Where weve been

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Where weve been where were going

• We can use data to address following questions
• 1. Question Is a mean some number? Large
  sample z-test and CI Small sample t-test and
  CI
• 2. Question Is a proportion some
  ? Proportion version of large sample
  z-test and CI

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Where weve been where were going

• 3. Question Is a diff between two means some
• Independent samples
• large sample z test and CI
• small sample t test and CI
• paired samples
• small sample paired t test and CI
• 4. Question Is diff between 2 proportions some
  
• Proportion version of large sample z test
  and CI

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Topics to be covered in remaining 8 classes
(including today)

• Analysis of Variance and Linear Regression
  (Chapters 11, 12 and 13)
• response b0 b1 covariate 1 bp
  covariate p error
• Categorical Data / Contingency Tables when
  response is discrete

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Back to Fabric DataTried to light 4 samples of
4 different (unoccupied!) pajama fabrics on
fire.
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Higher meanslessflamable
Mean16.85std dev0.94
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Mean10.95std dev1.237
Mean11.00std dev1.299
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Mean10.50std dev1.137
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Fabric
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Suppose we want to test

• H0 m1m2m3m4
• HA at least one mean is not equal.
• at level a 0.05.

Note that this is the probability ofmaking a
false claim (if they are all equal).
First idea for how to do this do four tests at
level a (m1m2, m2m3 etc) and reject H0 if at
least one is rejected.
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• Test 1
• H0 m1m2
• HA not equal
• Level a0.05

• Test 2
• H0 m2m3
• HA not equal
• Level a0.05

Reject all means equal if at least one test
fails. This will give you a decision, but whats
the overall probability of making a false claim
(if all means are equal) (a level) for this
procedure? gt,lt, or equal to a?
Test 3 H0 m3m4 HA not equal Level a0.05
Test 4 H0 m4m1 HA not equal Level a0.05
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• Overall a
• Pr(Falsely reject H0 m1m2m3m4)
• Pr( at least one test falsely rejects)
• 1-Pr(none falsely reject)
• 1-Pr( test 1 doesnt and and test 4 doesnt)
• 1-(0.954) 0.19(last step uses independence)

Point We thought we were doing a level 0.05
test, but its actually level 0.18! Thats a
problem!
Name for this problem multiple testing
problem. Whats one solution?
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Solution 1

• Do the 4 tests each at a level less than a
• Many methods to do this Bonferroni and Tukey
  are some common ones.
• We wont go into much mathematical detail, but
  these test are often conservative. (True a is
  smaller than the planned a and power is lower
  than planned.)
• For instance, divide a by of tests you do
• 1-(1-(a/4))4 1-(1-0.05/4)4 0.0491

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Solution 2 Analysis of Variance!

• Idea
• Variability in the fabric data occurs at two
  levels within fabric type and across fabric
  types.
• If across fabric type variability is large
  relative to variability within each fabric type,
  then the means are not equal.

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Fabric
Vertical spread of data points within each oval
is one type of variability. Vertical spread of
the ovals is another type of variability.
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To use the idea to test, we need a fact about
variances

• Suppose s12 s22
• If s12 is estimated by s12 from n1 data points
  and s22 is estimated with s22 from n2 data points
  (and the data are normal and independent), then
• s22 / s12 Fn2-1,n1-1

Another distribution. The F distribution. n2-1
numerator dfn1-1 denominator df
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Use the test to define large

• H0 s12 s22
• HA s22 gt s12
• Level a test reject H0 at level a if
• s22 / s12 gt F1-a,n2-1,n1-1

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• Test for fabric
• Formally
• At least one of the means is different if
• Variance among fabric types is greater than the
  variance within fabric types
• Variance among fabric types / Variance within
  fabric types gt F1-a,3-1,16-3
• When one does the test, one uses software that
  produce
• Analysis of variance or ANOVA tables.

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Suppose there are k treatments and n data
points.ANOVA table
ESTIMATE OF AMONG FABRIC TYPE VARIABILITY

• Source Sum of Meanof Variation df Squares Squa
  re F P
• Treatment k-1 SST MSTSST/(k-1) MST/MSE
• Error n-k SSE MSESSE/(n-k)
• Total n-1 total SS

ESTIMATE OF WITHIN FABRIC TYPE VARIABILITY
P-VALUEFOR TEST. (REJECT IF LESS THAN a)
SUM OF SQUARES IS WHAT GOES INTO NUMERATOR OF
s2 (X1-X)2 (Xn-X)2
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One-way ANOVA Burn Time versus Fabric Analysis
of Variance for Burn Time Source DF SS
MS F P Fabric 3
109.81 36.60 27.15 0.000 Error 12
16.18 1.35 Total 15
125.99 Explaining why ANOVA is an analysis of
variance MST 109.81 / 3 36.60 Sqrt(MST)
describes standard deviation among the
fabrics. MSE 16.18 / 12 1.35 Sqrt(MSE)
describes standard deviation of burn time within
each fabric type. (MSE is estimate of variance
of each burn time.) F MST / MSE 27.15 It
makes sense that this is large and p-value
Pr(F4-1,16-4 gt 27.15) 0 is small because the
variance among treatments is much larger than
variance within the units that get each
treatment. (Note that the F test assumes the
burn times are independent and normal with the
same variance.)