Chapter 9: Inferences for Two –Samplesmjun/lecture19.pdf · Chapter 9: Inferences for Two...
Transcript of Chapter 9: Inferences for Two –Samplesmjun/lecture19.pdf · Chapter 9: Inferences for Two...
Outline
1 Overview
2 Inferences about Two Means: Independent and Small Samples
3 Inferences about Two Means: Independent and Large Samples
4 Inferences about Two Proportions
5 Inferences about Two Means: Matched Pairs
OverviewThere are many important and meaningful
situations in which it becomes necessary to compare two sets of sample data.
Definitions
Two Samples: IndependentThe sample values selected from one population are not related or somehow paired with the sample values selected from the other population.
If the values in one sample are related to the values in the other sample, the samples are dependent. Such samples are often referred to as matched pairs or paired samples.
Example
Population of all female college students
Sample of n2 = 21 females report average of 85.7 mph
Population of all male college students
Sample of n1 = 17 males report average of 102.1 mph
Do male and female college students differ with respect to their fastest reported driving speed?
Graphical summary of sample data
75 85 95 105 115 125 135 145
Fastest Driving Speed (mph)
Gender
female
male
Numerical summary ofsample data
Gender N Mean Median TrMean StDevfemale 21 85.71 85.00 85.26 9.39male 17 102.06 100.00 101.00 17.05
Gender SE Mean Minimum Maximum Q1 Q3female 2.05 75.00 105.00 77.50 92.50male 4.14 75.00 145.00 90.00 115.00
The difference in the sample means is 102.06 - 85.71 = 16.35 mph
The Question in Statistical Notation
Let µµµµM = the average fastest speed of all male students.and µµµµF = the average fastest speed of all female students.
Then we want to know whether µµµµM ≠≠≠≠ µµµµF.
This is equivalent to knowing whether µµµµM - µµµµF ≠≠≠≠ 0
All possible questions in statistical notation
In general, we can always compare two averages by seeing how their difference compares to 0:
This comparison… is equivalent to …
µµµµ1 ≠≠≠≠ µµµµ2 µµµµ1 - µµµµ2 ≠≠≠≠ 0
µµµµ1 > µµµµ2 µµµµ1 - µµµµ2 > 0
µµµµ1 < µµµµ2 µµµµ1 - µµµµ2 < 0
Set up hypotheses
• Null hypothesis: – H0: µµµµM = µµµµF [equivalent to µM - µF = 0]
• Alternative hypothesis:– Ha: µµµµM ≠≠≠≠ µµµµF [equivalent to µM - µF ≠ 0]
Assumptions:
1. The two samples are independent.
2. Both samples are normal or the two sample sizes are small, n1 < 30 and n2 < 30
3. Both variances are unknown but equal. Assume variances are equal only if neither sample standard deviation is more than twicethat of the other sample standard deviation.
Pooled Two-Sample T Test and T Interval
Confidence IntervalsNormal Samples w/ Unknown Equal
Variance
1 2
2/ 2, 2 1 2
2 22 1 1 2 2
1 2
(1/ 1/ )
( 1) ( 1)
( 2)
n n p
p
E t S n n
n S n SS
n n
α + −= +
− + −=+ −
(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E
where
Leaded vs Unleaded
Each of the cars selected for the EPA study was tested and the number of miles per gallon for each was obtained and recorded (Leaded=1 and Unleaded=2).
Leaded (1) Unleaded(2)
n 11 10
x 17.2 19.9
S 2.1 2
95% Confidence Interval
1 2/ 2, 2 0.025,11 10 2
2 22
0.05, 2.093
(11 1)2.1 (10 1)2.04.216
(11 10 2)
1 117.2 19.9 2.093* 4.216( )
11 10
n n
p
t t
S
αα + − + −= = =
− + −= =+ −
− ± +
(-4.58, -0.82)
Pooled Two-Sample T TestsNormal Samples w/ unknown Variance
1 2 1 2
21 2
( )
(1/ 1/ )p
X Xt
S n n
µ µ− − −=+
P-value: Use t distribution with n1+n2-2 degrees of freedom and find the P-value by following the same procedure for t tests summarized in Ch 8.
Critical values: Based on the significance level αααα, use for upper tail tests, use
for lower tail tests and use for two tailed tests.
1 2, 2n ntα + −
1 2, 2n ntα + −−1 2/ 2, 2| |n ntα + −
Leaded vs Unleaded
Claim: µµµµ1 < µµµµ2
Ho : µµµµ1 = µµµµ2
H1 : µµµµ1 < µµµµ2
αααα = 0.01
t
Fail to reject H0Reject H0
-1.729
0.05,19 1.729t− = −
Leaded vs Unleaded
Pooled Two-Sample T Test
1 2 1 2
21 2
( ) 17.2 19.9 03.01
4.216(1/11 1/10)(1/ 1/ )p
X Xt
S n n
µ µ− − − − −= = = −++
Claim: µµµµ1 < µµµµ2
Ho : µµµµ1 = µµµµ2 H1 : µµµµ1 < µµµµ2 αααα = 0.05
Leaded vs Unleaded
Claim: µµµµ1 < µµµµ2
Ho : µµµµ1 = µµµµ2
H1 : µµµµ1 < µµµµ2
αααα = 0.01
t
Fail to reject H0Reject H0
-1.729sample data:t = - 3.01
Reject Null
There is significant evidence to support the claim that the leaded
cars have a lower mean mpg than unleaded cars
P-value=0.0077(=area of red region)
Assumptions:
1. The two samples are independent.
2. Both samples are normal or the two sample sizes are small, n1 < 30 and n2 < 30
3. Both variances are unknown but unequal
Two-Sample T Test and T Interval
Confidence IntervalsNormal Samples w/ Unknown Unequal
Variance
2 21 2
/ 2,1 2
v
S SE t
n nα= +
(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E
where
2 2 21 2 1 2
1 24 41 2 1 2
1 2
[( ) ( ) ], ,
( ) ( )1 1
se se S Sv se se
se se n nn n
+= = =+
− −
(round v down to the nearest integer)
Unpooled Two Sample T-TestNormal Samples w/ Known Variance
1 2 1 2
2 22 1 1 2
( )
/ /
X Xt
S n S n
µ µ− − −=+
P-value: Use t distribution with v degrees of freedom and find the P-value by following the same procedure for t tests summarized in Ch 8.
Critical values: Based on the significance level αααα, use for upper tail tests, use
for lower tail tests and use for two tailed tests.
, vt α
,vtα−/ 2,| |vtα
Example
We compare the density of two different types of brick. Assuming normality of the two densities distributions and unequal unknown variances, test if there is a difference in the mean densities of two different types of brick.
Type I brick Type 2 brick
n 6 5
x 22.73 21.95
S 0.10 0.24
1x
Unpooled Two-Sample T-Test
1 2 1 2
2 2 2 21 1 2 2
0.025,6
( ) 22.73 21.95 06.792
/ / 0.1 / 6 0.24 / 5
6,| | | 2.446 |
X Xt
S n S n
v t
µ µ− − − − −= = =+ +
≈ =
Ho : µµµµ1 = µµµµ2 H1 : µµµµ1 ≠≠≠≠ µµµµ2 αααα = 0.05
P-Value = 0.001; Reject the null and conclude that there is significant difference in the mean densities of the two types of brick
Two-sample t-test in Minitab
• Select Stat. Select Basic Statistics. • Select 2-sample t to get a Pop-Up window.• Click on the radio button before Samples in one
Column. Put the measurement variable in Samples box, and put the grouping variable in Subscripts box.
• Specify your alternative hypothesis.• If appropriate, select Assume Equal Variances.• Select OK.
Pooled two-sample t-test
Two sample T for Fastest
Gender N Mean StDev SE Meanfemale 21 85.71 9.39 2.0male 17 102.1 17.1 4.1
95% CI for mu (female) - mu (male ): ( -25.2, -7.5)T-Test mu (female) = mu (male ) (vs not =): T = -3.75
P = 0.0006DF = 36
Both use Pooled StDev = 13.4