Module6 Lecture 2(1)

7/28/2019 Module6 Lecture 2(1)

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Univariate Statistics

Lecture 2

Hypothesis Testing

Part 1


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Choosing the Appropriate

Statistical Technique Type of question to be answered

Number of variables

Univariate

Bivariate

Multivariate

Scale of measurement


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Univariate Statistics Now, we are entering the domain of

Inferential statistics

One of the tools for inferential statistics

Test of statistical significance

Hypothesis testing one variable at atime


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HypothesisAn unproven proposition or supposition

that tentatively explains certain facts or

phenomena Null hypothesis

Alternative hypothesis


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Null Hypothesis Statement about the status quo

No difference

For example

Compare superiority of two Brand: Airtel andReliance

Null Hypothesis: Brand Airtel=Brand RelianceAlternative Hypothesis: Brand Airtel is superior

to Brand Reliance


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Alternative Hypothesis Statement that indicates the opposite of

the null hypothesis

Is usually the one which one wishes toprove


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Level of Significance Very important in context of hypothesis

testing

Critical probability in choosing between thenull hypothesis and the alternative hypothesis

5% level of significance means that the nullhypothesis will get rejected if the result is lessthan 0.05 probability of occurring


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Level of SignificanceFactors that affect level of significance

Size of the sample

Magnitude of the difference betweensample means

Variability of measurements betweentwo samples


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An example Mainland China is concerned about its

image; one aspect is friendliness of the

service Survey with likert scale 1(unfriendly

service-5 very friendly service)

Sample mean calculated based on thesurvey was 3.78


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Example Contd. Ho:=3.0 (the hypothesis states that

the sample mean feels that the service

of Mainland China is neither friendly norunfriendly)

Ha:3.0

Level of significance: 5%

=3.78


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m=3.0

x

a=.025 a=.025

A Sampling Distribution


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LOWER

LIMIT

UPPER

LIMIT

m=3.0

A Sampling Distribution


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Critical values ofmCritical value - upper limit

n

SZZSX

= or mm

=

225

5.196.10.3


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1.096.10.3 =

196.0.3 =

196.3=

Critical values ofm


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Critical value - lower limit

n

SZZS

X-or- mm=

=

225

5.196.1-0.3

Critical values ofm


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1.096.10.3 =

196.0.3 =

804.2=

Critical values ofm


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Hypothesis Test m =3.0

2.804 3.196

m=3.03.78


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Results

Since, the sample mean 3.78 falls beyond thecritical value, then we reject Ho.

Meaning, that the customers believe that theservice is friendly

Also, it is unlikely that this result occurreddue to random sampling error

It also means that the management shouldlook for other factors that affect the image ofthe resturant


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Accept null Reject null

Null is true

Null is false

Correct-

no error

Type I

error

Type II

error

Correct-

no error

Type I and Type II Errors


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Type I and Type II Errors

in Hypothesis TestingState of Null Hypothesis Decision

in the Population Accept Ho Reject Ho

Ho is true Correct--no error Type I error

Ho is false Type II error Correct--no error


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Example

A company is engaged in packaging of asuperior quality tea in jars of 500gm

each. The company is of view that aslong as jars contain 500gm of tea, theprocess is in control. The standarddeviation is 50gm. A sample of 225jars

is taken at random and the sampleaverage is found to be 510gm. Has theprocess gone out of control?


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Example

A company manufacturing automobile tyres finds thattyre life is normally distributed with a mean of 40,000kms and standard deviation of 3,000kms. It is

believed that a change in the production process willresult in a better product and the company hasdeveloped a new tyre. A sample of 64 new tyres hasbeen selected. The company has found that the

mean life of these new tyres is 41,200kms. Can it beconcluded that the new tyre is significantly betterthan the old one?


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Example

A pharmaceutical company, engaged in themanufacture of a patient medicine claimed

that it was 80 percent effective in relieving anallergy for a period of 15 hrs. A sample of200 persons, who suffered from allergy, weregiven this medicine. It was found that the

medicine provided relief to 150 persons for atleast 12hrs. Do you think that the companysclaim is justified? Use 0.05 level ofsignificance.


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Types of Measurement

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Module6 Lecture 2(1)