hypothesis testing – two samples chapters 12 & 13
TRANSCRIPT
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Hypothesis Testing – Two Samples
Chapters 12 & 13
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Chapter Topics
Comparing Two Independent Samples Independent samples Z test for the difference in two
means Pooled-variance t test for the difference in two means
F Test for the Difference in Two Variances Comparing Two Related Samples
Paired-sample Z test for the mean difference Paired-sample t test for the mean difference
Two-sample Z test for population proportions Independent and Dependent Samples
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Comparing Two Independent Samples
Different Data SourcesUnrelated Independent
Sample selected from one population has no effect or bearing on the sample selected from the other population
Use the Difference between 2 Sample MeansUse Z Test or t Test for Independent Samples
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Independent Sample Z Test (Variances Known)Assumptions
Samples are randomly and independently drawn from normal distributions
Population variances are known
Test Statistic
1 2 1
2 2
1 2
( ) ( )X XZ
n n
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t Test for Independent Samples (Variances Unknown)Assumptions
Both populations are normally distributed Samples are randomly and independently
drawn Population variances are unknown but assumed
equal If both populations are not normal, need large
sample sizes
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Developing the t Test for Independent Samples
Setting Up the Hypotheses
H0: 1 2
H1: 1 > 2
H0: 1 -2 = 0
H1: 1 - 2
0
H0: 1 = 2
H1: 1 2
H0: 1
2
H0: 1 - 2 0
H1: 1 - 2 > 0
H0: 1 - 2
H1: 1 -
2 < 0
OR
OR
OR Left Tail
Right Tail
Two Tail
H1: 1 < 2
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Developing the t Test for Independent SamplesCalculate the Pooled Sample Variance as
an Estimate of the Common Population Variance
(continued)
2 22 1 1 2 2
1 2
21
21 2
22
( 1) ( 1)
( 1) ( 1)
: Pooled sample variance : Size of sample 1
: Variance of sample 1 : Size of sample 2
: Variance of sample 2
p
p
n S n SS
n n
S n
S n
S
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Developing the t Test for Independent SamplesCompute the Sample Statistic
(continued)
1 2 1 2
2
1 2
2 21 1 2 22
1 2
1 1
1 1
1 1
p
p
X Xt
Sn n
n S n SS
n n
Hypothesized
difference1 2 2df n n
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t Test for Independent Samples: Example
© 1984-1994 T/Maker Co.
You’re a financial analyst for Charles Schwab. Is there a difference in average dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:
NYSE NASDAQNumber 21 25Sample Mean 3.27 2.53Sample Std Dev 1.30 1.16
Assuming equal variances, isthere a difference in average yield (= 0.05)?
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Calculating the Test Statistic
1 2 1 2
2
1 2
2 21 1 2 22
1 2
2 2
3.27 2.53 02.03
1 11 1 1.51021 25
1 1
1 1
21 1 1.30 25 1 1.16 1.502
21 1 25 1
p
p
X Xt
Sn n
n S n SS
n n
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Solution
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (12)
= 0.05
df = 21 + 25 - 2 = 44
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = 0.05.
There is evidence of a difference in means.
t0 2.0154-2.0154
.025
Reject H0 Reject H0
.025
2.03
3.27 2.532.03
1 11.502
21 25
t
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p -Value Solution
p-Value 2
(p-Value is between .02 and .05) ( = 0.05) Reject.
02.03
Z
Reject
2.0154
is between .01 and .025
Test Statistic 2.03 is in the Reject Region
Reject
-2.0154
=.025
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Example
© 1984-1994 T/Maker Co.
You’re a financial analyst for Charles Schwab. You collect the following data:
NYSE NASDAQNumber 21 25Sample Mean 3.27 2.53Sample Std Dev 1.30 1.16
You want to construct a 95% confidence interval for the difference in population average yields of the stocks listed on NYSE and NASDAQ.
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Example: Solution
1 2
21 2 / 2, 2
1 2
1 1n n pX X t S
n n
1 13.27 2.53 2.0154 1.502
21 25
1 20.0088 1.4712
2 21 1 2 22
1 2
2 2
1 1
1 1
21 1 1.30 25 1 1.16 1.502
21 1 25 1
p
n S n SS
n n
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Independent Sample (Two Sample) t- Test in JMP
Independent Sample t Test with Variances Known Analyze | Fit Y by X | Measurement in Y box
(Continuous) | Grouping Variable in X box (Nominal) | | Means/Anova/Pooled t
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Comparing Two Related Samples
Test the Means of Two Related Samples Paired or matchedRepeated measures (before and after)Use difference between pairs
Eliminates Variation between Subjects
1 2i i iD X X
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Z Test for Mean Difference (Variance Known)Assumptions
Both populations are normally distributedObservations are paired or matched Variance known
Test Statistic
D
D
DZ
n
1
n
ii
DD
n
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t Test for Mean Difference (Variance Unknown)Assumptions
Both populations are normally distributedObservations are matched or paired Variance unknown If population not normal, need large samples
Test Statistic
D
D
Dt
S
n
2
1
( )
1
n
ii
D
D DS
n
1
n
ii
DD
n
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User Existing System (1) New Software (2) Difference Di
C.B. 9.98 Seconds 9.88 Seconds .10T.F. 9.88 9.86 .02M.H. 9.84 9.75 .09R.K. 9.99 9.80 .19M.O. 9.94 9.87 .07D.S. 9.84 9.84 .00S.S. 9.86 9.87 - .01C.T. 10.12 9.98 .14K.T. 9.90 9.83 .07S.Z. 9.91 9.86 .05
Dependent-Sample t Test: Example
Assume you work in the finance department. Is the new financial package faster (=0.05 level)? You collect the following processing times:
2.072
1 .06215
i
iD
DD
n
D DS
n
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Dependent-Sample t Test: Example Solution
Is the new financial package faster (0.05 level)?
.072D =
.072 03.66
/ .06215/ 10D
D
DtS n
H0: D H1: D
Test Statistic
Critical Value=1.8331 df = n - 1 = 9
Reject
1.8331
Decision: Reject H0
t Stat. in the rejection zone.Conclusion: The new software package is faster.
3.66
t
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Confidence Interval Estimate for of Two Dependent Samples
Assumptions Both populations are normally distributedObservations are matched or paired Variance is unknown
Confidence Interval Estimate:
D
100 1 %
/ 2, 1D
n
SD t
n
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User Existing System (1) New Software (2) Difference Di
C.B. 9.98 Seconds 9.88 Seconds .10T.F. 9.88 9.86 .02M.H. 9.84 9.75 .09R.K. 9.99 9.80 .19M.O. 9.94 9.87 .07D.S. 9.84 9.84 .00S.S. 9.86 9.87 - .01C.T. 10.12 9.98 .14K.T. 9.90 9.83 .07S.Z. 9.91 9.86 .05
Example
Assume you work in the finance department. You want to construct a 95% confidence interval for the mean difference in data entry time. You collect the following processing times:
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Solution:
2
/ 2, 1 0.025,9
/ 2, 1
D
.072 .062151
2.2622
.06215.072 2.2622
10
0.0275 0.1165
iiD
n
Dn
D DDD S
n nt t
SD t
n
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F Test for Difference in Two Population VariancesTest for the Difference in 2 Independent
Populations
Parametric Test Procedure
Assumptions Both populations are normally distributed
Test is not robust to this violation Samples are randomly and independently
drawn
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The F Test Statistic
= Variance of Sample 1
n1 - 1 = degrees of freedom
n2 - 1 = degrees of freedom
F 0
21S
22S = Variance of Sample 2
2122
SF
S
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Hypotheses H0:1
2 = 22
H1: 12 2
2 Test Statistic
F = S12 /S2
2
Two Sets of Degrees of Freedom df1 = n1 - 1; df2 = n2 - 1
Critical Values: FL( ) and FU( )
FL = 1/FU* (*degrees of freedom switched)
Developing the F Test
Reject H0
Reject H0
/2/2Do NotReject
F 0 FL FU
n1 -1, n2 -1 n1 -1 , n2 -1
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Easier Way Put the largest in the num.
Test Statistic F = S1
2 /S22
Developing the F Test
Reject H0
Do NotReject
F 0 F
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F Test: An Example
Assume you are a financial analyst for Charles Schwab. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data:
NYSE NASDAQNumber 21 25Mean 3.27 2.53Std Dev 1.30 1.16
Is there a difference in the variances between the NYSE & NASDAQ at the 0.05 level?
© 1984-1994 T/Maker Co.
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F Test: Example Solution
Finding the Critical Values for = .05
1 1
2 2
1 21 1 20
1 25 1 24
df n
df n
F. , , .0 5 2 0 2 4 2 0 3
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F Test: Example Solution
H0: 12 = 2
2
H1: 12 2
2
.05
df1 20 df2 24
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Do not reject at = 0.05.
0 F2.33
.05
Reject
2 212 22
1.301.25
1.16
SF
S
1.25
There is insufficient evidence to prove a difference in variances.
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F Test: One-Tail
H0: 12 2
2
H1: 12 2
2
H0: 12 2
2
H1: 12 > 2
2
Reject
.05
F0 F0
Reject
.05
= .05
or
Degrees of freedom switched
1 2
2 1
1, 11, 1
1L n n
U n n
FF
1 21, 1U n nF 1 21, 1L n nF
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Easier Way Put the largest in the num.
Test Statistic F = S1
2 /S22
F Test: One-Tail
Reject H0
Do NotReject
F 0 F
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Z Test for Differences in Two Proportions (Independent Samples)What is It Used For?
To determine whether there is a difference between 2 population proportions and whether one is larger than the other
Assumptions: Independent samples Population follows binomial distribution Sample size large enough: np 5 and n(1-p)
5 for each population
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Z Test Statistic
Z
p p
p q
n
p q
nPd Pd Pd Pd
1 2
1 2
pn p n p
n n
qn q n q
n n
Pd
Pd
1 1 2 2
1 2
1 1 2 2
1 2
Z
p p p p
p q
n
p q
n
1 2 1 2
1 1
1
2 2
2
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The Hypotheses for the Z Test
Research Questions
Hypothesis No DifferenceAny Difference
Prop 1 Prop 2Prop 1 < Prop 2
Prop 1 Prop 2Prop 1 > Prop 2
H0 p1 - p2 p1 - p2 0 p1 - p2 0
H1 p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0
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Z Test for Differences in Two Proportions: Example
As personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. At the 0.01 significance level, is there a difference in perceptions?
.p 1
6 3
7 80 8 0 7 7
.p 2
4 9
8 20 5 9 8
n 1 7 8
n 2 8 2
np 5 n p1 5
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Calculating the Test Statistic
Z
p p p p
p q
n
p q
nPd Pd Pd Pd
.
1 2 1 2
1 2
2 8 9 8
pn p n p
n n
qn q n q
n n
Pd
Pd
1 1 2 2
1 2
1 1 2 2
1 2
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Z Test for Differences in Two Proportions: Solution
H0: p1 - p2 = 0
H1: p1 - p2 0
= 0.01
n1 = 78 n2 = 82
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = 0.01.
There is evidence of a difference in proportions.
Z 290.
Z0 2.58-2.58
.005
Reject H0 Reject H0
.005
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Confidence Interval for Differences in Two ProportionsThe Confidence Interval for
Differences in Two Proportions 100 1 %
p p z
p p
n
p p
n1 21
2
1 1
1
2 2
2
1 1
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Confidence Interval for Differences in Two Proportions: Example
As personnel director, you want to find out the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. Construct a 99% confidence interval for the difference in two proportions.
.p 1
6 3
7 80 8 0 7 7
.p 2
4 9
8 20 5 9 8
n 1 7 8
n 2 8 2
np 5 n p1 5
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Confidence Interval for Differences in Two Proportions: Solution
We are 99% confident that the difference between two proportions is somewhere between 0.0294 and 0.3909.
. . .. . . .
. .
p p zp p
n
p p
n
p p
1 21
2
1 1
1
2 2
2
1 2
1 1
0 8 0 7 7 0 5 9 7 6 2 5 7 5 80 8 0 7 7 1 0 8 0 7 7
7 8
0 5 9 7 6 1 0 5 9 7 6
8 20 0 2 9 4 0 3 9 0 9
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Z Test for Differences in Two Proportions (Dependent Samples)What is It Used For?
To determine whether there is a difference between 2 population proportions and whether one is larger than the other
Assumptions: Dependent samples Population follows binomial distribution Sample size large enough: np 5 and n(1-p)
5 for each population
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Z Test Statistic for Dependent Samples
Za d
a d
Sample Two
Category 1 Category 2
Sample One Category 1 a b a+b
Category 2 c d c+d
a+c b+d N
This Z can be used when a+d>10 If 10<a+d<20, use the correction in the text
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Unit Summary
Compared Two Independent Samples Performed Z test for the differences in two
means Performed t test for the differences in two
means Performed Z test for differences in two
proportions Addressed F Test for Difference in Two Variances
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Unit Summary
Compared Two Related Samples Performed dependent sample Z tests for the
mean difference Performed dependent sample t tests for the
mean difference Performed Z tests for proportions using
dependent samples