# Deterministic and Probabilistic (Stochastic) Models

Document Sample

```					BIOL/STAT 335                                     LAB: Third Minitab Assignment

First, let’s get familiar with using Minitab to perform two-sample t-tests by working through the following two basic problems
(a) Work through exercise 7.32 on page 245-6 of our Samuls/Witmer text. For this problem, we can type the data into the spreadsheet by
typing 1.45, 1.19, 1.05 and 1.07 into column C1 (which we could name “Flooded”), and by typing 1.70, 2.04, 1.49, and 1.91 into the column
C2 (which we could name “Control”). This format of entering data (in separate columns) is called the “unstacked” format by Minitab.
Incidentally, although it is not requested to answer the exercise, we can obtain the relevant summary stats for the two groups by the sequence
“Stat – Basic Stats – Display Basic Statistics” and typing in C1 and C2. [When you do this, do the sample means appear nearly equal?] A
second aside is to obtain side-by-side box plots so as to visualize the data. Do this by first stacking the data, which is accomplished by the
sequence “Manip – Stack – Stack Columns” and highlighting C1 and C2, storing stacked data in C3 and subscripts in C4. The graph is
obtained by the sequence “Graph – Boxplot” Y is C3 and X is C4. The graph shows a big shift from one sample to the other, which is what
our two-sample t-test is testing. Now, to do the hypothesis test (and to answer the question), follow the sequence “Stat – Basic Statistics - 2-
sample t”, which performs hypothesis tests and finds confidence intervals for two independent sample situations (Chapter 7 material). In the
box, click next to “Samples in different columns” since we entered the data into two different columns. After “First”, type or click on “C1” or
“Flooded”, and after “Second”, type or click on “C2” or “Control”. Next, check the other conditions (under “Option”). The alternative is
non-directional, so next to “Alternative”, we choose “not equal” (this is the default). The alpha level is 5%, so “Confidence Interval” being
set at “95%” is correct. Finally, leave the box next to “assume equal variances” unchecked since we do not wish to make this assumption.
Then hit the “OK” button, and the results will appear in the session window. The 95% CI for the difference of the two population means is
given and does not contain zero. It is therefore no coincidence that the test statistic is “T = -3.92” and the P-value is “P = 0.011.”
Since the P-value is below 5% (our default alpha level), we reject the null hypothesis (the null states equality of population means) in favor of
the alternative (the alternative states that the effect of flooding results in a difference in the average ATP levels). Incidentally, this conclusion
relies very strongly on our normality assumption since the sample sizes are so small, so if we are interested in re-doing our analysis using the
relevant nonparametric test (the Wilcoxon Mann Whitney test), we could follow the sequence “Stat – Nonparametrics – Mann-Whitney” and
the first sample is C1and second sample is C2. Interestingly, for this test, the P-value turns out to equal 0.0304, which, since it is still less than
5%, leads us to the same conclusion. In summary, our conclusion (that the effect of flooding results in a difference in the average ATP levels)
is independent of the normality assumption in this example (which is somewhat comforting). Before continuing, remove the data for this
problem from the spreadsheet (highlight that which you want to remove with your left mouse button and hit “Edit – Delete Cells”).
(b) For exercise 7.102 on p.304-5, we can just import the data since it is on the data files of the Samuels/Witmer textbook (Lizards.mtp of the
Chapter 7 Minitab files). The analysis is again performed by the sequence “Stat – Basic Statistics - 2-sample t”, but since the data is only in
one column, we next click next to “Samples in one column”. Next to “Samples” click on C1, and next to “Subscripts” click on C2 or C3
(doesn’t matter which you choose). Here the alternative is “less than” since we are looking for a decrease in stamina (as indicated by the
distance an animal can run in two minutes). You see the P-value (0.030 or 3.0%) is less than alpha ( = 5%), so our data indicate that malaria
does indeed result in a decrease in stamina – at least using this data and the two-sample t-test (and thus assuming normality). Fortunately, the
corresponding Mann-Whitney (nonparametric) test gives a P-value of 4.07% (verify this after first unstacking the original data), which is also
less than alpha = 5%.

BIOL/STAT 335               Third Computer Assignment

1.   (Modification of Exercise 7.94) A developmental biologist removed the oocytes (developing egg cells) from the ovaries of 24 frogs
(Xenopus laevis). For each frog the oocyte pH was determined. In addition, each frog was classified according to its response to a
certain stimulus with the hormone progesterone. The pH values are on the Samuels/Witmer data files (Oocyte.mtp of the Chapter 7
Minitab files).
(a) Use a non-directional alternative and alpha = 5% to investigate the relationship of oocyte pH to progesterone response assuming the
measurements come from normal distributions. Write down the null and alternative hypothesis, and state your conclusion along with
the test statistic and P-value.

(b) Perform the same test as in (a) but do not assume normality – again give your conclusion, test statistic and P-value.
2.   (Modification of Exercises 7.99-7.103) In an investigation of the possible influence of dietary chromium on diabetic symptoms, 14 rats
were fed a low-chromium diet and 10 were fed a normal diet. One response variable was activity of the liver enzyme GITH, which was
measured using a radioactively labeled molecule. The results, expressed in thousands of counts per minute per gram of liver are on the
Samuels/Witmer data files (Chromium.mtp of the Chapter 7 Minitab files)”.
(a) Obtain the relevant graphs and obtain a rough guess of whether you feel that the sample means differ significantly. What problem do
you see with the data?

(b) We wish to compare the average GITH levels for the two diets by finding the 90% confidence interval for the difference of the two
means. Assuming normality and ignoring the problem in (a), what is this confidence interval?

(c) If possible, use the confidence interval obtain in (b) to test for a difference between the average GITH levels for the two diets at the
alpha = 10% level of significance. Again assume normality here. State your conclusion (in the context of the problem) and give

(d) Give the corresponding 90% confidence interval assuming we can’t make the normality assumption.

3.   (Modification of Exercise 7.102) In a study of the effect of amphetamine on water consumption, a pharmacologist injected four rats
with amphetamine and four with saline as controls. She measured the amount of water each rat consumed in 24 hours; the following are
the results, expressed as ml water per kg body weight:

Amphetamine group:          118.4    124.4    144.3     105.3
Control group:              122.9    162.1    184.1     154.9

(a) Use a t-test to compare the treatments at alpha = .05. Let the alternative hypothesis be that amphetamine tends to suppress water
consumption. Write down the null and alternative hypotheses, test statistic, P-value and decision (in clear English).

(b) Use a Wilcoxon-Mann-Whitney test to compare the treatments at alpha = .05, with the directional alternative that amphetamine tends
to suppress water consumption.

(c) How would the P-value and decision change in part (a) if the pharmacologist was looking for a difference in the water consumption
for the two groups?

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 15 posted: 9/11/2012 language: English pages: 2