Biostatistics for the Health Sciences: One-Way Analysis of Variance - Exercises questions answers
EXERCISES
13.1 Complete the following ANOVA table:
13.2 Complete the following ANOVA table:
13.3 Why does one use a Tukey’s HSD rather than a t test when comparing mean differences
in ANOVA?
13.4 Samples were taken of individuals with each blood
type to see if the average white blood cell count differed among types. Ten
individuals in each group were sampled. The results are given in the table
below:
Average White Blood Cell Count by Blood Type
Source: Modification to Exercise 10.9, page 171,
Kuzma and Bohnenblust (2001).
a. State the null hypothesis.
b. Construct an ANOVA table.
13.5 Using the data from the example in Exercise 13.4
and the ANOVA table from that exercise, determine the p-value for the test (use the F
statistic and the appropriate degrees of freedom based on the within and
between sum of squares). Is there a statistically significant difference in the
white blood cell counts among the groups?
13.6 Five individuals were selected at random from
three communities, and their ages were recorded in the table below. The
investigator was interested in de-termining whether these communities differed
in mean age.
Ages of Individuals (n = 5 in Each Group) in Three Communities
a. State the null hypothesis.
b. Construct an ANOVA table.
13.7 Using the data from the example in Exercise 13.6
and the ANOVA table from that exercise, determine the p-value for the test (use the F
statistic and the appropriate degrees of freedom based on the within and
between sum of squares). Is there a statistically significant difference in the
ages among the groups?
13.8 Researchers studied the association between birth
mothers’ smoking habits and the birth weights of their babies. Group 1
consisted of nonsmokers. Group 2 comprised smokers who smoked less than one
pack of cigarettes per day. Group 3 smoked more than one but fewer than two
packs per day. Group 4 smoked more than two packs per day.
Birth Weights of Infants (n = 11 in Each Group) by Mother’s Smoking Status
Use the above table to construct an ANOVA table for
the test of no mean differences in birth weight among the groups. What is the p-value for this test? What do you conclude
about the effect of smoking on birth weight?
13.9 Four brands of cereal are compared to see if they
produce significant weight gain in rats. Four groups of seven rats each were
given a diet of the respective cereal brand. At the end of the experimental
period, the rats were weighed and the weight was compared to the weight just
prior to the start of the cereal diet. Determine whether each brand has a
statistically significant effect on the amount of weight gain. The data are
provided in the table below.
Rat Weight by Brand of Cereal
13.10 A botanist wants to determine the effect of
microscopic worms on seedling growth.
He prepares 16 identical planting pots and then introduces four sets of worm
populations into them. There are four groups of pots with four pots in each
group. The worm population group sizes are 0 (intro-duced into the first group
of four pots), 500 (introduced into the second group of four pots), 1000
(introduced into the third group of four pots), and 4000 (introduced into the
fourth group of four pots). Two weeks after planting, he measures the seedling
growth in centimeters. The results are given in the table below.
Seedling Growth in Centimeters by Worm Population
Group
a. State the null hypothesis and determine the
ANOVA table.
b. What is the result of the F test?
c. Apply Tukey’s HSD test to see which means differ
if the ANOVA was significant at the 5% level.
13.11 Analysis of variance may be used in an industrial
setting. For example, man-agers of a soda-bottling company suspected that four
filling machines were not filling the soda cans in a uniform way. An experiment
on four machines doing five runs each gave the data in the following table.
Liquid Weight of Machine-Filled Cans in Ounces
Based on the analysis of variance, is there a
difference in the average num-ber of ounces filled by the four machines? Apply
Tukey’s HSD test to compare the mean differences if the overall ANOVA test is
significant at the 5% level.
13.12 The following table shows the home run production
of five of baseball’s greatest
sluggers over a period of 10 years. Each has hit at least 56 home runs in a
season and all but Griffey have had seasons with 60 or more. Sosa, Bonds, and
Griffey are still active, McGwire has retired, and Ruth is de-ceased, so this
time period constitutes the final 10 years of McGwire’s and Ruth’s respective
careers.
Home Run Production for Five Great Sluggers
a. Construct an ANOVA table to test whether or not
there are statistically significant differences in the home run production of
these sluggers over the ten-year period.
b. If the F
test indicates significant differences at the 0.05 significance level, apply
Tukey’s HSD to see if there is a slugger who stands out with the lowest
average. Is there a slugger with an average significantly higher than the rest?
Is Bonds at 42.5 significantly higher than Sosa at 37.0?
Answers:
13.1 Complete the following ANOVA table:
13.3 Since we are looking at more than one pair of mean
differences, there are multiple
hypothesis tests, each having its own type I error. We want to control
si-multaneously the type I errors that we could make. Tukey’s method guarantees
that the probability of making a type I error on any of the tests is controlled
to be less than α. A simple α-level t test on two or more
mean differences would not provide such a control.
13.11 We construct an ANOVA table based on the data in
the table below:
From above, the within-group sum of squares is
0.01376. The grand mean is 12.0125. So the between-group sum of squares is
5{(12.038 – 12.0125)2 + (12.02 – 12.0125)2 + (12.012 –
12.0125)2 + (11.98 – 12.0125)2} = 5(0.00065025 + 0.00005625
+ 0.00000025 + 0.00105625) = 5(0.001763) = 0.008815.
The result is significant at the 5% level since the
critical f with 3 and 16 degrees of freedom
is 3.24. So Tukey’s test is appropriate. Recall that HSD = q(α, k, N – k) √(MSw/n)
where n is the number of
observations per group, k is the
number of groups, N = kn is the total sample size, and q(α, k, N – k) is gotten from Tukey’s table for the studentized range. In
this case k = 4, n = 5, N = 20, and MSW = 0.00086. So HSD = q(α, 4, 16) 0.01311. We take α = 0.05 and from the table get q
= 4.05. So HSD = 0.0531. So we can
reject the hypothesis that the two means are equal if their differences are 0.0531 or more. The mean differences are 0.018 for
A minus B, 0.026 for A minus C, 0.058 for A minus D, 0.008 for B minus C, 0.040 for B minus D and 0.032 for C minus D. Note that only A minus D gives a value
greater than HSD. So we conclude that D is less than A but cannot be
confident about a differ-ence between any other pairs.
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