Lecture 14: Comparisons and Residuals
Test of Hypothesis
Conclusion
F Ratio = 40.9837
P-value: < 0.0001
Because the P-value is so
small we should reject the
null hypothesis.
There are statistically
significant differences
between some of the mean
volumes for the various
times.
1
Conclusion?
2
Compare Means
Confidence Intervals for the
difference in two means.
Least Significant
Difference.
We don’t know from the
analysis which times
produce the different mean
volumes.
3
Confidence Interval
4
Margin of Error
∗
∗
1
1
2.131 187.73
95% Confidence, df = df Error
2.131 7.9106
5
1
1
1
6
1
6
16.86
6
1
Lecture 14: Comparisons and Residuals
Confidence Intervals
Comparison
1.25 to 1.75
1.25 to 2.25
1.75 to 2.25
Interpretation
If a confidence interval contains
zero, then there could be no
difference in treatment means.
If a confidence interval does not
contain zero, then there is a
statistically significant difference
in treatment sample means.
95% Confidence Interval
–61 ± 16.86 (–77.86, –44.14)
–63 ± 16.86 (–79.86, –46.14)
–2 ± 16.86 (–18.86, 14.86)
7
8
Conclusion
Alternative Method
The differences in mean volumes for
1.25 compared to 1.75 and 1.25
compared to 2.25 are statistically
significant (no zero in the CI).
The difference in mean volumes for
1.75 compared to 2.25 is not
statistically significant (zero in the
CI).
The margin of error is also called
the Least Significant Difference
(LSD). An absolute difference
in treatment sample means
bigger than the LSD is
considered statistically
significant.
9
Least Significant Difference
Comparison
1.25 to 1.75
1.25 to 2.25
1.75 to 2.25
?
61 > 16.86 statistically significant
63 > 16.86 statistically significant
2 < 16.86 not statistically significant
11
10
Conclusion
The differences in mean volumes for
1.25 compared to 1.75 and 1.25
compared to 2.25 are statistically
significant (no zero in the CI).
The difference in mean volumes for
1.75 compared to 2.25 is not
statistically significant (zero in the
CI).
12
2
Lecture 14: Comparisons and Residuals
Oneway Analysis of Volume (cl)
centered by Time (min) By Time (min)
Analysis of Residuals
25
20
15
10
5
0
Plot residuals versus the
treatments.
Compute a standard deviation
of residuals for each of the
treatments.
-5
-10
-15
-20
-25
1.25
1.75
2.25
Time (min)
Means and Std Deviations
Level Number
13
Interpretation
1.25
1.75
2.25
6
6
6
Mean
Std Dev
0
0
0
16.4073
12.0996
12.1491
14
Analysis of Residuals
1.25 minutes has a slightly larger
standard deviation (16.41 cL)
than the other two times
(standard deviations around 12.0
cL).
The equal standard deviation
condition is satisfied.
Analyze the distribution of
residuals to see if they
could have come from a
normal distribution.
15
16
Distributions
Volume (cl) centered by Time (min)
Interpretation
1.64
1.28
0.67
0.0
-0.67
0.9
0.8
0.7
0.6
0.5
0.4
0.3
 Histogram is skewed right with two
mounds.
 Box plot skewed right.
 Normal quantile plot shows two groups.
 The normal distribution of errors
condition is probably not met.
However, this violation will not affect
our conclusions from the formal
analysis.
0.2
-1.28
0.1
-1.64
6
5
4
3
2
1
-30
-20
-10
0
10
20
30
17
18
3