1. TESTS OF HYPOTHESES

2. Engineers and Managers are dealing with the following CASES Frequently
A Car Tires Producer is claiming that the
Mean Life time of a tire is more than km
A Sample of 16 tires are RANDOMLY selected
and tested for life.
The results of the experiments are given below
Do these results support the claim
of the producer or not ?

3. A Diet and aerobic program in a club
is claimed to reduce the weight by at least 7 kg.
Ten persons participated in the program and
The following results have been obtained
Person 1 2 3 4 5 6 7 8 9 10
Before 87 95
110 89 83 93 96
131
138
After 80 84 98 76 82 88
123
127
On the basis of the test results,
Is there any evidence to support the club claim

4. A Sample of 50 components is taken from
a production run. It is believed (hypothesized)
that the failure probability of the component
is On test 3 of them are found faulty.
Is this enough evidence to disbelieve the Statement concerning the failure Probability?

5. TESTS OF HYPOTHESES
NORMAL POPULATIONS
NON-NORMAL POPULATIONS

6. NORMAL POULATIONS KNOWN POPULATION VARIANCE
UNKNOWN POPULATION VARIANCE

7. KNOWN POPULATION VARIANCE
Two Samples Z Tests
Single Sample Z Tests

8. UNKNOWN POPULATION VARIANCE
More than Two Samples ANOVA Tests
Single Sample T Tests
Two Samples T Tests
Nearly Equal Population Variances
Non Equal Population Variances

9. Statements of Tests Ho: The BASIC Hypothesis to be tested
(whether Accept or Reject)
H1: The Alternative Hypothesis
( Negation of Ho )

10. α = Probability of committing Type I errorHo
True
False
Reject
Type I Error α
No Error
1- α
Accept
1- β
Type II Error β
α = Probability of committing Type I error
Level of SIGNIFICANCE
β = Probability of committing Type II error
1 – β = Power of Test

11. Tests of Hypothesis on the MEAN
Statements of Test
Ho: μ = μo
H1: μ ≠ μo Two-Sided Test OR
: μ > μo Upper –Sided Test OR
: μ < μo Lower –Sided Test

12. Determination of zone of Acceptance
Step 3
Determination of zone of Acceptance
REJECT
REJECT
H1 :
ACCEPT
REJECT
H1 :
ACCEPT
REJECT
ACCEPT
H1 :

13. Example 1
A Car Tires Producer is claiming that the Mean Life time of a tire is more than km
A Sample of 16 tires are RANDOMLY selected and tested for life. The results of the experiments are given below and the standard deviation is known to be 400 km
Do these results support the claim of the producer or not ? Take α = 0.06
Ho : μ = km
H1 : μ > 6000 km
Step 1: Statement of Test
Step 2: Experimental Data Processing
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
Since, population variance is known and is a Normal variable, then
The Standard Normal Variable Z is the relevant Test Statistic:

14. Since we accept Ho, H1 (μ > 60000) is RREJECTED, then
Step 4: Define The Zone of ACCEPTANCE
H1 : μ > km
REJECT
The Alternative Hypothesis H1 is
UPPER-SIDED, then the Rejection Zone is defined
As shown in the figure to the right
ACCEPT Zα
α = 0.06, then the( area to the left of Zα )= 0.94
Therefore , Zα = (from tables or Excel) α
Step 5: Perform the hypothesis test by Comparing Zo with Zα
If Zo > Zα, then we are in the REJECT zone, then we REJECT Ho
Otherwise, we are not in a position to Reject Ho
Zo = < Zα = 1.555, Therefore we ACCEPT Ho
Step 6: CONCLUSION
Since we accept Ho, H1 (μ > 60000) is RREJECTED, then
THERE IS NO ENOUGH EVIDENCE TO STATE THAT
THE TIRE LIFE IS LONGER THAN km

15. Example 2
The yield of a chemical process is being studied. From past experience, the standard
deviation is known to be 3%. The past FIVE days of plant operation have resulted in
The following yields: 91.6%, 88.75%, 90.8%, 89.95%, 91.3%. (use α = 0.05)
a) Is there evidence that the yield is not 90%?
b) Is there evidence that the yield is less than 90%?
c) Evaluate the P-Value of the test.
d) What sample size would be required to detect a true mean yield of 85% with probability
0.95.
e) What is the type II error probability , if the true mean yield is 92%?
___________________________________________________
Ho : μ = 90
H1 : μ ≠ 90
Step1: Statement of Test
Step 2: Experimental Data Processing
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho
Step 4: Define The Zone of ACCEPTANCE
H1 : μ ≠ 90
Reject
Reject
The Alternative Hypothesis H1 is
TWO-SIDED, then the Rejection Zones are
Symmetrically placed as shown in the figure
ACCEPT

16. Since we accept Ho, H1 (μ ≠ 90) is RREJECTED, then
α = 0.05, then (the area in the center part)= 0.95
And the Rejection areas on both sides, each = 0.025
Therefore , Zα/2 = Z at an area = to the left
of Zα/2 = (from tables or Excel)
0.975
0.025
Zα/2 = 1.96
Step 5: Perform the hypothesis test by Comparing Z o with Z α/2
Zo = < Zα/2 = 1.96, Therefore we ACCEPT Ho
Step 6: CONCLUSION
Since we accept Ho, H1 (μ ≠ 90) is RREJECTED, then
THERE IS NO ENOUGH EVIDENCE TO STATE THAT
THE MEAN YIELD of the chemical process is NOT 90%

17. Example 3
A manufacturer is interested in the output voltage of a power supply used in a PC.
Output voltage is assumed to be normally distributed. With standard deviation 0.25 volts. The
The manufacturer wishes to test Ho : μ = 5 volts against H1 : μ ≠ 5. using a sample of 9 units.
If the acceptance zone is 4.85 <= <= Find the value of the type I error α.
If the manufacturer wants to keep α = 0.05, where should be the acceptance region located?
Find the power of test for detecting a true output voltage of 5.1 volts.
_______________________________________________________________________________
Ho : μ = 5 volts
H1 : μ ≠ 5 volts
a) Statement of Test
1 - α
Reject
Reject
α/2
Accept
α/2
Acceptance Zone:
4.85 <= <= 5.15
4.85
5.15
- Zα/2
Zα/2
Acceptance Zone in case of α = 0.05

18. β β β β is Probability of Accepting Ho when it is false
c) The power of test for detecting a true output voltage of 5.1 volts.
β is Probability of Accepting Ho when it is false β
Power of Test = 1 – β = = β μo μT β μo μT μT μo

19. type II errors if the critical region is and true mean found to be 195
Example 4
Foam height of a shampoo is normally distributed with standard deviation of 20 mm. Test the hypothesis
HO : μ = 175 and H1 : μ > 175 using results of n=10 samples. Evaluate probabilities of type I and
type II errors if the critical region is and true mean found to be 195
.Construct the operating characteristic curves using values of true mean of 178, 181, 184, 187, 190, 193, 196 and 199.
_____________________________________________________________________________________
The Critical Zone is
Then the Acceptance Zone is
Critical Zone
Reject
Accept Zone α
175
185

20. Operating Characteristic Curve
μTrue Z
beta
175
1.58
0.94
178
1.11
0.87
181
0.63
0.74
184
0.16
0.56
187
-0.32
0.38
190
-0.79
0.21
193
-1.26
0.10
196
-1.74
0.04
199
-2.21
0.01
202
-2.69
0.00
205
-3.16
208
-3.64
Operating Characteristic Curve