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Presented By:• Kapil Jain 2013PIE5199• Charan SIngh 2012PMM5004

Submitted to:Prof. Rakesh Jain

Dated: 20-Nov-2013

HYPOTHESES CONCERNING PROPORTIONS

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Introduction of Proportions

Hypotheses Concerning One Proportion with Example

Hypotheses Concerning Two Proportions with Example

Analysis of Contingency Table with Example

References

Overview

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For estimating a proportion of occurrences in a population, Statisticians use Sample.

Sample Discrete-Binomial Distribution.

Sample size increases, use Continuous Normal Distribution.

Introduction of Proportions

X

N

x

n

PopulationSample

Population, p = n

x

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Binomial Distribution:Mean, µ=n pStandard Deviation, σ= √(n p q)

Mean of sampling Distribution of the proportion

µ = (n p)/ n = pStandard Deviation, σ=

n is large, for large sample Confidence Interval for p :

Where α = level of significance

n= Number of trailsp= probability of successq= 1-p= probability of a failure

nnx

nx

Zn

xp

nnx

nx

Zn

x11

22

n

pq

n

npq

5

Now, Error, E= p-

So, Maximum Error, E max= p-

Physical SignificanceU.S.A. Government estimates proportion of

unemployed people by sampling procedure.It is quite beneficial in Acceptance Sampling.

n

ppZ

nnx

nx

Z)1(

12/2/

n

x

n

x

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In many methods, NULL Hypotheses Proportion equals some specified constant(po)

Statistic for large Sample Test Concerning p

Critical Region for testing p= po (Large Sample)

Hypotheses Concerning One Proportion

oo pnp

npXZ

o

1

Alternative Hypothesis

Reject Null Hypothesis if

p < p0 Z < -zα

p > p0 Z > zα

p ≠ p0 Z < -zα/2

P ≠ p0 Z > zα/2

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Example-A manufacturer of submersible pumps claims that at most 30% of the pumps require repairs within the first 5 years of operation. If a random sample of 120 of these pumps includes 47 which required repairs within the first 5 years, test the NULL hypotheses p=0.30 against the alternative hypotheses p > 0.30 at the 0.05 level of significance.

Solution- 1. NULL Hypotheses: p=0.30

Alternative Hypotheses: p > 0.30

2. Level of Significance: α = 0.05

3. Criterion: Reject the NULL hypotheses, If Z > 1.645 (from table)

Where

Hypotheses Concerning One Proportion

oo pnp

npXZ

o

1

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4. Calculations: Substituting x=47, n=120 and po=0.30 into the formula, we get

5. Decision: Since Z=2.19 is greater than 1.645, We reject the NULL hypotheses at level 0.05. In other words, there is sufficient evidence to conclude that the proportion of repairable pumps are greater than 0.30 or 30%.

19.230.0130.0120

)30.0(12047

.

Z

-4 -3 -2 -1 0 1 2 3 41.645

2.19

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When we compare Two different products, whether proportion of a given process remains constant from day by day.

Statistic for test concerning difference between Two Proportions:

Where Pooled Estimator

Large Sample Confidence Interval for the difference of two Proportions:

Hypotheses Concerning Two Proportion

21

2

2

1

1

11ˆ1ˆ

nnpp

nx

nx

Z

21

21

ˆnn

xxp

2

2

2

2

2

1

1

1

1

1

2/

2

2

1

1

11

nnx

nx

nnx

nx

Zn

x

n

x

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Example-A Study showed that 64 of 180 persons who saw a photocopying machine advertised during the telecast of a baseball game & 75 of 180 other persons who saw it advertised on a variety show remembered the brand name 2 hours later. Use the Statistic to test at the 0.05 level of significance whether the difference between the corresponding sample proportions is significant.

Solution- 1. NULL Hypotheses: p1=p2

Alternative Hypotheses: p1 ≠ p2 2. Level of Significance: α = 0.05

3. Criterion: Reject the NULL hypotheses, If Z > 1.96 & Z <-1.96(from table)

Where

Hypotheses Concerning Two Proportions

21

2

2

1

1

11ˆ1ˆ

nnpp

nx

nx

Z

11

4. Calculations: Substituting x1=64, n1=180, x2= 75and n2=180 into the formula, we get

5. Decision: Since Z=1.191 is lesser than 1.96, We are failed to reject the NULL hypotheses at level 0.05. In other words, there is sufficient evidence to conclude that the difference between the corresponding sample proportions is not significant.

191.1

1801

1801

614.0386.0

18075

18064

Z

-4 -3 -2 -1 0 1 2 3 41.96

386.0180180

7564ˆ

p

-1.96

1.191

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Analysis of Contingency TablePractically, We have to compare more than two sample

proportions, so we arrange the statistic in the form of Contingency Table

Expected Frequency fe/eij=

Sample 1

Sample 2

. . . . . . . . . .

Sample k Total

Success x1 x2 . . . . . . . . . .

xk x

Failure n1-x1 n2-x2 . . . . . . . . . .

nk-xk n-x

Total n1 n2 . . . . . . . . . .

nk n

nTotalGrand

TotalColumnXTotalRow

,.

....

x1

Observed Frequency,

fo/Oij

A Chi Square Statistic, e

eo

f

ff

e

eo

22

2

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Example-A large electronics firm that hires many workers with disabilities wants to determine whether their disabilities affect such workers performance. Use the level of significance α=0.05 to decide on the basis of the sample data shown in the following table whether it is reasonable to maintain that the disabilities have no effect on the worker’s performance:

Analysis of Contingency Table

Performance

Above average Average Below average

Blind 21 64 17

Deaf 16 49 14

No disability 29 93 28

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Solution- 1. NULL Hypotheses: p1=p2=p3

Alternative Hypotheses: p1 ≠ p2 ≠ p3 2. Level of Significance: α = 0.05

3. Criterion: Reject the NULL Hypotheses, If χ2 > 11.143 & χ2<-11.143, the value of for (3-1)(3-1)=4 degree of freedom, where χ2 is given by the formula.

4. Calculation: Calculating first the expected frequency for all the cells.

Total

21 64 17 102

16 49 14 79

29 93 28 150

66 206 59 331

e11=(102x66)/331= 20.33

e12=(102x206)/331=63.48

e13=(102x59)/331=18.18

e21=(79x66)/331= 15.75

e22=(79x206)/331=49.166

e23=(79x59)/331=14.08

e31=(150x66)/331= 29.90

e32=(150x206)/331=93.35

e33=(150x59)/331=26.73

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16.49

16.4949

75.15

75.1516

18.18

18.1817

48.63

48.6364

33.20

33.2021 222222

1958.0

73.26

73.2628

35.93

35.9393

90.29

90.2929

08.14

08.1414 2222

5. Decision: Since χ2=0.1958 before 11.143, we are failed to reject NULL hypotheses. We conclude that the disabilities have no effect on the worker’s performance.

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Probability & Statistics for Engineers, Eighth Edition: Richard A. Johnson

Chapter-10, Inferences Concerning Proportions.

Statistics for Management, Seventh Edition: Richard I. Levin, David S. RubinChapter-8,9-Testing Hypotheses: One Sample & Two Sample Test.

references

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