QSCI 482 Prof. Conquest TAs Kennedy, Malinick, Norman
REVIEW QUESTIONS FOR EXAM 3 [EXAM 3 IS THURSDAY DEC. 12, 2002]

1. This is a continuation of Problem #1 from HW 7. [Comparison
of 3 types of Timber Harvest: None, Moderate, Intensive.] 1e and 1f come
after having done a one-factor analysis of variance [ANOVA] on the data.

1e. Compute the best available 95% confidence interval for the INTENSIVE
mean.

1f. Compute the best available 95% confidence interval for the mean that
you get by pooling the two means from the NONE and the MODERATE groups.

2. The lengths of fish raised under 4 types of environmental conditions
are displayed below, along with the sample sizes. The data are normally
distributed.

TREATMENT GROUP

Group 1 Group 2 Group 3 Group 4
Mean 53.533 83.200 78.722 74.667
n 5 5 5 5


SOURCE OF VARIATION df SS

Total(corrected) 3125.880
Between Groups 2588.212
Within Groups (error) 537.668


GROUP q(Group 2) q(Group 3) q(Group 4)
Group 1 _________ _________ __________
Group 4 3.29 _________
Group 3 1.73

a. Fill in the analysis of variance table [alpha=.05] to test for equality
of mean fish lengths under the 4 types of environmental conditions. Write
out your null and alternative hypotheses, and include the final P-value
along with your results.

b. Fill in the 4 missing q-values that you need to do either a Tukey or
Student-Newman-Keuls (SNK) multiple comparison.

c. Now that you have all the computed q-values, do BOTH a Tukey multiple
comparison and a SNK multiple comparison; use significance level alpha =
0.05. Summarize your results for each method regarding which means can and
cannot be declared significantly different.

3. The following data set concerns an experiment in which the
growth increments of insect larvae (increases in length in mm) were
measured after one week of feeding on artifical diets under controlled
environmental conditions. There are 3 levels of Dietary Protein [Low,
Medium, and High]. There is also the Alkaloid factor [Absent in the diet,
or Present in the Diet]. There are 4 replicates in each combination of
Dietary Protein and Alkaloid.

NO ALKALOID WITH ALKALOID MEAN TOTAL

LOW 3,5,4,6 6,7,5,8 5.50 44
PROTEIN

MED. 3,2,4,5 7,8,6,7 5.25 42
PROTEIN

HIGH 7,6,5,4 5,2,3,2 4.25 34
PROTEIN

MEAN 4.5 5.5 5.00
TOTAL 54 66 120

The Sums of Squares for the different sources of variation for the above
data are as follows:

SOURCE OF VARIATION SS
Level of Protein 7.

Alkaloid

Interaction 39.

Error 28.

TOTAL 80.

a. At the .10 level of significance, test for the presence of INTERACTION.
Be sure to write out both your null and alternative hypotheses. Do a plot
of the means; use both the plot and the result of the statistical test to
interpret the results.

b. At the .05 level of significance, test for the overall effect
of PROTEIN LEVEL. Use both the plot of means [that you did in part 'a'
above] and the test to interpret the results.

c. At the .05 level of significance, test for the overall effect
of ALKALOID. Use both the plot of means and the test to interpret the
results.

d. Do the best available 95% confidence interval for the average growth
increment for the pooled {Low+Medium} Protein levels.

4. Scientists wish to test whether four Varieties of house plants reach
the same maximum height. They are also keenly aware that growing
conditions vary considerably in the greenhouse. Therefore, six green house
benches were set up as blocks (the assumption is that on each bench, the
growing conditions are the same). Each bench had the four Varieties
represented exactly once, assigned at random to a particular place on the
bench. The response variable is maximum plant height (cm.):

Block Variety 1 Variety 2 Variety 3 Variety 4
1 19.8 21.9 16.4 14.7
2 16.7 19.8 15.4 13.5
3 17.7 21.0 14.8 12.8
4 18.2 21.4 15.6 13.7
5 20.3 22.1 16.4 14.6
6 15.5 20.8 14.6 12.9

Test the null hypothesis (alpha = .05) that all four Varieties of plants
reach the same maximum height. If you end up concluding that there is an
equality somewhere among the Varieties, then follow up with a multiple
comparison technique to find out where the differences are.

Some needed SS are: SS(Varieties) = 188.54; SS(Blocks) = 19.79;
SS(Total, Corrected) = 214.86.

5. Some experimenters are trying to decide what is a good number of
treatment groups for their experiment. They know they can afford about 24
or 25 statistical replicates, total. They must decide between:

Plan A: 5 treatments with 5 replicates per treatment, for a total of N=25;

OR

Plan B: 4 treatments with 6 replicates per treatment, for a total of N=24.

Plan A includes more treatments in the experiment (and has a larger N),
and Plan B has more replicates per treatment (but a smaller N).

They decide to resolve the issue in the following manner: they will find
the Minimum Detectable Difference between the largest and smallest
population mean for 90% power, and they will choose the design that gives
them the "finer" [i.e., smaller] value for the Minimum Detectable
Difference. Using this rule, which design is better? IMPORTANT NOTE: In
your computations, you may assume that the response variable has been
standardized so that MSE = 1.0.