Variance-Test-1
Inferences about Variances (Chapter 7)
• Develop point estimates for the population variance• Construct confidence intervals for the population variance.• Perform one-sample tests for the population variance.• Perform two-sample tests for the population variance.
In this Lecture we will:
Note:
Need to assume normal population distributions for all sample sizes, small or large! If the population(s) are not normally distributed, results can be very wrong. Nonparametric alternatives are not straightforward.
Variance-Test-2
The point estimate for 2 is the sample variance:
n
1i
2i
2 yy1n
1s )(
What about the sampling distribution of s2? (I.e. what would we see as a distribution for s2 from repeated samples).
If the observations, yi, are from a normal (,) distribution, then the quantity
2
2s1n
)(
has a Chi-square distribution with df = n-1.
Point Estimate for 2
Variance-Test-3
y
x2
0 5 10 15 20 25 30
0.0
0.1
0.2
0.3
0.4
0.5
22df
25df
210df
215df
Non-symmetric.Shape indexed by one parameter called the degrees of freedom (df).
Chi Square (2) Distribution
Variance-Test-5
2
2s1n
)(
Ifhas a Chi Square Distribution, then a 100(1-)% CI can be computed by finding the upper and lower /2 critical values from this distribution.
1
s1n 2
2df2
22
21df)
)(Pr(
,,
.831212.83
.95
y
x5
0 5 10 15 20 25 30
0.0
0.0
50
.10
0.1
5
2
21df
,
df=5
3.24720.48
.95
y
x10
0 5 10 15 20 25 30
0.0
0.0
20
.04
0.0
60
.08
0.1
0
2
2df ,
df=10
Confidence Interval for 2
Variance-Test-6
1
s1n 2
2df2
22
21df)
)(Pr(
,,
2
21df
22
2
2df
2 s1ns1n
,,
)()(
Background Data:
13.1
48.2
7
s
y
n
A 95% CI for background population variance
297506
22
202506
2 13161316
.,.,
.)(.)(
193.653.0 2
Consider the data from the contaminated site vs. background.
s2 = 1.277
Variance-Test-7
What if we were interested in testing:
20
2
20
2
20
2
20
20
20
3
2
1
:
)specifiedis(:
aH
H
Test Statistic:
20
22 s1n
)(
Rejection Region:1. Reject H0 if 2 > 2
df,
2. Reject H0 if 2 < 2df,1-
3. Reject H0 if either 2 < 2df,1-/2 or 2 > 2
df,/2
Example:
13.1
48.2
7
s
y
n 6671
1316 22 .
.)( In testing Ha: 2 > 1:
Reject H0 if 2 > 26,0.05 =12.59
Conclude: Do not reject H0.
Hypothesis Testing for 2
Variance-Test-8
Objective: Test for the equality of variances (homogeneity assumption).
22
21
22
21
ss
has a probability distribution in repeated sampling which follows the F distribution.
y
x25
0 2 4 6 8 10
0.0
0.2
0.4
0.6
0.8
1.0
F(2,5)
F(5,5)
The F distribution shape is defined by two parameters denoted the numerator degrees of freedom (ndf or df1 ) and the denominator degrees of freedom (ddf or df2 ).
Tests for Comparing Two Population Variances
Variance-Test-9
• Can assume only positive values (like 2, unlike normal and t).• Is nonsymmetrical (like 2, unlike normal and t).• Many shapes -- shapes defined by numerator and denominator
degrees of freedom.
• Tail values for specific values of df1 and df2 given in Table 9.
df1 relates to degrees of freedom associated with s21
df2 relates to degrees of freedom associated with s22
F distribution:
Variance-Test-10
Note this table has three things to specify in order to get the critical value.
Numerator df = df1.
Denominator df = df2.
Probability Level
4.285.82
F Table
Table 9
Variance-Test-11
22
210H :
22
21
22
21a
2
1H
.
.:versus
Test Statistic:22
21
s
sF For one-tailed tests, define
population 1 to be the one with larger hypothesized variance.
Rejection Region:
1. Reject H0 if F > Fdf1,df2,.
2. Reject H0 if F > Fdf1,df2,/2 or if F < Fdf1,df2,1-/2.
In both cases, df1=n1-1 and df2=n2 -1.
Hypothesis Test for two population variances
,,1,, 12211
:10any For
dfdfdfdf FF
Variance-Test-12
131s
482y
7n
1
1
1
.
.
Background Samples
890s
824y
7n
2
2
2
.
.
Study Site Samples
612179210
27691
890
131
s
sF
2
2
22
21 .
.
.
.
.
= 0.05, F6,6,0.05 = 4.28 One-sided Alternative Hypothesis
T.S.
R.R.
Reject H0 if F > Fdf1,df2, where df1=n1-1 and df2=n2-1
Example
Reject H0 if F > Fdf1,df2,/2or if F < F df1,df2,1-/2 = 0.05, F6,6,0.025 = 5.82, F6,6,0.975 = 0.17 Two-sided Alternative
Conclusion: Do not reject H0 in either case.
Variance-Test-13
2dfdf22
21
22
21
2dfdf
22
21
12
21
Fs
s
F
1
s
s,,
,,
Note: degrees of freedom have been swapped.
Example (95% CI):
025.0,6,62
2
22
21
025.0,6,62
2
89.0
13.11
89.0
13.1F
F
2829277022
21 ..
82.5025.0,6,6 F
Note: not a argument!
(1-)100% Confidence Interval for Ratio of Variances
Variance-Test-14
Conclusion
While the two sample test for variances looks simple (and is simple), it forms the foundation for hypothesis testing in Experimental Designs.
Nonparametric alternatives are:• Levene’s Test (Minitab);• Fligner-Killeen Test (R).
Variance-Test-15
Software Commands for Chapters 5, 6 and 7
MINITABStat -> Basic Statistics -> 1-Sample z, 1-Sample t, 2-Sample t, Paired t,
Variances, Normality Test. -> Power and Sample Size -> 1-Sample z, 1-Sample t, 2-Sample t. -> Nonparametrics -> Mann-Whitney (Wilcoxon Rank Sum Test) -> 1-sample Wilcoxon (Wilcox. Signed Rank Test)
Rt.test( ): 1-Sample t, 2-Sample t, Paired t.power.t.test( ): 1-Sample t, 2-Sample t, Paired t.var.test( ): Tests for homogeneity of variances in normal populations.wilcox.test( ): Nonparametric Wilcoxon Signed Rank & Rank Sum tests.shapiro.test( ), ks.test( ): tests of normality.
Variance-Test-16
Example It’s claimed that moderate exposure to ozone increases lung capacity. 24 similar rats were randomly divided into 2 groups of 12, and the 2nd group was exposed to ozone for 30 days. The lung capacity of all rats were measured after this time.
No-Ozone Group: 8.7,7.9,8.3,8.4,9.2,9.1,8.2,8.1,8.9,8.2,8.9,7.5 Ozone Group: 9.4,9.8,9.9,10.3,8.9,8.8,9.8,8.2,9.4,9.9,12.2,9.3
• Basic Question: How to randomly select the rats?
• In class I will demonstrate the use of MTB and R to analyze these data. (See “Comparing two populations via two sample t-tests” in my R resources webpage.)