Chapter 12 Section 1
12.17 Calculating the ANOVA F test P-value, continued. For each of the following situations, find the F
statistic and the degrees of freedom. Then draw a sketch of the distribution under the null hypothesis and
shade in the portion corresponding to the P-value. State how you would report the P-value.
(a) Compare 5 groups with 9 observations per group, MSE = 50, and MSG = 127.
127
= 2.54. The degrees of freedom are: numerator is 4,
50
denominator is 5(9) – 5 = 40.
F=
P(F > 2.54) = 0.05463
(b) Compare 4 groups with 7 observations per group, SSG= 40, and
SSE = 153.
The degrees of freedom for sample mean differences is 4 – 1 = 3: MSG =
The degrees of freedom for the pooled standard deviation
153
is 4(7) – 4 = 24: MSE =
= 3.825
24
F=
13.33
= 3.485
3.825
P(F > 3.485) = 0.03134
40
= 13.33
3
FOR PROBLEMS 18 AND 19 YOU WILL NEED TO USE THE APPLET FOUND AT
WWW.WHFREEMAN.COM/IPS6E LOOK AT THE APPLETS SECTION, SCROLL DOWN TO THE BOTTOM; IT IS THE LAST
APPLET.
12.18 The effect of increased variation within groups. The One-Way ANOVA applet lets you see how
the F statistic and the P-value depend on the variability of the data within groups and the differences among
the means.
(a) The black dots are at the means of the three groups. Move these up and down until you get a
configuration that gives a P-value of about 0.01. What is the value of the F statistic?
I got an F value of about 5.537 which just about corresponded
to a p-value of 0.01. I can see from the graph, that actual pvalue is little larger than 0.01.
(b) Now increase the variation within the groups by dragging the mark on the pooled standard error scale to
the right. Describe what happens to the F statistic and the P-value. Explain why this happens.
The default value on the pooled standard deviation slider was all the way to the left, thus, I could only
move to the right. As I moved the slider to the right, increase the pooled standard deviation, the
individual group spread increased while the means of each group did not change.
As the spread increased the F statistic decreased to below 1, increasing the p-value near 1. Notice that
as F decreases in value, the p-value increases getting closer to the limit of 1.
Why does this happen? As the spread increases within each group, it makes the difference between
the means, insignificant, that is, I am willing to believe that the means of each group are the same and
the variation between the sample means that I am witnessing is due to random variation. Look at the
Situation where p-value is 0.01. The means for each group are not different. But now the spreads are
so close together that I can see clearly that at least one group is different; the green group I would
guess has a slightly smaller mean. WARNING: if we find a result statistically significant, it does NOT
mean the result difference is LARGE. We are merely stating that this is different.
So the picture shows that there is a difference in the means, from what I can see, it is hard to tell if the
difference is large (from the picture it does not appear that way) since there is no vertical scale, but in
any case the difference seems to be real.
12.19 The effect of increased variation between groups. Set the pooled standard error for the One-Way
ANOVA applet at a middle value. Drag the black dots so that they are approximately equal.
(a) What is the F statistic? Give its P-value.
The p-value is near 1, which corresponds to an F value of
0.0126.
The interpretation is that there is no difference between
means.
(b) Drag the mean of the second group up and the mean of the third group down. Describe the effect on the
F statistic and its P-value. Explain why they change in this way.
As the means change in difference then the F –value
starts to increase, which means that we are increasing
the amount of evidence to support the alternative
hypothesis that at least one of the means is different.
We are trying to find if there is a difference in the
means. If we continue to change the means so the
difference increases then the evidence will begin to
mount; larger F-value and smaller p-value. Also I
changed the group means in opposite direction
increasing the difference between all three. If I move the
last group mean upward then the F-value should
decrease.
12.20 Calculating the pooled standard deviation.
An experiment was run to compare four groups. The sample sizes were 25, 28, 150, and 21, and
the corresponding estimated standard deviations were 42, 38, 20, and 45.
(a) Is it reasonable to use the assumption of equal standard deviations when we analyze these data? Give a
reason for your answer.
Smaller s value = 20. 2(20) = 40. Now 40 is not greater than 45. So the rule of thumb is not met.
2smallest > slargest.
(b) Give the values of the variances for the four groups.
422 = 2025
382 =1444
202 = 400 452 = 2025.
(c) Find the pooled variance.
24(2025) + 27(1444) + 149(400) + 20(2025)
= 853.13
25 + 28 + 150 + 21 - 4
(d) What is the value of the pooled standard deviation?
853.13 ≈ 29.21
(e) Explain why your answer in part (d) is much closer to the standard deviation for the third group than to
any of the other standard deviations.
The third group accounted for 150 values out of the total of 224 values. The formula for calculating
pooled standard deviation accounts for the sample size by multiplying each individual standard
deviation by its degrees of freedom (n - 1). This type of calculation is called weighted. Thus the value
of 20 for the standard deviation was given more “weight” than the other values.
12.34 Air quality in poultry-processing plants. The air in poultry-processing plants often contains fungus
spores. If the ventilation is inadequate, this can affect the health of the workers. To measure the presence of
spores, air samples are pumped to an agar plate, and "colony-forming units (CPUs)" are counted after an
incubation period. Here are data from the "kill room" of a plant that slaughters 37,000 turkeys per day, taken
at four seasons of the year. The units are CFUs per cubic meter of air."
Fall
1231
1254
1088
Winter
987
778
852
Spring
2054
2092
1902
Summer
1452
1521
1352
(a) Examine the data using graphs and descriptive measures. How do airborne fungus spores vary with the
seasons?
The graphs clearly illustrate that there is a difference
between the means. The sample means look different,
the overlap between the groups is non-existence, so it is
hard to imagine that these 12 data values came from a
distribution with the same mean. The standard
deviations look about the same, and it is easy to believe
that this is the case. If appears that the spore count is
larger on average during the Spring, and by Winter it
decreases to its lowest point.
(b) Is the effect of season statistically significant?
When we run the formal test I am expecting to see a tiny p-value. As a matter of fact I would guess
about 8.2 X 10-8, clearly statistically significant.
Summary statistics:
Column n
Fall
3
Winter
3
Spring
3
Mean
Variance
1191
Std. Dev.
Std. Err.
8089 89.938866
51.92623
872.3333 11230.333 105.97327 61.183693
2016
10108 100.53855
Summer 3 1441.6666 7220.3335
Sample mean regardless of group x =
84.97254
58.04596
49.05892
1191(3)  872.33(3)  2016(3) 1441.67(3)
= 1380.25
12
SSG = 3(1191 – 1380.25)2 + 3(872.33 – 1380.25)2 + 3(2016 – 1380.25)2 + 3(1441.67 – 1380.25)2
= 2105234.92
SSE = 2(8089) + 2(11230.33) + 2(10108) + 2(7220.33) = 73,295.32 = sp2
ANOVA
Source of Variation
Between Groups (difference between means)
SS
2105234.92
df
3
called sometimes within groups
73295.32
8
Total variation regardless of group, s, calculated
by grouping entire data set as one.
2178530.25
11
MS
F
P-value
Error (Estimating the pooled standard deviation)
For the groups I = 4, thus the degrees of freedom is 4 – 1 3. MSG =
2105234.92
= 701744.97
3
For the pooling situation, ni = 4 for each group, thus N = 12. The degrees of freedom for pooling is N
73295.32
– 1, or 12 – 4 = 8. MSE =
= 9161.92
8
ANOVA
Source of Variation
Between Groups (difference between means)
SS
2105234.92
df
3
MS
701744.97
called sometimes within groups
73295.32
8
9161.92
Total variation regardless of group, s, calculated
2178530.25
11
Error (Estimating the pooled standard deviation)
F
P-value
by grouping entire data set as one.
Finally the F statistics is F =
701744.97
= 76.59.
9161.92
Finally P(F > 76.59) with 3 df and 8 d.f. is 3.1 x 10-6.
ANOVA
Source of Variation
Between Groups (difference between means)
SS
2105234.92
df
3
MS
701744.97
called sometimes within groups
73295.32
8
9161.92
Total variation regardless of group, s, calculated
by grouping entire data set as one.
2178530.25
11
Error (Estimating the pooled standard deviation)
F
76.59
P-value
3.1x10-6