QSCI 482  Prof. Conquest          TAs Kennedy, Malinick, Norman
            HW 7--DUE FRIDAY, DECEMBER 6, 2002

1. Scientists conducted a survey of stream segments in western
Washington to assess the effect of three different levels of commercial
timber harvest (none, moderate, intensive) on instream salmon spawning
habitat. One of the responses measured on each stream segment (the
sampling unit) was the percent of stream area comprised of pools. (Pools
are related to needed habitat for salmon spawning.) Data and
the summary statistics are as follows:

NONE: 43, 73, 36, 46, 42.  n = 5,  xbar=48.00,  s^2=208.50.
MODERATE: 39, 62, 20, 44.  n = 4,  xbar=41.25,  s^2=298.25
INTENSIVE: 37, 14, 21, 30, 07. n = 5, xbar=21.80, s^2=144.7.

1a. At the .05 level of significance [and assuming normality and equal
variances], test to see whether the three levels of timber harvest yield
the same mean pool fraction of stream area. Include the complete analysis
of variance (ANOVA) table. You may do this either by hand (it is indeed
do-able by hand) or by using statistical software like SPSS. NOTE: in
writing up your "conclusions"  statement, AVOID the words "accept",
"reject", and "hypothesis"; rather, express conclusions in terms of the
original research question.

1b. Are the three levels of timber harvest a "fixed effects" model or a
"random effects" model? Explain your answer. 

1c. If we were to use a sample size of n=6 for each group (total sample
size = 18), how far apart would the largest and smallest population
means have to be in order to reject the null hypothesis with 90% power
and level of significance = .05?

1d.  Now, at the alpha = 5 percent level of significance, do a parametric
(normality based) multiple comparison of means using the Student-Newman
Keuls method of multiple comparisons.

2.  A study comparing the effects of different toxic substances upon
aquatic communities yields the following results [response variable is:
a "toxicity index"--the higher the index, the worse the contamination].

CONTROLS: 105.0, 103.5, 84.2, 93.6, 113.6, 68.5, 124.7, 68.8
PCBs: 134.6, 140.1, 118.5, 122.3, 120.5
CADMIUM: 111.4, 112.0, 90.7, 103.9, 98.6,
MERCURY: 107.8, 132.0, 105.1, 149.0, 106.9

DESCRIPTIVE STATS:
CONTROLS: xbar =  95.24; std. dev. = 20.38
PCBs:     xbar = 127.20; std. dev. = 9.56
CADMIUM:  xbar = 103.32; std. dev. = 8.98
MERCURY:  xbar = 120.16; std. dev. = 19.54
pooled MSE = 269.62; pooled std. dev. = 16.42

Using analysis of variance, we have been able to reject the null
hypothesis of equality of the 4 treatment means. For multiple comparison
purposes, we are really only interested in whether or not each of the 3
non-control groups [PCBs, Cadmium, Mercury] differ from the control in
terms of the mean.  At the .05 level of significance, carry out the
appropriate test to see if each of the PCB mean, Cadmium mean, Mercury
mean differs from the control mean. Summarize your conclusions.

3. The following data are from an experiment to look at
weight gain [in gms] of lab mice under 3 different diets which have
low, medium, and high amounts of protein and other nutrients.  It was
also felt that male mice might respond to the diets differently than
female mice, so sex of the animal was also noted.
      FEMALES                 MALES

LOW   40.8              49.2
      40.5              43.8
      43.9              37.6

MED   51.5              50.9
      50.0              57.9
      51.9              59.9

HIGH  62.7              74.0
      56.4              72.1
      64.7              73.9

MEAN

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

      SOURCE OF VARIATION             SS     
      Diet                            1831.9
      Sex                             179.9 
      Diet x Sex Interaction          82.4   
      Error                           161.0  
      ----------------------          -----
      TOTAL                   2255.2

a. At the .10 level of significance, test for the presence of interaction
between Diet and Sex.  Why did the result of the test turn out the way it
did? Use a PLOT of the means [plot can be done by hand] to explain why.

b. At the .05 level of significance, test for the overall effect of Sex on
the mean weight gain.  Why did the result of the test turn out the way it
did?--refer to your plot from [a] to answer this.

c. At the .05 level of significane, test for the overall effect of the
three Diets on the mean weight gain.  Why did the result of the test turn
out the way it did?--refer to your plot from [a] to answer this.