Download presentation

Presentation is loading. Please wait.

Published byCayden Warhurst Modified about 1 year ago

1
TESTS OF HYPOTHESES 1

2
Engineers 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 60 000 km A Sample of 16 tires are RANDOMLY selected and tested for life. The results of the experiments are given below 60613 59836 59154 60252 59784 60221 60311 50040 60545 60257 60000 59997 69947 60135 60221 60523 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 Person12345678910 Before879511089839396110131138 After8084988476828898123127 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 0.01. 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 Single Sample Z TestsTwo Samples Z Tests

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

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

10
HoTrueFalse Reject Type I Error α No Error 1- α Accept No Error No Error 1- β Type II Error β α = Probability of committing Type I error Level of SIGNIFICANCE Level of SIGNIFICANCE β = Probability of committing Type II error β = Probability of committing Type II error 1 – β = P ower of Test

11
Tests of Hypothesis on the MEAN Statements of Test H o : μ = μ o H o : μ = μ o H 1 : μ ≠ μ o Two-Sided Test OR : μ > μ o Upper –Sided Test OR : μ < μ o Lower –Sided Test

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

13
A Car Tires Producer is claiming that the Mean Life time of a tire is more than 60 000 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 Example 1 60613 59836 59154 60252 59784 60221 60311 50040 60545 60257 60000 59997 69947 60135 60221 60523 Do these results support the claim of the producer or not ? Take α = 0.06 Step 1: Statement of Test Ho : μ = 60 000 km H1 : μ > 6000 km 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
Step 4: Define The Zone of ACCEPTANCE The Alternative Hypothesis H1 is UPPER-SIDED, then the Rejection Zone is defined As shown in the figure to the right H1 : μ > 60 000 km α ZαZα ACCEPT 0 REJECT α = 0.06, then the( area to the left of Zα )= 0.94 Therefore, Zα = 1.555 (from tables or Excel) 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 = 1.1475 < 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 60 000 km

15
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%? ___________________________________________________ Example 2 Step1: Statement of Test Ho : μ = 90 H1 : μ ≠ 90 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 The Alternative Hypothesis H1 is TWO-SIDED, then the Rejection Zones are Symmetrically placed as shown in the figure H1 : μ ≠ 90 ACCEPT Reject

16
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 = 0.975 to the left of Zα/2 = 1.96 (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 = 0.3578 < 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
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. a)If the acceptance zone is 4.85 <= <= 5.15. Find the value of the type I error α. b)If the manufacturer wants to keep α = 0.05, where should be the acceptance region located? c)Find the power of test for detecting a true output voltage of 5.1 volts. _______________________________________________________________________________ Example 3 a) Statement of Test Ho : μ = 5 volts H1 : μ ≠ 5 volts Acceptance Zone:4.85 <= <= 5.15 4.855.15 Accept Reject 1 - α α/2 Acceptance Zone in case of α = 0.05 - Zα/2Zα/2

18
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 Power of Test = 1 – β = 1 - 0.2741 = 0.7259 μoμo μTμT β μoμo μoμo μTμT μTμT β β

19
Foam height of a shampoo is normally distributed with standard deviation of 20 mm. Test the hypothesis H O : μ = 175 and H 1 : 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. _____________________________________________________________________________________ Example 4 The Critical Zone isThen the Acceptance Zone is Reject α Critical Zone Accept Zone 175185

20
μ True Zbeta 1751.580.94 1781.110.87 1810.630.74 1840.160.56 187-0.320.38 190-0.790.21 193-1.260.10 196-1.740.04 199-2.210.01 202-2.690.00 205-3.160.00 208-3.640.00 Operating Characteristic Curve

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google