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Chapter 10 10-1
Prof. Shuguang Liu
Chapter 10 Two-Sample Tests
Chap 10-1 Prof. Shuguang Liu
Learning Objectives
In this chapter, you learn how to use hypothesis testing for comparing the difference between:
• The means of two independent populations • The means of two related populations • The proportions of two independent
populations • The variances of two independent populations
by testing the ratio of the two variances
Chap 10-2
Prof. Shuguang Liu
Two-Sample Tests
Chap 10-3
Two-Sample Tests
Population Means,
Independent Samples
Population Means, Related Samples
Population Variances
Group 1 vs. Group 2
Same group before vs. after treatment
Variance 1 vs. Variance 2
Examples:
Population Proportions
Proportion 1 vs. Proportion 2
DCOVA
Prof. Shuguang Liu
Difference Between Two Means
Chap 10-4
Population means, independent
samples
Goal: Test hypothesis or form a confidence interval for the difference between two population means, µ1 – µ2
The point estimate for the difference is
X1 – X2
*
σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
DCOVA
Prof. Shuguang Liu
Difference Between Two Means: Independent Samples
§ Different data sources • Unrelated • Independent
u Sample selected from one population has no effect on the sample selected from the other population
Chap 10-5
Population means, independent
samples *
Use Sp to estimate unknown σ. Use a Pooled-Variance t test.
σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
Use S1 and S2 to estimate unknown σ1 and σ2. Use a Separate-Variance t test.
DCOVA
Prof. Shuguang Liu
Hypothesis Tests for Two Population Means
Chap 10-6
Lower-tail test:
H0: µ1 ≥ µ2 H1: µ1 < µ2
i.e.,
H0: µ1 – µ2 ≥ 0 H1: µ1 – µ2 < 0
Upper-tail test:
H0: µ1 ≤ µ2 H1: µ1 > µ2
i.e.,
H0: µ1 – µ2 ≤ 0 H1: µ1 – µ2 > 0
Two-tail test:
H0: µ1 = µ2 H1: µ1 ≠ µ2
i.e.,
H0: µ1 – µ2 = 0 H1: µ1 – µ2 ≠ 0
Two Population Means, Independent Samples
DCOVA
Chapter 10 10-2
Prof. Shuguang Liu
Hypothesis tests for µ1 – µ2
Chap 10-7
Two Population Means, Independent Samples
Lower-tail test:
H0: µ1 – µ2 ≥ 0 H1: µ1 – µ2 < 0
Upper-tail test:
H0: µ1 – µ2 ≤ 0 H1: µ1 – µ2 > 0
Two-tail test:
H0: µ1 – µ2 = 0 H1: µ1 – µ2 ≠ 0
α α/2 α/2 α
-tα -tα/2 tα tα/2
Reject H0 if tSTAT < -tα Reject H0 if tSTAT > tα Reject H0 if tSTAT < -tα/2 or tSTAT > tα/2
DCOVA
Prof. Shuguang Liu
Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and assumed equal
Chap 10-8
Population means, independent
samples
Assumptions: § Samples are randomly and independently drawn
§ Populations are normally distributed or both sample sizes are at least 30 § Population variances are unknown but assumed equal
* σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
DCOVA
Prof. Shuguang Liu
Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and assumed equal
Chap 10-9
Population means, independent
samples
• The pooled variance is:
• The test statistic is:
• Where tSTAT has d.f. = (n1 + n2 – 2)
(continued)
( ) ( )1)n()1(nS1nS1nS
21
222
2112
p −+−
−+−=
* σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
( ) ( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−−−=
21
2p
2121STAT
n1
n1S
µµXXt
DCOVA
Prof. Shuguang Liu
Confidence interval for µ1 - µ2 with σ1 and σ2 unknown and assumed equal
Chap 10-10
Population means, independent
samples
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛+±−
21
2p/221
n1
n1SXX αt
The confidence interval for µ1 – µ2 is:
Where tα/2 has d.f. = n1 + n2 – 2
* σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
DCOVA
Prof. Shuguang Liu
Pooled-Variance t Test Example
You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQ Number 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16
Chap 10-11
Assuming both populations are approximately normal with equal variances, is there a difference in mean yield (α = 0.05)?
DCOVA
Prof. Shuguang Liu
Pooled-Variance t Test Example: Calculating the Test Statistic
Chap 10-12
( ) ( ) ( ) ( )1.5021
1)25(1)-(211.161251.30121
1)n()1(n1n1n 22
21
222
2112 =
−+
−+−=
−+−
−+−=
SSSP
( ) ( ) ( )2.040
251
2115021.1
02.533.27
n1
n1S
µµXXt
21
2p
2121STAT =
⎟⎠
⎞⎜⎝
⎛+
−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−−−=
The test statistic is:
(continued)
H0: µ1 - µ2 = 0 i.e. (µ1 = µ2) H1: µ1 - µ2 ≠ 0 i.e. (µ1 ≠ µ2)
DCOVA
Chapter 10 10-3
Prof. Shuguang Liu
Pooled-Variance t Test Example: Hypothesis Test Solution
H0: µ1 - µ2 = 0 i.e. (µ1 = µ2) H1: µ1 - µ2 ≠ 0 i.e. (µ1 ≠ µ2) α = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 2.0154
Test Statistic:
Chap 10-13
Decision: Conclusion:
Reject H0 at α = 0.05
There is evidence of a difference in means.
t 0 2.0154 -2.0154
.025
Reject H0 Reject H0
.025
2.040
2.040
251
2115021.1
2.533.27tSTAT =
⎟⎠
⎞⎜⎝
⎛+
−=
DCOVA
Prof. Shuguang Liu
Excel Pooled-Variance t test Comparing NYSE & NASDAQ
Chap 10-14
Pooled-‐Variance t Test for the Difference Between Two Means (assumes equal popula=on variances)
Data Hypothesized Difference 0 Level of Significance 0.05
Popula*on 1 Sample Sample Size 21 =COUNT(DATA!$A:$A) Sample Mean 3.27 =AVERAGE(DATA!$A:$A) Sample Standard Devia=on 1.3 =STDEV(DATA!$A:$A)
Popula*on 2 Sample Sample Size 25 =COUNT(DATA!$B:$B) Sample Mean 2.53 =AVERAGE(DATA!$B:$B) Sample Standard Devia=on 1.16 =STDEV(DATA!$B:$B)
Intermediate Calcula*ons Popula=on 1 Sample Degrees of Freedom 20 =B7 -‐ 1 Popula=on 2 Sample Degrees of Freedom 24 =B11 -‐ 1 Total Degrees of Freedom 44 =B16 + B17 Pooled Variance 1.502 =((B16 * B9^2) + (B17 * B13^2)) / B18 Standard Error 0.363 =SQRT(B19 * (1/B7 + 1/B11)) Difference in Sample Means 0.74 =B8 -‐ B12 t Test Sta=s=c 2.040 =(B21 -‐ B4) / B20
Two-‐Tail Test Lower Cri=cal Value -‐2.015 =-‐TINV(B5, B18) Upper Cri=cal Value 2.015 =TINV(B5, B18) p-‐value 0.047 =TDIST(ABS(B22),B18,2)
Reject the null hypothesis =IF(B27<B5,"Reject the null hypothesis", "Do not reject the null hypothesis")
Decision: Conclusion:
Reject H0 at α = 0.05
There is evidence of a difference in means.
DCOVA
Prof. Shuguang Liu
Minitab Pooled-Variance t test Comparing NYSE & NASDAQ
Chap 10-15
Decision: Conclusion:
Reject H0 at α = 0.05
There is evidence of a difference in means.
Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 21 3.27 1.30 0.28 2 25 2.53 1.16 0.23 Difference = mu (1) - mu (2) Estimate for difference: 0.740 95% CI for difference: (0.009, 1.471) T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44 Both use Pooled StDev = 1.2256
DCOVA
Prof. Shuguang Liu
Pooled-Variance t Test Example: Confidence Interval for µ1 - µ2
( ) )471.1,009.0(3628.00154.274.0 n1
n1SXX
21
2p/221 =×±=⎟⎟
⎠
⎞⎜⎜⎝
⎛+±− αt
Since we rejected H0 can we be 95% confident that µNYSE > µNASDAQ? 95% Confidence Interval for µNYSE - µNASDAQ
Since 0 is less than the entire interval, we can be 95% confident that µNYSE > µNASDAQ
Chap 10-16
DCOVA
Prof. Shuguang Liu
Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown, not assumed equal
Chap 10-17
Population means, independent
samples
Assumptions: § Samples are randomly and independently drawn
§ Populations are normally distributed or both sample sizes are at least 30 § Population variances are unknown and cannot be assumed to be equal
*
σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
DCOVA
Prof. Shuguang Liu
Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and not assumed equal
1nnS
1nnS
nS
nS
2
2
2
22
1
2
1
21
2
2
22
1
21
−
⎟⎟⎠
⎞⎜⎜⎝
⎛
+−
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=ν
Chap 10-18
Population means, independent
samples
(continued)
*
σ1 and σ2 unknown, assumed equal
σ1 and σ2 unknown, not assumed equal
The test statistic is: ( ) ( )
2
22
1
21
2121STAT
nS
nS
µµXXt+
−−−=
tSTAT has d.f. ν =
DCOVA
Chapter 10 10-4
Prof. Shuguang Liu
Related Populations The Paired Difference Test
Tests Means of 2 Related Populations – Paired or matched samples – Repeated measures (before/after) – Use difference between paired values:
§ Eliminates Variation Among Subjects § Assumptions:
• Both Populations Are Normally Distributed • Or, if not Normal, use large samples
Chap 10-19
Related samples
Di = X1i - X2i
DCOVA
Prof. Shuguang Liu
Related Populations The Paired Difference Test
The ith paired difference is Di , where
Chap 10-20
Related samples
Di = X1i - X2i
The point estimate for the paired difference population mean µD is D : n
DD
n
1ii∑
==
n is the number of pairs in the paired sample
1n
)D(DS
n
1i
2i
D −
−=∑=
The sample standard deviation is SD
(continued) DCOVA
Prof. Shuguang Liu
The Paired Difference Test: Finding tSTAT
§ The test statistic for µD is:
Chap 10-21
Paired samples
nSµDtD
STATD−
=
n Where tSTAT has n - 1 d.f.
DCOVA
Prof. Shuguang Liu
The Paired Difference Test: Possible Hypotheses
Chap 10-22
Lower-tail test:
H0: µD ≥ 0 H1: µD < 0
Upper-tail test:
H0: µD ≤ 0 H1: µD > 0
Two-tail test:
H0: µD = 0 H1: µD ≠ 0
Paired Samples
α α/2 α/2 α
-tα -tα/2 tα tα/2 Reject H0 if tSTAT < -tα Reject H0 if tSTAT > tα Reject H0 if tSTAT < -tα/2
or tSTAT > tα/2 Where tSTAT has n - 1 d.f.
DCOVA
Prof. Shuguang Liu
The Paired Difference Confidence Interval
The confidence interval for µD is
Chap 10-23
Paired samples
1n
)D(DS
n
1i
2i
D −
−=∑=
nSD
2/αtD ±
where
DCOVA
Prof. Shuguang Liu
§ Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:
Chap 10-24
Paired Difference Test: Example
Number of Complaints: (2) - (1) Salesperson Before (1) After (2) Difference, Di C.B. 6 4 - 2 T.F. 20 6 -14 M.H. 3 2 - 1 R.K. 0 0 0 M.O. 4 0 - 4 -21
D = Σ Di
n
5.67
1n)D(D
S2
iD
=
−
−= ∑
= -4.2
DCOVA
Chapter 10 10-5
Prof. Shuguang Liu
§ Has the training made a difference in the number of complaints (at the 0.01 level)?
Chap 10-25
- 4.2 D =
1.6655.67/04.2
n/Sµt
DSTAT
D −=−−
=−
=D
H0: µD = 0 H1: µD ≠ 0
Test Statistic:
t0.005 = ± 4.604 d.f. = n - 1 = 4
Reject
α/2 - 4.604 4.604
Decision: Do not reject H0 (tstat is not in the reject region)
Conclusion: There is insufficient evidence there is significant change in the number of complaints.
Paired Difference Test: Solution
Reject
α/2
- 1.66 α = .01
DCOVA
Prof. Shuguang Liu
Paired t Test In Excel
Chap 10-26
Paired t Test
Data Hypothesized Mean Diff. 0 Level of Significance 0.05
Intermediate Calcula*ons Sample Size 5 =COUNT(I2:I6) Dbar -‐4.2 =AVERAGE(I2:I6) Degrees of Freedom 4 =B8 -‐ 1 SD 5.67 =STDEV(I2:I6) Standard Error 2.54 =B11/SQRT(B8) t-‐Test Sta=s=c -‐1.66 =(B9 -‐ B4)/B12
Two-‐Tail Test Lower Cri=cal Value -‐2.776 =-‐TINV(B5,B10) Upper Cri=cal Value 2.776 =TINV(B5,B10) p-‐value 0.173 =TDIST(ABS(B13),B10,2) Do not reject the null Hypothesis
=IF(B18<B5,"Reject the null hypothesis", "Do not reject the null hypothesis")
Data not shown is in column I
DCOVA
Since -2.776 < -1.66 < 2.776 we do not reject the null hypothesis. Or Since p-value = 0.173 > 0.05 we do not reject the null hypothesis. Thus we conclude that there is Insufficient evidence to conclude there is a difference in the average number of complaints.
Prof. Shuguang Liu
Paired t Test In Minitab Yields The Same Conclusions
Chap 10-27
DCOVA Paired T-Test and CI: After, Before Paired T for After - Before N Mean StDev SE Mean After 5 2.40 2.61 1.17 Before 5 6.60 7.80 3.49 Difference 5 -4.20 5.67 2.54 95% CI for mean difference: (-11.25, 2.85) T-Test of mean difference = 0 (vs not = 0): T-Value = -1.66 P-Value = 0.173
Prof. Shuguang Liu
Two Population Proportions
Chap 10-28
Goal: test a hypothesis or form a confidence interval for the difference between two population proportions,
π1 – π2
The point estimate for the difference is
Population proportions
21 pp −
DCOVA
Prof. Shuguang Liu
Two Population Proportions
Chap 10-29
Population proportions
21
21
nnXX
+
+=p
The pooled estimate for the overall proportion is:
where X1 and X2 are the number of items of interest in samples 1 and 2
In the null hypothesis we assume the null hypothesis is true, so we assume π1 = π2 and pool the two sample estimates
DCOVA
Prof. Shuguang Liu
Two Population Proportions
Chap 10-30
Population proportions
( ) ( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
−−−=
21
2121
11)1(nn
pp
ππppZSTAT
The test statistic for π1 – π2 is a Z statistic:
(continued)
2
22
1
11
21
21
nX , p
nX , p
nnXXp ==
+
+=where
DCOVA
Chapter 10 10-6
Prof. Shuguang Liu
Hypothesis Tests for Two Population Proportions
Chap 10-31
Population proportions
Lower-tail test:
H0: π1 ≥ π2 H1: π1 < π2
i.e.,
H0: π1 – π2 ≥ 0 H1: π1 – π2 < 0
Upper-tail test:
H0: π1 ≤ π2 H1: π1 > π2
i.e.,
H0: π1 – π2 ≤ 0 H1: π1 – π2 > 0
Two-tail test:
H0: π1 = π2 H1: π1 ≠ π2
i.e.,
H0: π1 – π2 = 0 H1: π1 – π2 ≠ 0
DCOVA
Prof. Shuguang Liu
Hypothesis Tests for Two Population Proportions
Chap 10-32
Population proportions
Lower-tail test:
H0: π1 – π2 ≥ 0 H1: π1 – π2 < 0
Upper-tail test:
H0: π1 – π2 ≤ 0 H1: π1 – π2 > 0
Two-tail test:
H0: π1 – π2 = 0 H1: π1 – π2 ≠ 0
α α/2 α/2 α
-zα -zα/2 zα zα/2
Reject H0 if ZSTAT < -Zα Reject H0 if ZSTAT > Zα Reject H0 if ZSTAT < -Zα/2
or ZSTAT > Zα/2
(continued)
DCOVA
Prof. Shuguang Liu
Hypothesis Test Example: Two population Proportions
Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?
§ In a random sample, 36 of 72 men and 35 of 50 women indicated they would vote Yes
§ Test at the .05 level of significance
Chap 10-33
DCOVA
Prof. Shuguang Liu
§ The hypothesis test is:
Chap 10-34
H0: π1 – π2 = 0 (the two proportions are equal) H1: π1 – π2 ≠ 0 (there is a significant difference between proportions)
n The sample proportions are: n Men: p1 = 36/72 = 0.50
n Women: p2 = 35/50 = 0.70
582012271
50723536
21
21 .nnXXp ==
+
+=
+
+=
§ The pooled estimate for the overall proportion is:
Hypothesis Test Example: Two population Proportions
(continued)
DCOVA
Prof. Shuguang Liu
Hypothesis Test Example: Two population Proportions
Chap 10-35
The test statistic for π1 – π2 is:
(continued)
.025
-1.96 1.96
.025
-2.20
Decision: Reject H0
Conclusion: There is evidence of a difference in proportions who will vote yes between men and women.
( ) ( )
( ) ( )202
501
7215821582
07050
11121
2121
.).(.
..
nn)p(p
ππppzSTAT
−=
⎟⎠
⎞⎜⎝
⎛+−
−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
−−−=
Reject H0 Reject H0
Critical Values = ±1.96 For α = .05
DCOVA
Prof. Shuguang Liu
Two Proportion Test In Excel
Chap 10-36
DCOVA Z Test for Differences in Two Propor=ons
Data Hypothesized Difference 0 Level of Significance 0.05
Group 1 Number of items of interest 36 Sample Size 72
Group 2 Number of items of interest 35 Sample Size 50
Intermediate Calcula*ons Group 1 Propor=on 0.5 =B7/B8 Group 2 Propor=on 0.7 =B10/B11 Difference in Two Propor=ons -‐0.2 =B14 -‐ B15 Average Propor=on 0.582 =(B7 + B10)/(B8 + B11) Z Test Sta=s=c -‐2.20 =(B16-‐B4)/SQRT((B17*(1-‐B17))*(1/B8+1/B11))
Two-‐Tail Test Lower Cri=cal Value -‐1.96 =NORMSINV(B5/2) Upper Cri=cal Value 1.96 =NORMSINV(1 -‐ B5/2) p-‐value 0.028 =2*(1 -‐ NORMSDIST(ABS(B18)))
Reject the null hypothesis =IF(B23 < B5,"Reject the null hypothesis", "Do not reject the null hypothesis")
Decision: Reject H0
Conclusion: There is evidence of a difference in proportions who will vote yes between men and women.
Since -2.20 < -1.96 Or Since p-value = 0.028 < 0.05 We reject the null hypothesis
Chapter 10 10-7
Prof. Shuguang Liu
Two Proportion Test In Minitab Shows The Same Conclusions
Chap 10-37
DCOVA
Test and CI for Two Proportions Sample X N Sample p 1 36 72 0.500000 2 35 50 0.700000 Difference = p (1) - p (2) Estimate for difference: -0.2 95% CI for difference: (-0.371676, -0.0283244) Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022
Prof. Shuguang Liu
Confidence Interval for Two Population Proportions
Chap 10-38
Population proportions
( )2
22
1
11/221 n
)p(1pn
)p(1pZpp −+
−±− α
The confidence interval for π1 – π2 is:
DCOVA
Prof. Shuguang Liu
Testing for the Ratio Of Two Population Variances
Chap 10-39
Tests for Two Population Variances
F test statistic
H0: σ12 = σ2
2 H1: σ1
2 ≠ σ22
H0: σ12 ≤ σ2
2 H1: σ1
2 > σ22
* Hypotheses FSTAT
= Variance of sample 1 (the larger sample variance)
n1 = sample size of sample 1
= Variance of sample 2 (the smaller sample variance)
n2 = sample size of sample 2
n1 –1 = numerator degrees of freedom
n2 – 1 = denominator degrees of freedom
Where:
DCOVA
21S
22S
22
21
SS
Prof. Shuguang Liu
The F Distribution
§ The F critical value is found from the F table
§ There are two degrees of freedom required: numerator and denominator
§ The larger sample variance is always the numerator
§ When
§ In the F table, • numerator degrees of freedom determine the column
• denominator degrees of freedom determine the row
Chap 10-40
df1 = n1 – 1 ; df2 = n2 – 1
22
21
SSFSTAT =
DCOVA
Prof. Shuguang Liu
Finding the Rejection Region
Chap 10-41
H0: σ12 = σ2
2 H1: σ1
2 ≠ σ22
H0: σ12 ≤ σ2
2 H1: σ1
2 > σ22
F 0 α
Fα Reject H0 Do not reject H0
Reject H0 if FSTAT > Fα
F 0
α/2
Reject H0 Do not reject H0 Fα/2
Reject H0 if FSTAT > Fα/2
DCOVA
Prof. Shuguang Liu
F Test: An Example
You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data: NYSE NASDAQ Number 21 25 Mean 3.27 2.53 Std dev 1.30 1.16
Is there a difference in the variances between the NYSE & NASDAQ at the α = 0.05 level?
Chap 10-42
DCOVA
Chapter 10 10-8
Prof. Shuguang Liu
F Test: Example Solution
§ Form the hypothesis test: (there is no difference between variances)
(there is a difference between variances)
Chap 10-43
n Find the F critical value for α = 0.05:
n Numerator d.f. = n1 – 1 = 21 –1 = 20
n Denominator d.f. = n2 – 1 = 25 –1 = 24
n Fα/2 = F.025, 20, 24 = 2.33
DCOVA
22
211
22
210
σ : σH
σ : σH
≠
=
Prof. Shuguang Liu
F Test: Example Solution
§ The test statistic is:
Chap 10-44
0
256.116.130.1
2
2
22
21 ===SSFSTAT
α/2 = .025
F0.025=2.33 Reject H0 Do not
reject H0
H0: σ12 = σ2
2 H1: σ1
2 ≠ σ22
n FSTAT = 1.256 is not in the rejection region, so we do not reject H0
(continued)
n Conclusion: There is insufficient evidence of a difference in variances at α = .05
F
DCOVA
Prof. Shuguang Liu
Two Variance F Test In Excel
Chap 10-45
DCOVA
F Test for Differences in Two Variables
Data Level of Significance 0.05
Larger-‐Variance Sample Sample Size 21 Sample Variance 1.6900 =1.3^2
Smaller-‐Variance Sample Sample Size 25 Sample Variance 1.3456 =1.16^2
Intermediate Calcula*ons F Test Sta=s=c 1.256 1.255945 Popula=on 1 Sample Degrees of Freedom 20 =B6 -‐ 1 Popula=on 2 Sample Degrees of Freedom 24 =B9 -‐ 1
Two-‐Tail Test Upper Cri=cal Value 2.327 =FINV(B4/2,B14,B15) p-‐value 0.589 =2*FDIST(B13,B14,B15)
Do not reject the null hypothesis =IF(B19<B4,"Reject the null hypothesis", "Do not reject the null hypothesis")
Conclusion: There is insufficient evidence
of a difference in variances at α = .05 because:
F statistic =1.256 < 2.327 Fα/2
or p-value = 0.589 > 0.05 = α.
Prof. Shuguang Liu
Two Variance F Test In Minitab Yields The Same Conclusion
Chap 10-46
DCOVA Test and CI for Two Variances Null hypothesis Sigma(1) / Sigma(2) = 1 Alternative hypothesis Sigma(1) / Sigma(2) not = 1 Significance level Alpha = 0.05 Statistics Sample N StDev Variance 1 21 1.300 1.690 2 25 1.160 1.346 Ratio of standard deviations = 1.121 Ratio of variances = 1.256 95% Confidence Intervals CI for Distribution CI for StDev Variance of Data Ratio Ratio Normal (0.735, 1.739) (0.540, 3.024) Tests Test Method DF1 DF2 Statistic P-Value F Test (normal) 20 24 1.26 0.589
Prof. Shuguang Liu
Chapter Summary
§ Compared two independent samples • Performed pooled-variance t test for the
difference in two means • Performed separate-variance t test for
difference in two means • Formed confidence intervals for the difference
between two means § Compared two related samples (paired
samples) • Performed paired t test for the mean
difference • Formed confidence intervals for the mean
difference Chap 10-47 Prof. Shuguang Liu
Chapter Summary
§ Compared two population proportions • Formed confidence intervals for the difference
between two population proportions • Performed Z-test for two population
proportions
§ Performed F test for the ratio of two population variances
Chap 10-48
(continued)