Post on 18-Jan-2016
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
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
548.1
28
1046.115121031.1031.1
2
21
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 > 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
Prof. Dr. Ahmed Farouk Abdul Moneim
Tests on The Difference of MEANS
The Two Populations are of NEARLY Equal Variances
The Two Populations are of Non Equal Variances
Prof. Dr. Ahmed Farouk Abdul Moneim
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
Prof. Dr. Ahmed Farouk Abdul Moneim
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
Step 2: Experimental Data Processing
68
10
112121
21
21
21
SS
nn
XX
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
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:
Prof. Dr. Ahmed Farouk Abdul Moneim
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
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
Prof. Dr. Ahmed Farouk Abdul Moneim
846.2OT 734.118,05.0 T From Tables
The Two Populations are of NONEQUAL Variances
Prof. Dr. Ahmed Farouk Abdul Moneim
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
Step 2: Experimental Data Processing
928.144.41
100912.067.242
2
VVV
MMM
NSX
NSX
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
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:
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
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
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
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