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# STAT 292 Assignment 3 solution

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## 1. Skink Temperatures

Skinks are tested for their preferred daytime temperature. Each one is placed in a
long tank which is warmer at one end, cooler at the other. The temperature at the
position where it settles is recorded. There are four different species of skink, and
we wish to test (at the 5% level of significance) whether the species differ in their
preferred temperature.

The following table gives the data.
Species Preferred temperatures (oC) Total Mean
A 18 21 22 18 20 19 19 23 17 22 199 19.9
B 24 18 19 21 20 17 23 22 22 19 205 20.5
C 22 21 24 19 25 18 23 21 24 22 219 21.9
D 21 19 26 24 25 21 20 20 27 25 228 22.8
SAS output is given on pages 3 and 4.

(a) When running the experiment, other possible factors such as time of day, light,
amount of food recently eaten, are kept as near constant as possible. Why?
(b) The skinks are not put in the tank together. Why?
(c) Give values of n and p (the number of treatments) for this experiment. How
many degrees of freedom are in the Treatments row, the Error row and the
Total row of the ANOVA table? (Give the algebraic expressions and the actual
values for this experiment.)

(d) Use the output to write up the ANOVA in the style suggested in the Assignment
Guidelines on page 1. You should include a statement of the (complete) model
equation, and also comments on whether the assumptions are satisfied. Use a
5% significance level for the ANOVA test.

One-Way Analysis of Variance
Dependent Variable: Temperature
Source DF Sum of Squares Mean Square F Value Pr > F
Model 3 52.0750000 17.3583333 3.06 0.0402
Error 36 203.9000000 5.6638889
Corrected Total 39 255.9750000
R-Square Coeff Var Root MSE Temperature Mean
0.203438 11.18633 2.379893 21.27500
One-Way Analysis of Variance

Levene’s Test for Homogeneity of Temperature Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Species 3 86.1428 28.7143 1.36 0.2691
Error 36 757.4 21.0391
Means and Descriptive Statistics
Species Mean of
Temperature
Std. Dev. of
Temperature
Minimum of
Temperature
Maximum of
Temperature

21.275 2.5619253577 17 27
A 19.9 2.0248456731 17 23
B 20.5 2.2730302828 17 24
C 21.9 2.2335820757 18 25
D 22.8 2.8982753492 19 27

## 2. Nasal Sprays

Improvement in breathing airflow is measured for twenty-five people suffering from
nasal congestion. They were treated with either a saline spray (A) or one of four
nasal sprays (B, C, D, E) available over the counter in pharmacies.
Spray Airflow improvement Total Mean
A 15 10 16 14 8 63 12.6
B 25 41 37 44 26 173 34.6
C 21 6 9 15 14 65 13.0
D 16 7 24 22 15 84 16.8
E 24 15 39 34 30 142 28.4
527 21.08

Relevant SAS output follows, on pages 5 to 7.
Write a report that compares the five treatments using the guidelines on page 1.
Ensure that you comment on all the included SAS output. Use a 5% significance
level for all statistical tests. Make a recommendation for either a single best nasal
spray, or a group of best choices which are similar in their effects; refer to the Tukey
One-Way Analysis of Variance
Results: Nasal Spray Example
The ANOVA Procedure
Class Level Information
Class Levels Values
Spray 5 A B C D E

Number of Observations Used 25
Dependent Variable: Improvement
Source DF Sum of Squares Mean Square F Value Pr > F
Model 4 1959.440000 489.860000 9.73 0.0002
Error 20 1006.400000 50.320000
Corrected Total 24 2965.840000

R-Square Coeff Var Root MSE Improvement Mean
0.660669 33.65113 7.093659 21.08000
Source DF Anova SS Mean Square F Value Pr > F
Spray 4 1959.440000 489.860000 9.73 0.0002
STAT 292, 2020 5 Assignment 3
The ANOVA Procedure
Nasal Spray Example

Levene’s Test for Homogeneity of Improvement Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Type 4 11893.9 2973.5 1.59 0.2156
Error 20 37393.0 1869.6
Level of
Type N
Improvement
Mean Std Dev

A 5 12.6000000 3.43511281
B 5 34.6000000 8.67755726
C 5 13.0000000 5.78791845
D 5 16.8000000 6.68580586
E 5 28.4000000 9.28977933
Tukey’s Studentized Range (HSD) Test for Improvement
Note: This test controls the Type I experimentwise error rate, but it generally has a higher
Type II error rate than REGWQ.
Alpha 0.05
Error Degrees of Freedom 20
Error Mean Square 50.32

Critical Value of Studentized Range 4.23186
Minimum Significant Difference 13.425
Means with the same letter
are not significantly different.
Tukey Grouping Mean N Type
A 34.600 5 B
A
B A 28.400 5 E
B
B C 16.800 5 D
C

C 13.000 5 C
C
C 12.600 5 A

Nonparametric One-Way ANOVA
The NPAR1WAY Procedure: Nasal Spray Example
Wilcoxon Scores (Rank Sums) for Variable Improvement
Classified by Variable Type
Type N
Sum of
Scores
Expected
Under H0

Std Dev
Under H0
Mean
Score
A 5 36.50 65.0 14.682756 7.30
B 5 108.00 65.0 14.682756 21.60
C 5 35.00 65.0 14.682756 7.00
D 5 55.50 65.0 14.682756 11.10
E 5 90.00 65.0 14.682756 18.00

Average scores were used for ties.
Kruskal-Wallis Test
Chi-Square 15.8695
DF 4
Pr > Chi-Square 0.0032

3. Forensic dental X-rays
The extent to which X-rays can penetrate tooth enamel has been suggested as a suitable mechanism for differentiating between females and males in forensic medicine
(e.g., think about shows like ‘CSI’ and parts of ‘NCIS’). The table below gives spectropenetration gradients for one tooth from each of eight females and eight males.
Gender Y = spectropenetration gradient Mean Std. dev.
Female 4.8 5.3 3.7 4.1 5.6 4.0 3.6 5.0 4.5125 0.7605

Male 4.9 5.4 5.0 5.5 5.4 6.6 6.3 4.3 5.4250 0.7440
Note that a high reading reflects a fast drop-off in X-ray penetration, with less
penetration by X-rays.
(a) Explain why the teeth have been sampled from eight different people of each
sex, and not eight teeth from one female and eight from one male.
(b) Given that the researcher could afford to test n = 16 subjects, explain the
advantages of choosing eight from each group.

(c) SAS output from an ANOVA is on pages 9 and 10. Write a report, following
the guidelines on page 1.
(d) Explain why there is no point doing a Tukey test with this data.
STAT 292, 2020 8 Assignment 3
One-Way Analysis of Variance
The ANOVA Procedure
Class Level Information
Class Levels Values

Gender 2 Female Male
Number of Observations Used 16
Source DF Sum of Squares Mean Square F Value Pr > F
Model 1 3.33062500 3.33062500 5.88 0.0294
Error 14 7.92375000 0.56598214
Corrected Total 15 11.25437500

R-Square Coeff Var Root MSE Xray_grad Mean
0.295940 15.14099 0.752318 4.968750
Source DF Anova SS Mean Square F Value Pr > F
Gender 1 3.33062500 3.33062500 5.88 0.0294

The ANOVA Procedure
Levene’s Test for Homogeneity of Xray_grad Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
Gender 1 0.00189 0.00189 0.01 0.9305
Error 14 3.3504 0.2393
Level of
Gender N

Mean Std Dev
Female 8 4.51250000 0.76052144
Male 8 5.42500000 0.74402381

4. Personality types
In psychology, there are tests to classify people into one of many personality types.
An experiment is run to find the extent of the influence of personality type on the
subject’s score in a certain test. A random sample of four personality types is taken,
and within each type a random sample of ten subjects is taken. Each subject is given
the test, and the score Y is recorded, with data as follows:
Type Test Score, Y

T1 50 52 44 49 60 51 40 41 54 39
T2 63 45 48 49 65 55 47 58 57 56
T3 50 52 47 48 44 56 55 39 51 53
T4 39 38 51 50 53 53 59 41 45 48
(a) Explain why this is a random effects design, rather than a fixed effects design.
(b) Some SAS output is given on pages 12 and 13. Note that the boxplots do not
include estimates of group means, since any differences in population means
are not the focus of this investigation.

whether the relevant assumptions seem satisfied. Give your estimated components of variance, plus the percentage of the total variance of Y that is due to
personality, along with the percentage unexplained,
Do you think personality type is important in determining the score on this
particular test?

SAS Output for Personality Type Example
Box Plot
One-Way Analysis of Variance
Results
The ANOVA Procedure
Class Level Information
Class Levels Values
PersType 4 T1 T2 T3 T4

Number of Observations Used 40
Dependent Variable: Score
Source DF Sum of Squares Mean Square F Value Pr > F
Model 3 279.675000 93.225000 2.23 0.1017
Error 36 1506.700000 41.852778
Corrected Total 39 1786.375000

R-Square Coeff Var Root MSE Score Mean
0.156560 12.97117 6.469372 49.87500
Source DF Anova SS Mean Square F Value Pr > F
PersType 3 279.6750000 93.2250000 2.23 0.1017
STAT 292, 2020 12 Assignment 3
SAS Output for Personality Type Example
Levene’s Test for Homogeneity of Score Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
PersType 3 2400.7 800.2 0.49 0.6926
Error 36 59004.7 1639.0

5. Phytoremediation
Phytoremediation (New Scientist, 20 Dec 1997, p.26) is a process by which plants
are used to remove toxic metals from the soil. For example, sunflowers were used
around Chernobyl, where there was radioactive contamination from a nuclear power
station accident.
Certain plants take up toxic metals (e.g. zinc, cadmium, uranium) and accumulate
them in their vacuoles as protection against chewing insects and infection.
Suppose that four species of plant were tested, at lower and higher soil pH, for their
uptake of zinc, Y , measured in parts per million (ppm) of dry plant weight at the
end of the trial.

Uptake of zinc, Y, (ppm):
Soil pH
Plant Name 5.5 (acid) 7 (neutral)
Lettuce 250 470 330 400 310 430
Martin red fescue 2850 2380 3130 1070 960 1300
Alpine pennycress 6340 4280 5170 2880 4330 3050
Bladder campion 3690 4750 5100 2360 1990 2140
(a) What kind of design is this? Give the model equation, including an interaction
term.

(b) SAS analysis of the data using the model from part (a) was tried on both raw
data Y and transformed data log Y . Diagnostic graphs from both analyses
are given on pages 15 and 16. Explain, with reasons, whether it is better to
analyse Y or log Y .
(c) Further SAS output is given on pages 17 to 19. Present a report and your
conclusions, following the usual guidelines. Use a 5% significance level.

SAS Output for Phytoremediation Example
STAT 292, 2020 15 Assignment 3
SAS Output for Phytoremediation Example
STAT 292, 2020 16 Assignment 3
SAS Output for Phytoremediation Example
Linear Models
The GLM Procedure
Class Level Information
ClassLevelsValues
pH 2acid neutral

Number of Observations Used24
Dependent Variable: logZinc
Source DFSum of SquaresMean SquareF Value Pr > F
Model 7 24.03027190 3.43289599 92.71<.0001
Error 16 0.59245521 0.03702845

Corrected Total 23 24.62272711
R-Square Coeff Var Root MSElogZinc Mean
0.9759392.589699 0.192428 7.430507
Source DF Type I SSMean SquareF Value Pr > F
pH 1 1.46932364 1.46932364 39.68<.0001
Plant 321.65807393 7.21935798 194.97<.0001
pH*Plant 3 0.90287433 0.30095811 8.130.0016

Source DF Type III SSMean SquareF Value Pr > F
pH 1 1.46932364 1.46932364 39.68<.0001
Plant 321.65807393 7.21935798 194.97<.0001
pH*Plant 3 0.90287433 0.30095811 8.130.0016