DATA ANALYSIS
Problem:
Courtney Leisner (MS Botany, 2009) designed an experiment was designed to explore the growth of the
plant (Bienertia sinuspersici), a species with C4 photosynthesis and is salt tolerant. Four levels of salt
concentration were of interest to the researcher (0, 50, 100 and 200 mM NaCl). The experiment consisted of
16 tubes identically planted to the plant species and then randomly assigned to the four salt treatments (4
replicate tubes per treatment). The tubes were placed in individual hydroponic growth systems at the
specified concentration of NaCl in the water and the plants allowed to grow for 8 weeks. At the end of the
growth period, the plants were harvested and the dry weight of the roots measured. The following table
presents the data from this experiment.
NaCl Concentration (mM)
REPLICATE 0 50 100 200
1 2.4 40.9 70.5 127.2
2 10.6 47.7 96.7 134
3 12 31.9 26.4 28.1
4 4.6 36.3 76 78.3
MEAN 7.4 39.2 67.4 91.9
Show your work for each of following. All parts must be included.
For parts 2, 3 and 4 the necessary SAS code and analyses are available on pages 4 - 7 of the homework notes.
1. Construct, by hand, the entire ANOVA table for this analysis. Test the hypothesis of equal
means and show all five parts for the test. Use α = 0.05 for the test.
= 26022.31
= 15924.27
SS Error = SS Total – SS Error = 26022.31 - 15924.27
SS Error = 10098.04
ANOVA TABLE
Hypothesis Formula for Mean Square:
Null Mean Sum of Squares (ERROR & TREATMENT)
Ho: µ1+ = µ2 = µ3 = µ4 SS ERROR
MSERROR =
Alternative n−k
Ha: At least one µi ≠ µj for i ≠ j
SS TREATMENTS
MSTREATMENTS=
= 0.05 k−1
F Value
5308.09
F value=
841.503 MSTREATMENT
F=
MSS ERROR
F = 6.31
SS ERROR SS TREATMENTS
MSERROR = MSTREATMENTS =
n−k k−1
10098.04
= 15924.27
16−4 =
4−1
10098.04
= 15924.27
12 =
3
MSERROR = 841.503
MSERROR = 5308.09
Therefore, Ho (Variance are equal) is rejected and have a sufficient evidence that mean root
dry weight differs across four replications of NaCl concentration.
2. Assess the assumption of normality using:
a) The Anderson-Darling test.
TEST FOR NORMALITY RESULTS IN SAS CODE UNIVARIATE PROCEDURE
Ho: Errors are normally distributed
Ha: Errors are not normally distributed
P- Value = 0.0471 (Anderson-Darling)
IF P-value < 0.05 Null hypothesis will be rejected.
Therefore, Ho (Errors are normally distributed) is rejected and conclude that errors are not
normally distributed
3. Asses the assumption of equal variance using:
a). Lavene’s Test
TEST FOR NORMALITY RESULTS IN SAS CODE GLM PROCEDURE
TREATMENT RESPONSE MEDIAN ABSOLUTE VALUE
MEDIAN BASED
RESDIUALS
0 2.41 7.6 5.19
0 10.6 7.6 3
0 12 7.6 4.4
0 4.6 7.6 3
50 40.9 38.6 2.3
50 47.7 38.6 9.1
50 31.9 38.6 6.7
50 36.3 38.6 2.3
100 70.5 73.25 2.75
100 96.7 73.25 23.45
100 26.4 73.25 46.85
100 76 73.25 2.75
200 127.2 102.75 24.45
200 134 102.75 31.25
200 28.1 102.75 74.65
200 78.3 102.75 24.45
AVONA FOR ABSOLUTE VALUE MEDIAN BASED RESDIUALS
0 50 100 200
1 5.19 2.3 2.75 24.45
2 3 9.1 23.45 31.25
3 4.4 6.7 46.85 74.65
4 3 2.3 2.75 24.45
MEAN 3.8975 5.1 18.95 38.7
GRAND MEAN 16.661875
Ho: Variances are equal
Ha: Variances are not equal
F = 4.04 and P-value = 0.0335
Reject Ho if F > F (0.05, 3, 12) = 3.490 or P-value < 0.05
Therefore, Ho (Variance are equal) is rejected and have a sufficient evidence that variances
are not equal.
4. Interpret the results shown in the graph of the residuals plotted against the treatment
levels.
Based on the graph of the residuals against the treatment levels, at Zero (0) Concentration
salt level reached 10+, at 50 salt level (50), at 100 salt level (96) and at Salt Level 200 is
about 130 that is 13 times much bigger compare to Salt Level 0. That concludes, the obtain
variances of 4 replication in each 4 different Salt Level is not Equal.
5. Summarize the results of parts 2 and 3. Include in your summary whether any remedial
efforts should be undertaken and what form those efforts should take.
Both P-VALUE in part 2 and part 3 are being rejected because it doesn’t meet the condition. The
test for Normality results is equal to 0.0471 (Anderson-Darling) which is less than 0.05 and
Assumptions for equal variance have 4.04 F critical and 0.0335 P-value which is F critical is larger
than 0.05 and P-value lower than 0.05. Remedial changes are needed for the analysis that leads to
the Invalidation of the One way-Anova results and raw data. Either a transformation of the
response should be tried or proc mixed might be used and the variances modeled
6. By hand, produce the residual, standardized residual and studentized residual for the last
two observations (28.1 and 78.3) in the 200 mM NaCl treatment. Do you think either of these
observations might represent an extreme value? Explain your answer.
OBSERVATION RESPONSE VALUE STANDARDIZED STUNDENTIZED
RESIDUAL RESIDUAL RESIDUALS
28.1 -63.8 -2.199 -2.539
78.3 -13.6 -0.4688 -0.5414
The obtain value for standardized and studentized residuals for the observation 28.1 is -2.99 and
-2.539 while in 78.3 is -0.4688 and -0.5414. The observation in 28.1 is less than negative two (-2)
which considered as extreme value and for observation 78.3 are too low and not greater than or less
than negative two which doesn’t consider as extreme value. Generally, if conditions are satisfied
across all observation do not expect that all values are not extreme which is greater than two or less
than negative 2 and expect to have extreme values.
