1. TESTS OF HYPOTHESES

2. type II errors if the critical region is and true mean found to be 195
Example
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

3. 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

4. Tests on The VARIANCE

5. المعيار الأحصائى للأختبار
Statement of Test
Example 1
Consider the life data of the tires in the previous example. Can you conclude ,using α = 0.05, that the
standard deviation of the tire life exceeds 2500 km.? Find the P-Value of the test.
_______________________________________________________________________________________
Step 1
Statement of Test
Step 2
Process Data from experiments
المعيار الأحصائى للأختبار
Step 3: Build A Relevant TEST STATISTIC by means of which we can Test Hypothesis Ho

6. Since we Reject Ho, H1 (σ2> 6 250 000) is ACCEPTED, then
Step 4: Define The Zone of ACCEPTANCE
REJECT
1-α α
ACCEPT
Step 5: Perform the hypothesis test by Comparing with
Since, then we are in the REJECT zone,
then we REJECT Ho.
Step 6: CONCLUSION
Since we Reject Ho, H1 (σ2> ) is ACCEPTED, then
THERE IS ENOUGH EVIDENCE TO STATE THAT
THE TIRE LIFE STANDARD DEVIATION
IS GREATER THAN 2500 km

7. β 1-β Then, we Reject Ho absolutely
What is the P-Value of the test in the above example
P-VALUE
Then, we Reject Ho absolutely
_____________________________________________________________________
If it is , by some means, known that the true standard deviation of the life time of the tires
Is 4000 km. What is the power of the experiment with sample size 16 to DETECT
that difference between the hypothetical and the true.
N=16
N=24 β
1-β
Power of Test (with sample size =16 (#of degrees of Freedom=16-1))= 1 – β = 0.64
Increasing the sample size to 24 the Power of test will be: (#of degrees of Freedom=24-1=23
1 – β =

8. CONFIDENCE INTERVAL ON THE VARIANCE
α/2
α/2
1- α

9. Tests on The PROPORTIONS

10. Exceed 1% (proportion of non defective circuits exceeds 99%.)
Example 1
The Fraction of Defective Integrated Circuits produced in a process is studied.
A Random Sample of 50 circuits is tested, revealing 3 defectives. Is that result
Support the claim that the Proportion of defective circuits in the said process does Not
Exceed 1% (proportion of non defective circuits exceeds 99%.)
______________________________________________________________________
Here, the Number of Defective Circuits in a Sample of 50 Circuits is a Random
Variable distributed according to BINOMIAL distribution.
Step 1
Statement of test
Ho : p = 0.01
H1 : p > 0.01
Step 2
Experimental Data
Number of Defectives X = 3
Sample size N = 50
Step 3
Test Statistic
Since X is a Binomial Variable, we use the Binomial Distribution
RULE of DECISION by use of P-Value
Step 4
If P-Value ≤ REJECT Ho absolutely
If P-Value ≥ ACCEPT Ho absolutely
Otherwise it is up to concerned parties

11. Step 5: Perform the hypothesis test by applying the rule of P-Value
P-Value = which is near 0.01, then we REJECT Ho
Hence ACCEPT H1 : p > 0.01
Step 6: Conclusion
Since we Reject Ho, [ H1 : p > 0.01 ] is ACCEPTED, then
THERE IS ENOUGH EVIDENCE TO STATE THAT the PROPORTION of Defective
Circuits is MORE THAN 0.01.
__________________________________________________________________________________________________________
What are the Upper and Lower Limits of Defective Proportions
that render the Test INSIGNIFICANT (Accept Ho and thus Reject H1)?
The Upper and Lower Limits of p are as follows:

12. Tests on The RATIOS of VARIANCES

13. F - DISTRIBUTION Consider the following RANDOM VARIABLE: α =0.05
The numerator and the denominator are CHI-Square Variables with
Degrees of Freedom m and n respectively . α α
α =0.05 n m 2 5 10 20 19

14. Therefore, is a Chi-square variable with (n-1) degrees of freedom
Example 1
A fuel economy study was conducted on two German automobiles, Mercedes and Volkswagen
One 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
23.9
41.6
39.9
39.5
40.9
24.5
41.9
29.6
Is there any Evidence to support the claim that the variability in
Mileage performance is greater for a Volkswagen than for a Mercedes.
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
Therefore, is a Chi-square variable with (n-1) degrees of freedom
This is an F- Variable with N1 – 1 and N Degrees of Freedom

15. Since we Reject Ho, is ACCEPTED, then
Step 4: Define The Zone of ACCEPTANCE
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 Ho
Otherwise, we are not in a position to Reject Ho
= >> = 3.388, Therefore we REJECT Ho
Step 6: CONCLUSION
Since we Reject Ho, is ACCEPTED, then
THERE IS STRONG EVIDENCE TO STATE THAT The Variation in Fuel Mileage of Volkswagen is GREATER than that of MERCEDES