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Unformatted text preview: BIO318M Answer Key Section 1: (2 pts)
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F % Correct
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30 Comment Section 2: (5 pts)
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59 Comment BIO318M Answer Key Section 3
Question 21 (30 pts) 21. Question of interest – You are interested in the effect of 3,4
Methylenedioxymethamphetamine on the levels of serotonin in the brains of college
students between the ages of 18 – 25.
N.B.: If necessary, use the liberal degrees of freedom
Points were deducted for using the wrong test, i.e. the independent samples ttest or
the CI of the difference of two means.
A: Describe the steps you would carry out to understand information (6 pts)
Step 1 – Understand the data structure & define the question of interest (2 pts)
We are interested in the effect of ecstasy on serotonin levels & will compare
treated subjects with untreated subjects.
We have 2 samples based upon the question of interest (control & test)
The output variable is quantitative continuous (serotonin level /ng/mm3)
Step 2 – Describe & summarize the sample data (2 pts)
Produce summary tables & charts
As the variable is quantitative continuous then a table of n, mean & SD would
be good.
As the variable is quantitative continuous with two samples then a paneled
histogram would show the distribution of cases in the two samples (on the
same xaxis), which would show the audience what is going on in the data.
Step 3 – Inferential statistics (2 pts)
Determine if the data shown in the sample reflects true population differences
or if any sample differences are just due to random sampling error.
We can assess if the sample data reflects a true population difference using
hypothesis tests & sometimes confidence intervals. BIO318M Answer Key B: Explain which inferential test you would select (5 pts)
Question of interest & Data structure: (1 pts)
We want to determine if the serotonin levels of ecstasy treated students is different
from the serotonin levels of control treated students.
Based upon our question of interest we have two samples with a quantitative
continuous output variable
Determination of inferential statistics: (1 pts)
To determine the inferential test which is appropriate we have to see the distribution
to see which index best describes the data. Given the nature of the samples and
output variable a paneled histogram would help to answer this question.
(2 pts for correctly drawing and fully labeling the paneled histograms)
Both histograms are right skewed. To control group distribution is fairly right skewed
while the test group distribution is less right skewed. Therefore, a single
transformation would not transform both distributions to normality. Therefore, we
can NOT consider an independent ttest as we have nonnormal case distributions and
small sample sizes – therefore the sampling distribution cannot be normal (by the
central limit theorem). As the distributions are both right skewed the WMW test
would be a perfect choice of test to run. (1 pts) BIO318M Answer Key C: Conduct the most appropriate directional analysis at = 0.1 (5 pts)
As we had two samples of quantitative data which was right skewed we have chosen
to run the WMW test (2 pts deducted for running the independent ttest or CI)
H0*: There is no difference in the serotonin levels between the control & ecstasy
groups
H0: Ycontrol = Yecstasy or K1 = K2
HA*: The control group has higher serotonin levels than the ecstasy group
HA: Ycontrol > Yecstasy or K1 > K2 = 0.01
(1 point was deducted for not writing EACH correct hypothesis or stating alpha)
2
6
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K1 = 41 Control Group Ecstasy Group
18.20
10.80
26.00
16.90
28.60
19.50
40.30
20.80
83.20
22.10
87.20
23.40
98.30
48.30 0
0
1
1
1
1
4
K2 = 8 As there was a bit of manipulation of the data we confirm that we have carried out
the protocol correctly by comparing n1xn2 = K1 + K2 => 7x7 = 41 + 8 = 49.
From our analysis K1 > K2, this deviation is in the correct orientation from that
specified by the alternative, therefore we can proceed with our test.
Us = K1 = 41, n = 7, n’ = 7
From table 6 – 0.01 < P < 0.025 (For a directional test)
Comparing P to – (0.01 < P < 0.025) > 0.01
Therefore, do not reject the null
We infer that based upon the sample data at an a of 0.01 there is insufficient
evidence to say that the serotonin levels of college students between the ages of 18 –
25 years differs between control treatment and ecstasy treatment. BIO318M Answer Key NOTE: Points were deducted for not organizing your answer logically and not stating
your inference; accordingly to the grading rubric. There is no point conducting an
inferential test if you are not going to make the inference.
D: Conduct most appropriate nondirectional inferential analysis at = 0.05 (5
pts)
As we had two samples of quantitative data which was right skewed we have chosen
to run the WMW test (2 pts deducted for running the independent ttest or CI)
H0*: There is no difference in the serotonin levels between the control & ecstasy
groups
H0: Ycontrol = Yecstasy or K1 = K2
HA*: The control group has higher serotonin levels than the ecstasy group
HA: Ycontrol < = > Yecstasy or K1 < = > K2 = 0.05
(1 point was deducted for not writing EACH correct hypothesis or stating alpha) 2
6
6
6
7
7
7
K1 = 41 Control Group Ecstasy Group
18.20
10.80
26.00
16.90
28.60
19.50
40.30
20.80
83.20
22.10
87.20
23.40
98.30
48.30 0
0
1
1
1
1
4
K2 = 8 As there was a bit of manipulation of the data we confirm that we have carried out
the protocol correctly by comparing n1xn2 = K1 + K2 => 7x7 = 41 + 8 = 49.
Us = K1 = 41, n = 7, n’ = 7
From table 6 – 0.02 < P < 0.05 (For a nondirectional test)
Comparing P to – (0.02 < P < 0.05) < 0.05
Therefore, we reject the null BIO318M Answer Key We infer that based upon the sample data at an of 0.05 there is sufficient evidence
to say that the serotonin levels of college students between the ages of 18 – 25 years
differs between control treatment and ecstasy treatment.
Once we have rejected the null we can also have directional conclusions and we have
sufficient evidence to say that the cerebellum serotonin levels of college students
between the ages of 18 – 25 years treated with ecstasy are lower than control treated
levels.
NOTE: Points were deducted for not organizing your answer logically and not stating
your inference. There is no point conducting an inferential test if you are not going to
make the inference.
E: What is your best possible estimate of power (5 pts)
In this question we are asked to calculate power, however, we have only learned to
calculate power for the ttest. As the distribution of cases in both samples is not
normally distributed our possible power calculation will be inaccurate.
To calculate power we need to know: = 0.05 (for a nondirectional test)
N.B.: Remember if alpha was not stated you should also assume it to be 0.05
n(control) = 7
n(ecstacy) = 7
mean(control) = 54.54
mean(ecstacy) = 23.11
s(control) = 33.69 (largest so use for power calculation)
s(ecstacy) = 11.86
(Note SE was given & so you had to calculate s yourself – you should not calculate it
from SE = s/root n! & should ALWAYS use the raw data for your calculations to
prevent rounding errors!)
To calculate power for the ttest from table 5 we need to know:
Effect size = mean(control) – mean(ecstasy)/s(control) = 54.54 – 23.11/33.69 = 0.932
We can now look up the power (for the ttest, our best guess) from table 5.
The power of our nondirectional test at an alpha of 0.05 is, therefore, lower than
50%  as our sample sizes are less than 11 or 10 (our effect size is between 0.95 &
0.90). BIO318M Answer Key Again this is our best guess of power as we have not learned how to calculate power
for the WMW test. The power for the WMW test is not expected to be that different
from that we obtain from the ttest.
F: What would you have to question regarding this study? (4 pts)
In order for our inferences to be correct we have to ensure that we have met the
WMW inferential test requirements:
1) It must be reasonable to regard the cases as being randomly selected from a
large population (2 pts)
a. It is not possible to test this from the data once it has been collected, so
we have to review the sampling strategy very carefully to ensure that
every case in the population truly had the same probability of being in
the sample.
2) It must be reasonable to regard the cases as being independent of one another.
(2 pts)
a. It is not possible to test this from the data once it has been collected, so
we have to think if there are any connections between cases to ensure
that the cases can not influence each other and affect either our type I
or type II error rates. ...
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This note was uploaded on 02/22/2012 for the course BIO 318m taught by Professor Staff during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Staff

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