# TESTS OF HYPOTHESES.

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TESTS OF HYPOTHESES

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 ?

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

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?

TESTS OF HYPOTHESES NORMAL POPULATIONS NON-NORMAL POPULATIONS

NORMAL POULATIONS KNOWN POPULATION VARIANCE
UNKNOWN POPULATION VARIANCE

KNOWN POPULATION VARIANCE
Two Samples Z Tests Single Sample Z Tests

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

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

α = Probability of committing Type I error
Ho 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

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

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

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:

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 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

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

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%

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

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

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

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