Transcript 11-w11-stats250-bgunderson-chapter-16-more

Author(s): Brenda Gunderson, Ph.D., 2011
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ANOVA for 3 Drugs
ANOVA
RESPONSE
Between Groups
Within Groups
Total
Sum of
Squares
21.981
41.988
63.969
df
2
16
18
Mean Square
10.991
2.624
F
4.188
Sig.
.034

Give estimate of common population standard deviation

Give observed test statistic value
ANOVA for 3 Drugs
ANOVA
RESPONSE
Between Groups
Within Groups
Total
Sum of
Squares
21.981
41.988
63.969
df
2
16
18
Mean Square
10.991
2.624
F
4.188
Sig.
.034

What is the distribution of the test statistic if the three drugs are
equally effective in terms of the mean response?

What is the corresponding p-value?

At the 5% level, what is your (decision and) conclusion?
Think about this and be ready to click in …
What is your decision and conclusion?
ANOVA
RESPONSE
Between Groups
Within Groups
Total
A)
B)
C)
D)
Sum of
Squares
21.981
41.988
63.969
df
2
16
18
Mean Square
10.991
2.624
F
4.188
Sig.
.034
Reject H0 and conclude that the three drugs do appear to be
equally effective in terms of the average response.
Reject H0 and conclude that at least one of the three drugs
appears to be different in terms of the average response.
Fail to reject H0 and conclude that the three drugs do appear to
be equally effective in terms of the average response.
Fail to reject H0 and conclude that at least one of the three
drugs appears to be different in terms of the average response.
We Reject H0 … What is Next? Multiple Comparisons page 178

At least one population mean is different? Which one(s)?

Pairwise Comparisons:
Drug 1 vs Drug 2, Drug 1 vs Drug 3, Drug 2 vs Drug 3

Does CI contain 0?
Are two population means significantly different?


Use a multiple comparisons method that controls for
overall Type I error rate (or for overall confidence level).
TUKEY’S Method (via SPSS only)
Try It! Comparing 3 Drugs

Multiple
Comparisons
We rejected H0 … which
drug(s)
significantly differ?
Dependent Variable: TIME
Tukey HSD
(I) DRUG
1
2
3
(J) DRUG
2
3
1
3
1
2
Mean
Difference
(I-J)
-1.0800
1.4200
1.0800
2.5000*
-1.4200
-2.5000*
Std. Error
.9485
.9485
.9485
.8659
.9485
.8659
Sig.
.505
.318
.505
.027
.318
.027
95% Confidence Interval
Lower Bound Upper Bound
-3.5276
1.3676
-1.0276
3.8676
-1.3676
3.5276
.2657
4.7343
-3.8676
1.0276
-4.7343
-.2657
*. The mean difference is significant at the .05 level.
a. Use output to report about the three comparisons:
 Does the confidence interval for m1 – m2 contain 0?

Does the confidence interval for m1 – m3 contain 0?

Does the confidence interval for m2 – m3 contain 0?
Try It! Comparing 3 Drugs
b. State conclusions regarding the differences between
the mean response for 3 drug groups from the
Tukey multiple comparison method.
The population mean responses appear to differ for …
but not differ for …
Individual Confidence Intervals
for the Population Means

We assume population standard deviations are all
equal, so estimate of that common population standard
deviation s p  MSE is used in forming individual CIs.

DF for t* multiplier = N – k.
From Utts, Jessica M. and Robert F. Heckard. Mind on Statistics, Fourth Edition. 2012.
Used with permission.
Try It! Comparing 3 Drugs




k = 3 groups based on N = 19 obs
Pooled standard deviation sp = 1.62
Drug 1:
Sample mean = 8.22
Sample size = 5
Drug 2:
Sample mean = 9.30
Sample size = 7
Drug 3:
Sample mean = 6.80
Sample size = 7
DF for t* multiplier is N – k = ______.
For level 0.95, t* = __________________
Drug 3 was descriptively the best. Compute a 95% CI
for the mean response for all subjects taking Drug 3.
Try It! Memory Experiment

Three groups of subjects –
each group given list of words to remember.

Lengths: 1st group = 10 words (short), 2nd group =
20 words (medium), 3rd group = 40 words (long).

% of words recalled for each subject recorded.

Sample mean percentage of words recalled:
Short: 68.3%, Medium: 48%, Long: 39.2%

One-way ANOVA used to assess whether length
of list had significant effect on % of words recalled.
Try It! Memory Experiment
Between
Groups
Within
Groups
Total
SS
df
2668.8
2
Mean
Square
F
Sig.
.0003
84.6
3852.9
16
a. Complete the ANOVA table.
b. State hypotheses that the above F statistic is testing.
H0: _______________________________________
Ha: _________________________________________
Click in for part (c)
Click in: Yes or No
c. Using a 5% level, does it appear that the
average % words recalled is the same for the
three different lengths of list?
SS
df
Mean Square
F
Sig.
Between Groups
2668.8
2
1334.4
15.77
.0003
Within Groups
1184.1
14
84.6
Total
3852.9
16
Try It! Memory Experiment
Multiple Comparisons
d. Multiple Comparisons:
Dependent Variable: Percentage of words recalled
Circle
all pairs significantly different (at 1% level).
Tukey HSD
(I) GROUP
s hort lis t
medium list
long list
(J) GROUP
medium list
long list
short list
long list
short list
medium list
short vs medium
Mean
Difference (I-J)
20.33
29.17
-20.33
8.83
-29.17
-8.83
short vs long
99% Confidence Interval
Lower Bound Upper Bound
1.06
39.61
10.79
47.54
-39.61
-1.06
-10.44
28.11
-47.54
-10.79
-28.11
10.44
medium vs long
e. Calculate a 99% CI for mean % words recalled for long
list group, where sample mean based on 6 subjects
in long list group was 39.2 percent.
SS
df
Mean Square
F
Sig.
Between Groups
2668.8
2
1334.4
15.77
.0003
Within Groups
1184.1
14
84.6
Total
3852.9
16
What if conditions do not hold?
Pg 181

Section 16.3: two methods used when one or both
assumptions about equal popul std devs and normal
distributions are violated.

When data skewed, or extreme outliers present, better
to analyze median rather than mean.

Tests for comparing medians:


Kruskal-Wallis Test: Compare relative rankings (sizes) of the
data in the observed samples, called a rank test or
nonparametric test (no assumptions made about a specific
distrib for population of measurements.
Mood’s Median Test
Two-Way ANOVA

One-way ANOVA: only one explanatory variable
and one quantitative response variable.

Section 16.4 = overview of two-way ANOVA

Possibility of interaction …effect of one factor
on mean response depends on specific level of other factor.
Want to learn more?
Consider taking for Stats
401 next term 
Another great stat class (no prereq) = Stats 408 (winter only)