea is to test whether there is significant difference between the means of these populations

Let μ1 and μ2 be mean of two population

σ1 and σ2 be std deviation of two population

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HYPOTHESIS TESTING FOR POPULATION PARMETERS WITH LARGE SAMPLES

Z statistic = [(x1-x2) – (μ1-μ2)]/ σx1 –x2

= [(x1-x2) – (μ1-μ2)]/ √σ₁²/n1 + σ₂²/n2

The null and alternate hypothesis are stated as Null hypothesis

Ho :μ1 – μ2 = 0

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Alternate hypothesis

One tailed test

H1:(μ1 –μ2) > 0

H1 : (μ1 –μ2) < 0

Two tailed test

H1: (μ1 –μ2) ≠ 0

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HYPOTHESIS TESTING FOR POPULATION PARMETERS WITH LARGE SAMPLES

Decision Rule

• Reject Ho at a specified level of significance when

One Tailed Test

• Z(cal) >z(α) [or when z<-z(α)]

Two tailed test

• Z(cal)>z(α/2) or z(cal) <-z(α/2)

3-63EXAMPLE 10

A firm believes that the tyres produced by process A on an average last longer than tyres produced by process B. To test this belief, random supplies of tyres produced by the two processes were tested and the result are

• Is there evidence at a 5 % level of significance that the firm is correct in its belief?

Process Sample size Average lifetime(km)

Standard deviation(km)

A 50 22400 1000

B 50 21800 1000

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EXAMPLE 10Let us take the null hypothesis that there is no significant difference in the average life of tyres produced by processes A and B

Ho : μ1=μ2 or μ1-μ2 = 0

H1 : μ1=μ2

Given x1(mean) = 22400, x2 (mean) =21800

Σ1=σ2=1000, n1=n2=50

Z= [(x1-x2) – (μ1-μ2)]/ σx1 –x2

= [(x1-x2) – (μ1-μ2)]/ √σ₁²/n1 + σ₂²/n2

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EXAMPLE 10

= 22400 – 21800 /√(1000)²/50 + (1000)²/50

= 600/√ 20000+20000

= 600/200=3

Since the calculated value z(cal )=3 is more than its critical value z (α/2) =+- 1.645at α =0.5 level of significance therefore Ho is rejected

Hence we conclude that the tyre produced by process A last longer than that of B

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HYPOTHESIS TESTING FOR SINGLE POPULATION PROPORTION

The three forms of null hypothesis and alternative hypothesis pertaining to the hypothesized population proportion Po are as follows

Null Hypothesis Alternative Hypothesis

Ho: P = Po H1: P = Po ( two tailed test)

Ho :P ≥ Po H1: P < Po (left tailed test)

Ho : P ≤ Po H1 : P >Po (right tailed test

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HYPOTHESIS TESTING FOR SINGLE POPULATION PROPORTION

Test statistic

z = p-po/σp

= P- Po /√Po(1-Po)/n

One tailed test Two tailed test

Z(cal) > z(α) or Z(cal) < - z(α) When H1 ; p< po

Z(cal) > z(α/2) orZ(cal) < - z(α/2)

P value <α

Documents

3-1 HYPOTHESIS TESTING. 3-2 INTRODUCTION One of the ways of estimating the true value of population parameters is to test the validity of the claim(assertion