STAT500 HW#6_solutions
Homework 6 Solutions
1.
a.
This poses an interesting question since we never discussed “intuitive” decisions prior
to performing an analysis.
So here is an intuitive thought process.
Since one assumption is that the each population follows a normal distribution,
then one can consider the application of 689599 rule.
That is, 68%, 95% and 99% of all
observations fall within 1, 2, or 3 standard deviations respectively.
Using 95% and the 2
SD rule, we can take the lowest mean, 0.1605 and highest mean, 0.2735, and build
“loose” 95% confidence intervals around each [for ease, use sqrt(n) = sqrt(6) = 2]:
0.1605 +/ (2*0.1767/2) =  0.3372 to 0.0162
0.2735 +/ (2*0.2492/2) =  0.0243 to 0.5227
Notice these intervals do not overlap.
Without an overlap we can deduce that there is a
difference between these two means, and thus not all means are equal.
b.
Below is the output of the ANOVA.
The pvalue of 0.011, compared to alpha of
0.05, suggests that our sample provides statistical evidence that not all means are equal.
This is consistent with our intuitive decision in part a.
At least one device provides Ph
readings that differ, on average, than the other three devices.
Oneway ANOVA: Ph versus Device
Source
DF
SS
MS
F
P
Device
3
0.5837
0.1946
4.85
0.011
Error
20
0.8029
0.0401
Total
23
1.3866
S = 0.2004
RSq = 42.10%
RSq(adj) = 33.41%
c.
The pvalue of our F test is 0.011
d.
The three conditions are: 1. Independent random samples (assumed by what was stated
in the problem).
2. Equal variances (assumed since the test of equal variances in
MINITAB produce a Bartlett’s pvalue of 0.748 and Levene’s pvalue of 0.926, both
greater than alpha of 0.05), and 3. the normal distribution assumption is satisfied by the
probability plot of the residuals (pvalue of 0.550).
e.
Random sampling of a wide range of Ph values could result in one device receiving
much more variable sample than the remaining devices.
Having a greater variability in
samples may cause this device to have more variability in its readings compared to
another device(s) that receive samples that are more similar.
This sampling bias could
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 '08
 Chow
 Regression Analysis, Errors and residuals in statistics

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