testing of hypothesis-agbs

8/8/2019 Testing of Hypothesis-AGBS

http://slidepdf.com/reader/full/testing-of-hypothesis-agbs 1/74

Testing of Hypothesis



2

Statistics in Decision Making

Getting Observations

of the real world

Keeping Records

based on

observations

Organizing Data

according to records

Data

Analysis

Decision

-Making

© 1984-1994 T/Maker Co.



3

Statistics

A collection of principles andmethods concerned with extractinguseful information from a set of data to help managers makedecisions.

The subject of statistics can besub-divided into two basic areas:

Descriptive statistics

Inferential statistics



4

Descriptive Statistics

Collect data

e.g. Survey

Present data

e.g. Tables and graphs

Characterize datae.g. Sample mean =

i X

n

§



5

Inferential Statistics Estimation

e.g. Estimate the

population mean weight

using the sample meanweight

Drawing conclusions and/or making decisionsconcerning a population based on sample results.

Hypothesis testing

e.g. Test the claim that the population mean

weight is 120 pounds



Problem Under Study

q

Data

Survey Experiment

Once sample data has been gathered, statistical

inference allows to assess evidence in favor or some

claim about the population from which the sample

has been drawn.

The method of inference used to support or reject

claims based on sample data is known as testing of

hypothesis.



What is a Hypothesis?

A hypothesis is a claim(assumption) about apopulation parameter:

± population mean

± population proportion

Example: The mean monthly cell phone bill of

this city is = $42

Example: The proportion of adults in this city

with cell phones is = 0.68



The Null Hypothesis, H0

States the claim or assertion to be tested

Example: The average number of TV sets in

U.S. Homes is equal to three ( )

Is always about a population parameter,

not about a sample statistic

3:H0 !

3:H0 ! 3X:H0 !



The Null Hypothesis, H0

Begin with the assumption that the nullhypothesis is true

± Similar to the notion of innocent untilproven guilty

Refers to the status quo

Always contains ³=´ , ³´ or ³u´ sign May or may not be rejected

(continued)



The Alternative Hypothesis, H1

Is the opposite of the null hypothesis

± e.g., The average number of TV sets in

U.S. homes is not equal to ( H1: )

Challenges the status quo

Never contains the ³=´ , ³´ or ³u´ sign

May or may not be proven

Is generally the hypothesis that theresearcher is trying to prove



Population

Claim: the

population

mean age is 50.

(Null Hypothesis:

REJECT

Suppose

the samplemean age

is 20: X = 20

SampleNull Hypothesis

20 likely if = 50?!Is

Hypothesis Testing Process

If not likely,

Now select a

random sample

H0: = 50 )

X



To compare the effectiveness of different methods of

teaching

To know whether average self-confidence score of college

students is equal to some specified value.

To know whether average yield of a crop in a certain district

is equal to a specified value

To compare the effects of stress management programs on

self-esteem.

Testing of Hypothesis: Objectives



To know whether intelligence level measured through

intelligence quotient is up to the standard

To find if a new drug is really effective for the particular

ailment, say, in reducing blood pressure or inducing sleep

To compare two processes with regard to production of

certain items

To know if the genetic fraction of the total variation in a

strain is more than a given value

Testing of Hypothesis: Objectives



Statistical test is a procedure governed by certain rules,

that leads to take a decision about the hypothesis for its

acceptance or rejection on the basis of the sample values

These tests have wide applications in agriculture,

medicine, industry, social sciences, psychology,

etc.

Tests of Significance



Definitions

Statisticp Function of sample values, like sample mean,

sample variance

Parameterp Function of population values, like

population mean, population variance

Statistical Hypothesisp A definite statement about the

population parameters

If all the parameters are completely specified, the

hypothesis is called a simple hypothesis, otherwise it is

a composite hypothesis.



Definitions...

H0p The hypothesis under test for a sample study

H1p The hypothesis tested against the null

hypothesis

H0: Q = Q

o

H1: Q { Qo (Two-Tailed Test)

Q < Qo (Left-Tailed Test)

Q > Qo (Right-Tailed Test)

Level of Significance (E) p The maximum size of the error

(rejecting H0 when it is true) which we are

prepared to risk. The higher the value of E, less

precise is the result.



Definitions...

Test Statistic

A quantity calculated from sample of data.

Its value is used to decide whether or not the null

hypothesis should be rejected in the hypothesis test

Critical value(s)

The critical value(s) for a hypothesis test is a value to

which the value of the test statistic in a sample is compared

to determine whether or not the null hypothesis is rejected. The critical value for any hypothesis test depends on the

significance level at which the test is carried out, and

whether the test is one-sided or two-sided.



6 Steps in

Hypothesis Testing1. State the null hypothesis, H0 and the

alternative hypothesis, H1

2. Choose the level of significance, E, andthe sample size, n

. Determine the appropriate test statistic

and sampling distribution

4. Determine the critical values that dividethe rejection and nonrejection regions



6 Steps in

Hypothesis Testing5. Collect data and compute the value of

the test statistic

6. Make the statistical decision and statethe managerial conclusion. If the teststatistic falls into the nonrejection region,do not reject the null hypothesis H0. If

the test statistic falls into the rejectionregion, reject the null hypothesis.Express the managerial conclusion inthe context of the problem

(continued)



Tests of Significance

Normal Test

t - Test

Chi - Square Test

F - Test



The Normal Distribution

X

f(X)

®

µ

Changing shift s the

distribution left or right.

Changing increasesor decreases the spread.



Normal distribution is defined byits mean and standard dev.

E(X)=Q

Var(X)=W2

Standard Deviation(X)=W

),( N~x 2



68-95-99.7 Rule

68% of

the data

95% of the data

99.7% of the data



**The beauty of the normal curve:

No matter what Q and W are, the area between Q-W and

Q+W is about 68%; the area between Q-2W and Q+2W isabout 95%; and the area between Q-3W and Q+3W is

about 99.7%. Almost all values fall within 3 standard

deviations.



The Standard Normal Distribution

(Z)All normal distributions can be converted into

the standard normal curve by subtracting the

mean and dividing by the standard deviation:

W

Q!

X Z



Normal test

Test for the Mean of a Normal Population

If xi ( i =1,~,n) is a r.s of size n from N(Q, W2), then

H0 : Q = Q0 or

H0 : the sample has been drawn from the population

with mean Q0

H1 : Q { Q0 (two-tailed) or Q > Q0 (right-tailed) or

Q < Q0 (left-tailed)

Test Statistic: )1(0,N~

n

xZ

!

)n

,( N~x

2

Population Variance is Known



Test for the Mean«

Test Criteria

Depending on the alternative hypothesis selected, the test

criteria is as follows:

H1

Test RejectH

0 at levelof significance Eif

Q { Q0Two-tailed test Z> ZE /2

Q < Q0

Left-tailed test Z < -ZE

Q > Q0 Right-tailed test Z > ZE

ZE is the table value of Z at level of significance E.



Some Critical Values of Z

Level of

Significance

Critical value of Z

Two-tailed

test ZE /2Single tailed test

(ZE)

10% 1.645 1.280

5% 1.960 1.645

1% 2.580 2.330



Test for the Mean«

Population Variance is Unknown

2n

1i

i2 )x(x

1n

1s

! §

!

Large Sample (n>30)

W2 is estimated by sample variance i.e.,

= s2,

Normal test is then applied

Ö W

where



Hypothesis Testing Example

Test the claim that the true mean # of TV sets in

Indian homes is equal to 3.

(Assume = 0.8)

1. State the appropriate null and alternative

hypotheses

H0: = 3 H1: 3 (This is a two-tail test)

2. Specify the desired level of significance and the sample

size

Suppose that E = 0.05 and n = 100 are chosen for this

test



2.0.08

.16

100

0.8

32.84

n

XZ !

!

!

!

Hypothesis Testing Example

3. Determine the appropriate technique

is known so this is a Z test.

4. Determine the critical values

For E = 0.05 the critical Z values are±1.965. Collect the data and compute the test statistic

Suppose the sample results are

n = 100, X = 2.84 ( = 0.8 is assumed known)

So the test statistic is:

(continued)



Reject H0 Do not reject H0

6. Is the test statistic in the rejection region?

E = 0.05/2

-Z= -1.96 0

Reject H0 if

Z < -1.96 or

Z > 1.96;otherwise do

not reject H0

Hypothesis Testing Example(continued)

E = 0.05/2

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so the test

statistic is in the rejection region



6(continued). Reach a decision and interpret the result

-2.0

Since Z = -2.0 < -1.96, we reject the null hypothesis and

conclude that there is sufficient evidence that the mean

number of TVs in Indian homes is not equal to 3

Hypothesis Testing Example(continued)

Reject H0 Do not reject H0

E = 0.05/2

-Z= -1.96 0

E = 0.05/2

Reject H0

+Z= +1.96



From a class, 36 students were selected at random and their marks

in a subject out of 20 were observed. The mean and standard

deviation are 18.7 and 1.25. Test whether the mean marks of

students is 19.

Solution:

H0: The sample of students has been drawn from thepopulation with mean marks Q = 19

H1: Q { 19

n = 36, Q = 19, W = 1.25

Under H0,

440.1

3625.1

0.197.18!

!Z

H0

is accepted at 5% level of significance.

.x !

Test for the Mean«ANOTHER Example



Test for Difference of Means

Popn. I p N(Q1, ) Popn. II p N(Q2, ),

q q

n1 n2

H0 : Q1 = Q2

Test Statistic: Normal test

n

n

)(xxZ

2

22

1

21

2121

1x

2

1

2

2

2x

Population Variances are Known



Test for Difference of Means«Example

Information on two sets of samples regarding the expenditure

in Rs. per month per family

43.1581652

1!

61264132

2!

744!1x

78.516!2

x

n1= 42

n2= 32

Test whether the average expenditure per month per family is

equal.

H0 : Q1 = Q2 H1: Q1 { Q2

Z = 3.36

Z > 1.96, H0 is rejected. The average expenditure per month

per family in the two populations is not equal.



Under H0

222

21

21

21

2

2

2

1

2

1

21 if

n

1

n

1

xx

n

n

xxZ !!

!

!

Test for Difference of Means«

Population Variances are known & Equal



Test for Single Proportion

n p Sample size

x p Persons possessing the given attribute

p Observed proportion of successes

P p Population proportion, Q = 1- P

H0: P = P0

H1: P { P0 or P > P0 or P < P0


(0,1) N~ PQ/n

P- p Z !

pn

x!



Test for Single Proportion«

Test Criteria:

H1

TestReject H

0at level of

significance E if P { P

0Two-tailed test Z> ZE /2

P < P0

Left-tailed test Z < -ZE

P > P0

Right-tailed

test

Z > ZE



In order to test the conjecture of the management that 60%employees favour a new bonus scheme, a sample of 150

employees was taken. 55 employees favoured the new bonus

scheme.

Solution: n = 150, x = No. of employees favoured = 55

p = = 0.367

H0: P = 0.60

H1:P

{ 0.60

Z>2.58, Ho is rejected and it is concluded that 60% employees

do not favour the new bonus scheme.

n

x

5.825-0.40/150x0.60

0.60-0.367

PQ/n

P- p Z !!!

Test for Single Proportion«Example



Test for Difference of Proportions

Let x1 (x2) be the number of persons possessing a givenattribute A in random sample of size n1 (n2) from 1st (2nd)

population. Then sample proportions will be

Let P1 and P2 be the population proportions

H0: P1 = P2

H1: P1 { P2 or P1 > P2 or P1 < P2


2

22

1

11

n

x p ,

n

x p !!

( ,1)

n

QP

n

QP

)PP( p pZ

2

22

1

11

2121

!



Test for difference of proportions«

Test Statistic: Under H0: P1 = P2 = P (say)

Q =1-P

)n

1

n

1(PQ

p pZ

21

21

21

2211

nn

pn pnP



In a district, 450 persons were regular consumers of tea out of asample of 1000 persons. In another district, 400 were regular

consumers of tea out of a sample of 800 persons. Is there a

significant difference between the two districts as far as tea

drinking habit is concerned?

Solution: H0: P1 = P2 =P

H1: P1 { P2

Significant at 5% as calculated value is more than table value

(1.96). Reject H0

08.2

)8001

10001(0.47x0.53

5.045.0Z !

!

Test for difference of proportions«Example



A r.s x1,«,xn p N(Q, W2)

H0 : Q = Q0

H1 : Q { Q0 or Q > Q0 or Q < Q0

Test Statistic: t - test

The null hypothesis is accepted or rejected accordingly.

1n0 t~ns/

xt

§

n

1iix

n

1x 2

n

1ii

2 )x(x1n

1s

§

¡

Test for the Mean of a Normal PopulationSmall Sample (n < 30) and Population Variance is Unknown

t - tests



Test for the Mean«

Test Criteria:

H1

TestReject H

0at level

of significance E if

Q { Q0

Two-tailed test t> tn-1(E /2)

Q < Q0

Left-tailed test t < -tn-1(E)

Q > Q0

Right-tailed

test

t > tn-1(E)



Suppose the claim has been made that the height of adult males in a

college is different from what it used to be and we wish to test this

hypothesis. A campus wide survey made 20 years ago found that the

mean height of males was 69.5 in. To study this, a random sample of

15 males of the same age from current students was taken and their

height recorded.Solution:

H0: The average height is 69.5 in.

H1: The average height is more than 69.5 in.

Since |t| < 2.14 (value of t at 5% and 14 d.f), the mean height is 69.5 in.

4.07x ! 71.0s !

14t~1.27150.71/

5.694.07!

!t

Test for the Mean«Example

65.0, 67.5, 68.0,

68.5, 69.0, 69.5,

69.5, 70.0, 71.0,

71.5, 71.5, 72.5,

72.5, 74.5, 75.5



t-table

'



Test for the Difference of Two Population Means

Let be the sample mean of a sample of size n1 (n2) from

a population with mean Q1 (Q2).

H0 : Q1 - Q2 = H0

Test Statistic: Under H0

W2 is estimated from the sample

)x(x 21

2nn

21

021

21t~

n

1

n

1

xxt

2nn

1)s(n1)s(ns

21

222

2112

Population Variances are Unknown but Equal



In order to compare the reading ability of dyslexic children by

placing a blue plastic overlay on reading material with a clear

overlay, an experiment was conducted with 12 children in each

group and the scores are as follows:

Blue Overlay ± 70, 80, 90, 80, 50, 80, 70, 80, 70, 80, 80, 70

Clear Overlay ± 50, 40, 50, 50, 60, 60, 60, 40, 60, 70, 60, 80

Assess whether reading ability of dyslexic children improved by

placing a blue plastic overlay?

Test for the Difference«Example



Solution:

H0 : Q1 = Q2 H1 : Q1 > Q2

Since t > 2.074 (value of t at 5% and 22 d.f), The null hypothesis

is significant and thus not accepted. This concludes that blue

plastic overlays on reading material improves visual processing

and provides immediate improvement in the reading ability of

dyslexic children.

,57x1! ,7.56x

2!

,100s2

1!

22t~4.15t !

Test for the Difference«Example

3.133s2

2!



Paired t-test for Difference of Means

When n1 = n2 = n

Two samples are not independent (paired)

Let (xi, yi), i=1,..,n be a r.s from a B.N. population

Let di = xi - yi

H0 : Q1 - Q2 = Q0


1-n0 t~ ns/

dt

§¢

n

1iid

n

1 d §

¢

n

1i

2i

2 )dd1n

1s



In a study to know the effect of training to 8 researchers on a

particular subject, following are the pre and post training

scores:

Before training: 49 53 51 52 47 50 52 53

After training: 52 55 52 53 50 54 54 53

Can we conclude that training has improved the performance

of the researchers?

Paired t-test«Example



Solution:

H0 : QB = QA H1 : QB < QA

Researcher No. Before After di = xi - yi

(xi

) (yi

)

1. 49 52 -3 9

2. 53 55 -2 4

3. 51 52 -1 1

4. 52 53 -1 1

5. 47 50 -3 9

6. 50 54 -4 16

7. 52 54 -2 4

8. 53 53 0 0

2

id




t (5%,7 d.f.) = 1.90

H0 is rejected, so it is concluded that training has improved

the performance of the researchers.

-28

16- n

1 n

1ii !!! §

!

714.1)d(d

7

1s

n

1i

2i

2!! §

!

7t~4. 21.714/

2

ns/

|d||t| !!!




Let x1, x2,«,xn (nu2) be a r.s from N(Q, W2).

H0 :


when Q is known

when Q is unknown

20

2 !

2

n

2

0

in

1i

2

~

x

G¹¹ º

¸

©©ª

¨

!G §!

2

1-n

2

0

in

1i

2 ~

xxG¹¹

º

¸©©ª

¨ !G §

!

Chi-Square tests

Test for the variance of a normal population



Chi-Square tests« Example

The precision of an instrument, measured in terms of variance, is not less than 0.16. Given the 11 measurements

(2.3, 2.5, 2.3, 2.4, 2.7, 2.5, 2.6, 2.5, 2.7, 2.6, 2.5) on the

instrument, test the claim:

16.02!

16.02"

51.2!

Solution: H0:

H1:

1.1 2

2

01

2

!¹¹ º

¸

©©ª

¨

!G §!Tabulated value of G2 with 10 d.f. at 1% level is 2.5. Since

calculated value < tabulated, H0 is not significant and hence the

precision of the instrument is 0.16.



To test the discrepancy between the observed and the expected

frequency

H0 : the fitted distribution is a good fit

H1 : not a good fit

Test Statistic:

Oip Observed frequency of ith class

Ei p Expected frequency of ith class, i =1,«,n.

21-r -n

i

2

iin

1i

2 ~E

EOG

!G §!

Test of Goodness of Fit



Suppose four brands of cola are equally preferred by consumers.

It is of interest to know whether in the population of consumers,

the proportion of individuals preferring each brand is ¼.

To test this null hypothesis, 100 individuals were randomly

selected.

They were asked to taste the four brands without disclosing the

brand name and then declare their preference.

H0 : PA =PB = PC = PD = 0.25

Test of Goodness of Fit«Example



Since 3.44 < 7.82 (value of chi-square at 5% and 3 d.f), the result is

not significant and hence the proportion is same which is 1/4.

Frequenc

O served (Oi) pected ( i) Oi - i

2 25 -5

31 25

28 25 3

21 25 -4

3.44E

)E(O2

1i i

2

ii2

!

!G §!

Test of Goodness of Fit«Example



Contingency Table

Class A1 A2 A3

B1

n11

n21

n31

B2 n12 n22 n32

B3 n13 n23 n33

Test of Independence



H0: The attributes are independent

H1: They are not independent

Test Statistic:

Oij p Observed frequency

Eij p Expected frequency i =1,«,r j =1,«,s

H0 is rejected at level E if

21)-(c1)-(r

c

1 j

r

1i ij

2ijij2 ~

E

)E-(O G!G § §

! !

21)-(c1)-(r

2 G"G

Test of Independence«



From the following table, test the hypothesis that the test

result is related to the sex of the student:

Sex Pass Fail Total

Male 99 36 135

Female 20 5 25

Total 119 41 160

Test of Independence«Example 1



84.3%( 2 !

22 .492 !


Sex Pass Fail Total

Male 99

(100.4)

36

(34.6)

135

Female 20

(18.6)

5

(6.4)

25

Total 119 41 160

H0: Test result is independent of the sex of the student



The educational standard of adoptability of new innovationsamong 500 farmers are given below. Test whether the educational

standard has any impact on their adoptability of innovation.

H0: Adoptability is independent of the educational standard

Educational Standard

Illiterate Matric Graduat

e

Post

Graduate

Adopted 60

(110.2)

70

(38.0)

35

(20.9)

25

(20.9)

Not

Adopted

230

(179.8)

30

(62.0)

20

(34.1)

30

(34.1)

34.11%)1( 23 !82.7%)5( 23 !23

2 ~96.99 !




F- tests

Test for the comparison of two population variances

Popn. I p N(Q1, ) Popn. II p N(Q2, ),

q q

n1

n2

2

1

2

2

1

x2

1

s2

x2

2

s

§§

21 n

1 j

22 j

2

22

n

1i

21i

1

21 )x(x

1n

1s ,)x(x

1n

1s

§§21 n

1 j j

2

2

n

1ii

1

1 xn

1x ,x

n

1x



H0 :

Test Statistic:

The computed value of F is compared with the tabulated

value and the inference is drawn accordingly.

22

21 !

1n,1n1

1F ~

s

sF !

F- test«



The nicotine contents in mgm in two samples of tobacco

were found to be as follows:

Sample A: 24 27 26 21 25

Sample B: 27 30 28 31 22 36

Test whether the two samples have been taken from thepopulation with the same variability.

Solution: n1 = 6 , n2 = 5

H0

: , H1

:

Since F < 5.19 (value of F at 5% and 5, 4 d.f), the two

samples have the same variability

2

2

2

1

!2

2

2

1

W{W

4,1 F ~ 4.08s

sF !!

Example



Example: Z Test for



Example: Z Test for Proportion

A marketing companyclaims that it receives8% responses from its

mailing. To test thisclaim, a random sampleof 500 were surveyedwith 25 responses. Test

at the E = 0.05significance level.

Check:

n = (500)(.08) = 40

n(1-) = (500)(.92) = 460



Z Test for Proportion: Solution

E = 0.05

n = 500, p = 0.05

Reject H0 at E = 0.05

H0: = 0.08

H1: { 0.08

Critical Values:± 1.96

Test Statistic:

Decision:

Conclusion:

z 0

Reject Reject

.025.025

1.96

-2.47

There is sufficient

evidence to reject the

company¶s claim of 8%

response rate.

2.47

500

.08).08(1

.08.05

n

)(1

pZ !

!

!

T T

T

-1.96

Example: Two Tail Test



Example: Two-Tail Test(W Unknown)

The average cost of ahotel room in New Yorkis said to be $168 per

night. A random sampleof 25 hotels resulted inX = $172.50 and

S = $15.40. Test at theE = 0.05 level.( Assume the population distribution is normal)

H0: = 168

H1: { 168

E l S l ti



E= 0.05

n = 25

W is unknown, so

use a t statistic Critical Value:

t24 = ± 2.0639

Example Solution:Two-Tail Test

Do not reject H0: not sufficient evidence that true

mean cost is different than $168

Reject H0Reject H0

E/2=.025

-t n-1,/2

Do not reject H0

0

E/2=.025

-2.0639 2.0639

1.46

2515.40

168172.50

nS

Xt 1n !

!

!

1.46

H0: = 168

H1: { 168

t n-1,/2

testing of hypothesis-agbs

Documents