1. Where we’ve been & where we’re 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

2. Where we’ve been & where we’re 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

3. Minitab Output (this is for paired test, but all are similar)
Minitab: basic statistics: paired t-test:
Paired T for Country - City
N Mean StDev SE Mean
Country
City
Difference
95% CI for mean difference: (-7.71, -0.56)
T-Test of mean difference = 0 (vs not = 0): T-Value = P-Value = 0.027
(add notes about what everything means)

4. Topics to be covered in remaining 9 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…”

5. Back to Fabric Data: Tried to light 4 samples of 4 different (unoccupied!) pajama fabrics on fire.
Higher # means less flamable 18
Mean=16.85 std dev=0.94 17 16 e 15 m i 14 T n r 13 u B 12
Mean=10.95 std dev=1.237
Mean=11.00 std dev=1.299 11
Mean=10.50 std dev=1.137 10 9 1 2 3 4
Fabric

6. Suppose we want to test:
H0: m1=m2=m3=m4
HA: at least one mean is not equal.
at level a = 0.05.
Note that this is the probability of making a false claim (if they are all equal).
First idea for how to do this: do four tests at level a (m1=m2, m2=m3 etc) and reject H0 if at least one is rejected.

7. Reject all means equal if at least one test fails.
H0: m1=m2
HA: not equal
Level a=0.05
Test 2
H0: m2=m3
HA: not equal
Level a=0.05
Reject all means equal if at least one test fails.
This will give you a decision, but what’s the overall probability of making a false claim (if all means are equal) (a level) for this procedure?
>,<, or equal to a?
Test 3
H0: m3=m4
HA: not equal
Level a=0.05
Test 4
H0: m4=m1
HA: not equal
Level a=0.05

8. Overall a =
Pr(Falsely reject H0: m1=m2=m3=m4)
=Pr( at least one test falsely rejects)
=1-Pr(none falsely reject)
=1-Pr( test 1 doesn’t and … and test 4 doesn’t)
=1-(0.95^4) = 0.19 (last step uses independence…)
Point: We thought we were doing a level test, but it’s actually level 0.18! That’s a problem!
Name for this problem: multiple testing problem.
What’s one solution?

9. 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 won’t go into much mathematical detail, but these methods 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 = …

10. 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.

11. Vertical spread of the ovals is another type of variability.18 17 16 e 15 m i 14 T n r 13 u B 12 11 10 9 1 2 3 4
Fabric
Vertical spread of data points within each oval is one type of variability.
Vertical spread of the ovals is another type of variability.

12. 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 df n1-1 = denominator df
(see picture)

13. Use the test to define “large”
H0: s12 = s22
HA: s22 > s12
Level a test: reject H0 at level a if
s22 / s12 > F1-a,n2-1,n1-1

14. 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 > F1-a,3-1,16-3
When one does the test, one uses software that produce:
Analysis of variance or ANOVA tables.

15. Suppose there are k treatments and n data points. ANOVA table:
ESTIMATE OF “AMONG FABRIC TYPE” VARIABILITY
Source Sum of Mean of Variation df Squares Square F P
Treatment k-1 SST MST=SST/(k-1) MST/MSE
Error n-k SSE MSE=SSE/(n-k)
Total n-1 total SS
ESTIMATE OF “WITHIN FABRIC TYPE” VARIABILITY
P-VALUE FOR TEST.
(REJECT IF LESS THAN a)
“SUM OF SQUARES” IS WHAT GOES INTO NUMERATOR OF s2: “(X1-X)2 + … + (Xn-X)2”

16. One-way ANOVA: Burn Time versus Fabric
Analysis of Variance for Burn Time
Source DF SS MS F P
Fabric
Error
Total
Explaining why ANOVA is an analysis of variance:
MST = / 3 = 36.60
Sqrt(MST) describes standard deviation among the fabrics.
MSE = / 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 > 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.)