tests of hypotheses of populations with known variance two sample tests on the difference of means...

TESTS OF HYPOTHESESOF POPULATIONS

WITH KNOWN VARIANCE

TWO SAMPLE Tests on The Difference of Means Tests on The Ratio of Variances

Tests on The Difference of MEANS

Example 1Two formulations of paints are tested for Drying Time. Formulation1 is the standard chemistry, and formulation2 has a new drying ingredient that reduce the drying time. The standard deviationOf the drying time is 8 minutes. Ten specimens are taken from each formulation. The Two Sample Average drying time are 121 and 112 minutes respectively. What conclusions can be drawn aboutThe effectiveness of the new ingredient. Evaluate the P-Value of the test____________________________________________________________________________________

Step 1: Statement of Test Ho : μ1 = μ2 H1 : μ1 > μ2

Step 2: Experimental Data Processing

8

10

112121

21

21

21

nn

XX

Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho

516.2

28

10112121

2

22

1

21

2121

nn

XXZO

Step 4: Define The Zone of ACCEPTANCE

The Alternative Hypothesis H1 is UPPER-SIDED, then the Rejection Zone is definedAs shown in the figure to the right

H1 :μ1 > μ2

α

ACCEPT

0

REJECT

The P-Value of a Test

Prof. Dr. Ahmed Farouk Abdul Moneim

In order to decide whether ACCEPT or REJECT Ho WITHOUT HAVING αThe P-Value should be calculated as follows depending on the Statement of the Test

H1 : μ ≠ μo

H1 : μ > μo

H1 : μ < μo

Z o

½ P-ValueΦ(Z o)

OZValueP 12

Z o

P-Value

OZValueP

OZValueP 1

Z o

P-Value

RULE of DECISION by use of P-Value

If P-Value ≤ 0.01 REJECT Ho absolutely

If P-Value ≥ 0.1 ACCEPT Ho absolutely

Otherwise it is up to concerned partiesProf. Dr. Ahmed Farouk Abdul Moneim

Step 5: Perform the hypothesis test by evaluating of the P-Value of the test

01.0006.0994.01)516.2(1 1 OZValueP

Step 6: CONCLUSION

Since we Reject Ho, H1 (: μ1 > μ2 )) is ACCEPTED, thenTHERE IS ENOUGH EVIDENCE TO STATE THATTHE NEW INGERIEDNT IS EFFECTIVE IN REDUCING DRYING TIME OF THE PAINT

P-Value

Zo = 2.516

Example 2If in Example 1, the measure of Effectiveness of the new ingredient is preset to be that the drying time

of new formulation2 is 97% of that of standard formulation1. Do the experimental result support the claim that the new ingredient is effective in reducing the drying time.If by some means the true difference between the two drying times is 1.5 minutes, Evaluate the type II error. What sample size that render the power of test in detecting the true differenceMore than 90%.___________________________________________________________________________________

Step 1: Statement of Test Ho : μ1 = (1/0.97) μ2 Ho: μ1 = 1.031 μ2 H1 : μ1 > 1.031 μ2 Step 2: Experimental Data Processing

8

10

112121

21

21

21

nn

XX


548.1

28

1046.115121031.1031.1

2

21

1

21

2121

nn

XXZO



H1 :μ1 > 1.031μ2

P-Value0

REJECT

Z o

0.1 061.0939.01 1 OZValueP

We may Reject Ho with a significance error of 0.061

Step 5: Perform the hypothesis test We REJECT Ho and then ACCEPT H1

Step 6: CONCLUSION

Since we Reject Ho, H1 (: μ1 > 1.031 μ2 )) is ACCEPTED, then THERE IS ENOUGH EVIDENCE TO STATE THAT THE NEW INGERIEDNT IS EFFECTIVE IN REDUCING DRYING TIME OF THE PAINT

δ > 0 2

2/

ZZN

δ < 0 2

12/

ZZN

N

Z 2/1

N

Z 2/

OT

Type II Error β

3203.31

028.4

8281.1524.1||

22

N

ZZN O

low! very 473.0 test ofpower 527.0

8

10028.4524.111

N

Z

028.4112*031.11215.1

differenceedHypothesizdifferenceTrue

In order to raise the power of test to be 90% , the Sample size should be:

δ < 0 2

12/

ZZN

N

Z 2/1

TESTS OF HYPOTHESESOF POPULATIONS

WITH UNKNOWN VARIANCE

TWO SAMPLE Tests on The Difference of Means


Tests on The Difference of MEANS

The Two Populations are of NEARLY Equal Variances

The Two Populations are of Non Equal Variances


2

11

21

222

2112

NN

SNSNSP

2 21 NN

2

11

2

22

1

2

1

21

2

2

22

1

21

N

NS

N

NS

N

S

N

S

M

M

The Two Populations are of Nearly Equal

Variances


Example 1Two formulations of paints are tested for Drying Time. Formulation1 is the standard chemistry, and formulation2 has a new drying ingredient that reduce the drying time. . Ten specimens are taken from each formulation. The Two Sample Average drying time are 121 and 112 minutes respectively. The Two Samples standard deviations are 8 and 6 minutes respectively.What conclusions can be drawn about The effectiveness of the new ingredient. If we assume that variances of populations are nearly equal (α=0.05) ____________________________________________________________________________________

Step 1: Statement of TestHo : μ1 = μ2 H1 : μ1 > μ2


68

10

112121

21

21

21

SS

nn

XX


Since the Two populations are of Nearly equal variances then, evaluating the Pooled Variance and Pooled degrees of Freedom of the two samples as follows:

182

07.7505018

)3664(9

2

11

21

21

222

2112

NN

SNN

SNSNS PP

846.2

10

207.7

112121

11

21

2121

nnS

XXT

P

O

The Test Statistic in this case will take the form:




H1 :μ1 > μ2

α

ACCEPT

0

REJECT

Tα, ν

Step 5: Perform the hypothesis test by Comparing To with Tα, ν

To = 2.846 > T0.05, 18 =1.734 Therefore the Test Result is SIGNIFICANT

we REJECT Ho

Step 6: CONCLUSION

Since we Reject Ho, H1 (: μ1 > μ2 )) is ACCEPTED, thenTHERE IS ENOUGH EVIDENCE TO STATE THATTHE NEW INGERIEDNT IS EFFECTIVE IN REDUCING DRYING TIME OF THE PAINT


846.2OT 734.118,05.0 T From Tables

The Two Populations are of NONEQUAL Variances


Example 1A fuel economy study was conducted on two German automobiles, Mercedes and VolkswagenOne vehicle of each brand was selected, and the mileage performance was observed for 10 tanks of fuel in each car. The data are as follows:

Mercedes Volkswagen

24.7 24.9 41.7 42.8

24.8 24.6 42.3 42.4

24.9 23.9 41.6 39.9

24.7 24.9 39.5 40.9

24.5 24.8 41.9

Do the data support the claim that the mean mileage for the Volkswagen is at least 15 Km pg higher than that for the Mercedes (α = 0.04)

Step 1: Statement of Test


928.144.41

100912.067.242

2

VVV

MMM

NSX

NSX


15:

15:

1

MV

MVO

H

H

Since the Two populations are of NONEQUAL variances then the combined Degrees of freedom will be:

1028.92

110

100912.0

19

928.1

10

0912.0

9

28.1

2

11

22

2

2222

222

M

M

M

V

V

V

M

M

V

V

N

NS

N

NS

N

S

N

S

56.4

10

0912.0

9

28.1

1567.2444.4122

M

M

V

V

MVMV

O

n

S

n

S

XXT

The Test Statistic in this case will take the form:



α

ACCEPT

0

REJECT

Tα, νStep 5: Perform the hypothesis test by Comparing To with Tα, ν

To = 4.56 > > T0.025, 10 = 2.228 Therefore the Test Result is SIGNIFICANT

we REJECT Ho

Step 6: CONCLUSION

15:1 MVH

Since we Reject Ho, H1 (: μ1 > μ2 )) is ACCEPTED, thenTHERE IS ENOUGH EVIDENCE TO STATE THATTHE VOLKSWAGEN HAS GREATER MILEAGE THAN MERCEDES BY 15 MPG

Tests on The RATIOS of VARIANCES

F - DISTRIBUTION

Consider the following RANDOM VARIABLE:

n

mfn

m

nm 2,

2,

,,

The numerator and the denominator are CHI-Square Variables with

Degrees of Freedom m and n respectively .

nmf ,, nmf ,,1

αα

nmnm ff

,,,,1

1

α =0.05 n

m 2 5 10 20

2 19 5.786135 4.102821 3.492828

5 19.29641 5.050329 3.325835 2.71089

10 19.3959 4.735063 2.978237 2.347878

20 19.44577 4.558131 2.774016 2.124155

Example 1A fuel economy study was conducted on two German automobiles, Mercedes and VolkswagenOne vehicle of each brand was selected, and the mileage performance was observed for 10 tanks of fuel in each car. The data are as follows:

Mercedes Volkswagen

24.7 24.9 41.7 42.8

24.8 24.6 42.3 42.4

24.9 23.9 41.6 39.9

24.7 24.9 39.5 40.9

24.5 24.8 41.9 29.6

Is there any Evidence to support the claim that the variability inMileage performance is greater for a Volkswagen than for a Mercedes.

Step 1: Statement of Test

Step 2: Experimental Data Processing10 28.1

10 0912.02

2

VV

MM

NS

NS


221

22

:

:

MV

MVO

H

H

035.140912.0

28.12

2

M

VO S

Sf

This is an F- Variable with N1 – 1 and N2 - 1 Degrees of Freedom

1

1

11

11

/

/2

1,

21,

2

2

2

2

22

22

M

NM

V

NV

MM

MM

VV

VV

MM

VVO

N

N

NSN

NSN

S

Sf

M

V

222

2

1

2

2

2

2

2

2

2

2

22

2

2

2

22

11

2

2

2

22

1 1

22

1

2

1

22

1

22

)1()1(

22)1(

But 2)1(

2)1(

1

1

ZnZnZ

n

xxSn

n

x

n

xxxn

n

xxSn

xnxxx

n

xxSn

xxxxxxxxSn

xxn

S

n

i

i

ii

n

ii

n

ii

i

n

i

n

iii

n

ii

n

ii

n

ii

Therefore, is a Chi-square variable with (n-1) degrees of freedom

2

2)1(

Sn

Step 4: Define The Zone of ACCEPTANCE 221 : MVH

388.3 8,9,05.0 f

α =0.05

Step 5: Perform the hypothesis test by Comparing f o with

If f o > , then we are in the REJECT zone, then we REJECT HoOtherwise, we are not in a position to Reject Ho

= 14.035 >> = 3.388, Therefore we REJECT Ho

Step 6: CONCLUSION

Since we Reject Ho, is ACCEPTED, thenTHERE IS STRONG EVIDENCE TO STATE THAT The Variation in Fuel Mileage of Volkswagen is GREATER than that of MERCEDES

8,9,05.0 f

8,9,05.0 fOf

221 : MVH

8,9,05.0 f

tests of hypotheses of populations with known variance two sample tests on the difference of means...

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