2
Review Point estimation: calculate the estimated standard error to accompany the point estimate of a population.
Interval estimation whatever the population, when the sample size is large, calculate the
100(1-a)% confidence interval for the mean
When the population is normal, calculate the 100(1-a)% confidence interval for the mean
Where is the obtained from t-distribution with n-1 degrees of
freedom.
ns /x
/ 2 / 2x z x zn n
/ 2 / 2
s sx t x t
n n
/ 2t
3
Review con.Test of Hypothesis
5 steps totally. Formulate the assertion that the experiment seeks to confirm as the alternative hypothesis
P-value calculation
the smallest fixed level at which the null hypothesis can be rejected.
4
Outline
Inference concerning two means
Design Issues – Randomization and Pairing
5
7.8 Inference concerning two means
In many statistical problems, we are faced with decision about the relative size of the means of two or more populations.
Tests concerning the difference between two means
Consider two populations having the mean
and and the variances of and
and we want to test null hypothesis1 2 1 2
1 2 Random samples of size 1 2n and n
6
Two Population Tests
TwoPopulations
Z Test(Large
sample)
t Test(Pairedsample)
Z Test
Proportion Variance
F Testt Test(Small
sample)
Paired
Indep.
Mean
Testing Two Means
Independent Sampling& Paired Difference Experiments
8
Two Population Tests
TwoPopulations
Z Test(Large
sample)
t Test(Pairedsample)
Z Test
Proportion Variance
F Testt Test(Small
sample)
Paired
Indep.
Mean
TwoPopulations
Z Test(Large
sample)
t Test(Pairedsample)
Z Test
Proportion Variance
F Testt Test(Small
sample)
Paired
Indep.
Mean
9
Independent & Related Populations
IndependentIndependent RelatedRelated
10
Independent & Related Populations
1. Different Data Sources
Unrelated
Independent
IndependentIndependent RelatedRelated
11
1. Different Data Sources
Unrelated
Independent
1. Same Data Source
Paired or Matched
Repeated Measures(Before/After)
IndependentIndependent RelatedRelated
Independent & Related Populations
12
1. Different Data Sources
Unrelated
Independent
2. Use Difference Between the 2 Sample Means
X1 -X2
1. Same Data Source
Paired or Matched
Repeated Measures(Before/After)
IndependentIndependent RelatedRelated
Independent & Related Populations
13
1. Different Data Sources
Unrelated
Independent
2. Use Difference Between the 2 Sample Means
X1 -X2
1. Same Data Source
Paired or Matched
Repeated Measures(Before/After)
2. Use Difference Between Each Pair of Observations
Di = X1i - X2i
IndependentIndependent RelatedRelated
Independent & Related Populations
14
Two Independent Populations Examples
1. An economist wishes to determine whether there is a difference in mean family income for households in 2 socioeconomic groups.
2. An admissions officer of a small liberal arts college wants to compare the mean SAT scores of applicants educated in rural high schools & in urban high schools.
15
Two Related Populations Examples
1. Nike wants to see if there is a difference in durability of 2 sole materials. One type is placed on one shoe, the other type on the other shoe of the same pair.
2. An analyst for Educational Testing Service wants to compare the mean GMAT scores of students before & after taking a GMAT review course.
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Thinking Challenge
1. Miles per gallon ratings of cars before & after mounting radial tires
2. The life expectancy of light bulbs made in 2 different factories
3. Difference in hardness between 2 metals: one contains an alloy, one doesn’t
4. Tread life of two different motorcycle tires: one on the front, the other on the back
Are They Independent or Paired?Are They Independent or Paired?
Testing 2 Independent Means
18
Two Population Tests
TwoPopulations
Z Test(Large
sample)
t Test(Pairedsample)
Z Test
Proportion Variance
F Testt Test(Small
sample)
Paired
Indep.
Mean
TwoPopulations
Z Test(Large
sample)
t Test(Pairedsample)
Z Test
Proportion Variance
F Testt Test(Small
sample)
Paired
Indep.
Mean
19
Test The test will depend on the difference between the sample means and if both samples come from normal population with known variances, it can be based on the statistic
1 2X X
1 2
1 2
( )X X
X XZ
20
TheoremIf the distribution of two independent random variables have the mean and
and the variance and , then the distribution of their sum (or difference) has the mean (or ) and the variance
1 21 2
1 2 1 2 2 21 2
1
22 1
1X n
2
22 2
2X n
Two different sample of size
21
Statistic for test concerning different between two means
1 2
2 21 2
1 2
( )X XZ
n n
Is a random variable having the standard normal distribution.
Or large samples
1 2
2 21 2
1 2
( )X XZ
S Sn n
22
Criterion Region for testing 1 2
Alternative hypothesis
Reject null hypothesis if
1 2 Z z
Z z
/ 2 / 2Z z or Z z
1 2
1 2
23
EX.To test the claim that the resistance of electric wire can be reduced by more than 0.05 ohm by alloying, 32 values obtained for standard wire yielded ohm and ohm , and 32 values obtained for alloyed wire yielded
ohm and ohm
Question: At the 0.05 level of significance, does this support the claim?
1 0.136x 1 0.004s
2 0.083x 2 0.005s
24
Solution1. Null hypothesis: 1 2 0.05
Alternative hypothesis 1 2 0.05
2. Level of significance: 0.05
3. Criterion: Reject the null hypothesis if Z > 1.645
4. Calculation: 2 2
0.136 0.083 0.052.65
(0.004) (0.005)
32 32
z
5. The null hypothesis must be rejected.
6. P-value: 1-0.996=0.04 < level of significance
25
Critical values
One-sided alternatives
Two-sided alternatives
-1.645
1.645
-1.96
1.96
-2.33
2.33
-2.575
2.575
0.05
0.01
26
Type II errorsTo judge the strength of support for the null hypothesis when it is not rejected.
Check it from Table 8 at the end of the textbook
2 21 22 21 2
1 2
n
n n
The size of two examples are not equal
27
Small sample size2-sample t test.
2 221 2 1 1 2 2
1 2
1 2
( ) ( 1) ( 1),
21 1p
p
X X n S n St where S
n nS
n n
28
Criterion Region for testing (Statistic for small sample )
Alternative hypothesis
Reject null hypothesis if
1 2
T t
T t
/ 2 / 2T t or T t
1 2
1 2
1 2
29
EX
Mine 1 Mine 28260 79508130 78908350 79008070 81408340 7920
7840
The following random samples are measurements of the heat-producing capacity of specimens of coal from two mines
Question: use the 0.01 level of significance to test where the difference between the means of these two samples is significant.
30
Solution1. Null hypothesis: 1 2 0
Alternative hypothesis 1 2 0
2. Level of significance: 0.01
3. Criterion: Reject the null hypothesis if t > 3.25 or t< -3.25
4. Calculation:
5. The null hypothesis must be rejected. 6. P-value: 0.004 < level of significance 0.01
21 2 1
2 22
63408230, 7940, 15750
454600 63000 54600
10920, 13066.75 5 4
8230 7940114.31, 4.19
1 1114.31
5 6
p
p
x x s
s s
s t
31
Calculate it in Minitab
32
OutputTwo-sample T for Mine 1 vs Mine 2
SE N Mean StDev MeanMine 1 5 8230 125 56Mine 2 6 7940 104 43
Difference = mu (Mine 1) - mu (Mine 2)Estimate for difference: 290.00099% CI for difference: (133.418, 446.582)T-Test of difference = 0 (vs not =): T-Value = 4.19 P-Value = 0.02 DF = 9
33
SE mean: (standard error of mean) is calculated by dividing the standard deviation by the square root of n.
StDev: standard deviation .1s
34
Confidence interval100(1-a)% confidence interval for
2 21 1 2 2
1 2 / 21 2 1 2
( 1) ( 1) 1 1
2
n s n sx x t
n n n n
Where is based on degrees of freedom./ 2t 1 2 2n n
35
CI for large sample
2 21 2
1 2 / 21 2
s sx x z
n n
36
Matched pairs comparisonsQuestion: Are the samples independent in the application of the two sample t test?For instance, the test cannot be used when we deal with “before and after” data, where the data are naturally paired. EX: A manufacturer is concerned about the loss of weight of ceramic parts during a baking step. Let the pair of random variables denote the weight before and weight after baking for the i-th specimen.
( , )i iX Y
37
Statistical analysisConsidering the difference
This collection of differences is treated as random sample of size n from a population having mean
i i iD X Y
D: indicates the means of the two responses are the same0D
Null hypothesis:0 ,0: D DH
2
,0 21 1
( ), ,
1/
n n
i iD i i
D
D
D D DD
where D Sn nS n
38
EXThe following are the average weekly losses of worker-hours due to accidents in 10-industrial plants before and after a certain safety program was put into operation:
Before 45 73 46 124 33 57 83 34 26 17
After 36 60 44 119 35 51 77 29 24 11
Question: Use the 0.05 level of significance to test whether the safety program is effective.
39
Solution1. Null hypothesis: 0D
Alternative hypothesis 0D
2. Level of significance: 0.05
3. Criterion: Reject the null hypothesis if t > 1.833
4. Calculation: 5.2 04.03
4.08 / 10t
5. The null hypothesis must be rejected at level 0.05.
6. P-value: 1-0.9985=0.0015 < level of significance
40
Confidence intervalA 90% confidence interval for the mean of a paired difference.
Solution: since n=10 difference have the mean 5.2 and standard variance 4.08,
/ 2 / 2
s sx t x t
n n
4.08 4.085.2 1.83 5.2 1.83
10 10
4.0 6.4
D
Dor
41
7.9 Design issues: Randomization and Pairing
Randomization: of treatments prevents uncontrolled sources of variation from exerting a systematic influence on the response
Pairing: according to some variable(s) thought to influence the response will remove the effect of that variable from analysis
Randomizing the assignment of treatments within a pair helps prevent any other uncontrolled variables from influencing the responses in a systematic manner.