test of hypotheses: two sample
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Test of Hypotheses: Two Sample. Outlines: Inference on the difference in means of two normal distributions, variance known Inference on the difference in means of two normal distributions, variance unknown Paired t-test Inference on the variances of two normal distributions - PowerPoint PPT PresentationTRANSCRIPT
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Test of Hypotheses: Two Sample.Outlines: Inference on the difference in means of two
normal distributions, variance known Inference on the difference in means of two
normal distributions, variance unknown Paired t-test Inference on the variances of two normal
distributions Inference on the two population proportions
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Hypothesis testing
Engineers and scientist are often interested in comparing two difference conditions to determine whether either conditions produce a significant effect on the response.
Condition => Treatment Cause and effect relationship: the difference in
treatments resulted in the difference in response.
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Case I
Inference on the difference in means of two normal distributions, variance known
Hypothesis
Test Statistic
We should reject H0 if
0211
0210
:
:
H
H
2
22
1
21
0210
nn
XXZ
2/02/0 ZZorZZ
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Case I
Ex. A product developer is interested in reducing the drying time of a primer paint. Two formulations (old, new) of the paint are tested. The sd of drying time is 8 mins. Ten specimens are paint with formulation 1, and another 10 specimens are painted with formulation 2; the 20 specimens are paint in random order. The two sample average drying times are
What conclusion can be drawn about the effectiveness of the new ingredient, using =0.05
1. Parameter of interest is the difference in mean drying time,
2.
3. =0.05
4. Test statistic ,
5. reject H0 if
6. Calculate Z0 = 2.52
7. Conclusion : since , we reject H0 at 0.05 significance level. Adding new ingredient to the paint significantly reduces the drying time.
112,121 21 xx
645.12/0 ZZ
21
211210 :,0: HH
2
22
1
21
0210
nn
XXZ
112,121,10,8,0 2121210 xxnn
2/0 ZZ
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Case I
Sample size Using Operating Characteristic Curve (OC Curve)
If the value of , we can use the formula to calculate the value of n1 when n2 is fixed,
22
21
0
22
21
021 ||||
d n=n1=n2
21 nn
2221
21
22
21
// nnn
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Case I
Sample Size formulas
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Case I
Confidence Interval
The error in estimating µ1-µ2 by will be less than E at 100(1- )% confidence. Therequired sample size from each population is
21 xx
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Case I
Ex. Tensile strength tests were performed on two different grades of aluminum spars. From past experience with the spar manufacturing process and testing procedure, the standard deviations of the tensile strengths are assumed to be known. The data obtained are as follows:
Find a 90% confidence interval on the difference in mean strength µ1-µ2
5.1,5.74,12,1,6.87,10 222111 xnxn
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Case I.I
We can use the concept of Case I for the cases that we don’t know exactly about the population distribution (may be not normal distribution) and the number of sample size are large. (n1, n2 >=40)
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Inference on the difference in means of two normal distributions, variance unknown
Case 1: Hypothesis
Test Statistic
We should reject H0 if
Case II.I
0211
0210
:
:
H
H
2
)1()1(,
11 21
222
2112
21
0210
nn
SnSnS
nnS
XXT p
p
2,2/02,2/0 2121 nnnn ttortt
Pooled Estimator of variance
22
21
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Ex. Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table. Is there any difference between the mean yields? Use =0.05, and assume equal variances.
Case II.I
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Case II.I
1. Parameter of interest: µ1 and µ2, the mean process yield using C1, and C2
2. H0: µ1-µ2=0 or H0: µ1=µ2 , H1: µ1≠µ2
3. =0.05
4. Test statistic is
5. Reject H0 if
6. Calculate t0;
7. Conclusion H0 cannot be rejected. At the 0.05 level of significant, we do not have a strong evidence to conclude that C2 results in a mean yield differ from C1
2
)1()1(,
11 21
222
2112
21
0210
nn
SnSnS
nnS
XXT p
p
145.2145.2 14,025.0014,025.00 ttortt
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Inference on the difference in means of two normal distributions, variance unknown
Case 2: Hypothesis
Test Statistic
We should reject H0 if
Case II.II
0211
0210
:
:
H
H
2
2
1
1
021*0
nS
nS
XXT
,2/0,2/0 ttortt
Degree of freedom
22
21
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Case II.II
Ex. Arsenic concentration in public drinking water supplies is a potential health risk. An article in the Arizona Republic reported drinking water arsenic concentration in parts per billion (ppb) for 10 metropolitan Phoenix communities and 10 communities in rural Arizona.
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Case II.II
22
21
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Case II.II
1. Parameter of interest: µ1 and µ2, the mean arsenic concentration of two regions
2. H0: µ1-µ2=0 or H0: µ1=µ2 , H1: µ1≠µ2
3. =0.05
4. Test statistic:
5. We should reject H0 if
6. Compute t0
2
2
1
1
021*0
nS
nS
XXT
,025.00,025.00 ttortt
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Case II.II
7. Conclusion: t0<t0.025,13, we reject H0. There is evidence to conclude that the mean arsenic concentration in the drinking water in rural Arizona is differ from the mean arsenic concentration in metropolitan Phoenix. Furthermore, the mean arsenic is higher in rural of Arizona. P value is approximate P=0.016
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Case II
Sample size can be approximated by OC curves Only for the case that 1= 2
Where and
Ex.
2|| 0
d 12* nn
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Case III: Paired t- test
A special case of the two-sample t-test. This test is used when the observations on the two populations of interest are collected in pairs.
Each pair of observations is taken under homogeneous condition.
Ex. We are interested in comparing two different types of tips for a hardness-testing machine.Tip1 Tip2
Sheet Metal
Comparing the depth of the depression caused by the tips
Tip1 Tip2Pair t-test
2 sample test
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Case III: Paired t-test
Paired t-test Hypothesis
Test Statistic
We should reject H0 if
0211
0210
:
:
H
H
01
00
:
:
D
D
H
H
nS
DT
D
00
1,2/01,2/0 nn ttortt
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Case III: Paired t-test
Ex. An article in the journal of Strain Analysis compares several methods for predicting the shear strength for steel plate girders. Data for two of these methods, Karlsruhe and Lehigh procedures, when applied to nine specific girders are shown in table. We wish to determine whether there is any difference between the two methods.
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Case III: Paired t-test
1. Parameter of interest: the difference in mean shear strength between the two methods µD=µ1-µ2
2. H0: µD=0, H1: µD≠0
3. =0.05
4. Test statistic:
5. We should reject H0 if
6. Calculate t0
7. T0 =6.08>2.306, we conclude that the strength prediction methods yield different results. Specifically, the data indicate that the Karlsruhe method procedures, on the average, higher strength predictions than does the Lehigh method. P value for t0 = 6.08 is P=0.0003
nS
DT
D
00
306.2306.2 8,025.008,025.00 ttortt
08.691351.0
02739.00
t
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Case III: Paired t-test
Confidence Interval
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Case III: Paired t-test
Ex
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Inference on the variance two normal distribution Hypothesis
Test Statistic
We should reject H0 if
211
210
:
:
H
H
22
21
0 S
SF
1,1,2/101,1,2/0 2121 nnnn fforff
uvfvu
f,,
1,,1
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Inference on the variance two normal distribution Ex
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Inference on the variance two normal distribution
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Inference on the variance two normal distribution Sample size: can be approximated by OC
curve Only for the case that n1=n2=n Where Ex
2
1
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Inference on the variance two normal distribution Confidence Interval on the ratio of two
variances
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Inference on the variance two normal distribution Ex.
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Inference on the variance two normal distribution
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Test on two population proportion Hypothesis
Test Statistic
We should reject H0 if
211
210
:
:
ppH
ppH
21
21
21
210
ˆ,
)11
)(ˆ1(ˆ
ˆˆ
nn
XXP
nnPP
PPZ
2/02/0 zzorzz
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Test on two population proportionEx.
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Test on two population proportion
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Test on two population proportion Type II error
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Test on two population proportion Sample size
For one sided, replace /2 by
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Test on two population proportion Confidence Interval on the difference in
Population proportions
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Test on two population proportion Ex
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Test on two population proportion
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Homework
Using Minitab program to find the conclusion of these problems.
1. An article in solid state technology describes an experiment to determine the effect of the C2F6 flow rate on the uniformly of the etch on a silicon wafer used in integrated circuit manufacturing. Data for two flow rate are as follows:
a) Does the C2F6 flow rate affect average etch uniformity? Use =0.05
b) What is the P-value for the test in a)
c) Does the C2F6 flow rate affect the variability in etch uniformity? Use =0.05
d) Draws box plots to assist in the interpretation of the data from this etch uniformity.
C2F6 flow rate
observation
1 2 3 4 5 6
125 2.7 4.6 2.6 3.0 3.2 3.8
200 4.6 3.4 2.9 3.5 4.1 5.1
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Homework
2. A computer scientist is investigating the usefulness of two different design languages in improving programming task. Twelve expert programmers, familiar with both languages, are asked to code a standard function in both language, and the time (in minutes) is recorded. The data follow:
a) Is the assumption that the difference in coding time is normally distributed reasonable?
b) Find P-value for the test in a)
c) Find a 95% confidence interval on the difference in mean coding times. Is there any indication that one design language is preferable?