QSCI 482/Dr. Conquest TAs Kennedy, Malinick
HW1--DUE *THIS*
CLASS SURVEY SHEET ON DAY 1, PLEASE DO SO BY THE END OF THE WEEK.
1. Attach a copy of a recent photo of yourself; it will help me learn
everyone's names. I will return it if you wish. You don't have to be
smiling ["mug shots" are OK] but photo must be in good taste [no pictures
of your last skinny-dipping party, please. No "Full Monty's".]. Thanks!
2. In a 1995 the Puget Sound Gilnetters Association conducted a field test
to test the effects of different kinds of fishing gear on alcid bird
entanglement (diving seabirds like rhinocerous auklet and common murre,
covered under the Migratory Bird Treaty Act). First, a test was done to
see if the total bird entanglements followed a Poisson distribution with
mean mu=.09, since this is what was found from previous field seasons.
(That is, the overall entanglement rate is about .09 birds per net set.)
Out of 449 net sets (a fishing net set in the water and fishing for the
same specified period of time), 418 of them had 0 birds caught, 30 sets
had 1 bird caught, and 1 of the sets experienced 2-or-more birds caught.
At the .05 level of significance, do a goodness-of-fit test to test
whether the observed entanglements follow a Poisson distribution with mean
mu=.09, against the alternative hypothesis that they do not follow such a
distribution. What do you conclude? [NOTE. The Poisson probability for a
particular count, X, is on p. 571 in Zar (in the 3rd ed. it's p. 569) and
is: exp(-mu)*(mu^X)/X!.]
[Why do we care about whether the data follow this distribution or not? In
order to plan experimental designs for future field tests, it was
necessary to check out some specific distributional assumptions first. The
eventual outcome of field studies done in 1995 and 1996 was a new
minimize accidental bird entanglement.]
3. This exercise involves computing probabilities for the binomial
distribution (in Zar, the chapter titled "More on Dichotomous Variables",
or look under "binomial distribution" in any elementary statistics book).
According to a mathematician who works there, The Anchor Gaming Company is
the largest supplier of slot machines in the world. Let's consider a
simple slot machine and compute some EXPECTED VALUES OF OUTCOMES FOR A
BINOMIAL PROBABILITY DISTIRIBUTION. Suppose a slot machine has 4 slots,
each of which shows a picture of either a cherry or a pear, and the slots
work independently. For a given slot, the probability that a cherry shows
up is 0.4, and the probability that a pear shows up is 0.6. So for a given
"pull" of the machine, one can observe either: 4 cherries; 3 cherries and
1 pear; 2 cherries and 2 pears; 1 cherry and 3 pears; or 4 pears. Suppose
a given machine is "pulled" 10,000 times. How many times would we expect
to see:
a. 4 cherries
b. 3 cherries and 1 pear
c. 2 cherries and 2 pears
d. 1 cherry and 3 pears
e. 4 pears
The following problems (4-6, on the back side of this sheet) fall under the heading, "algebra review". If you have not done algebra for awhile, this will give you needed practice. If this stuff comes easily to you, give a classmate some help--we're all in this together.
4. You may recall that the formula for the standard error for a sample
mean X-bar is: Standard Error of X-bar = sqrt(sample variance/n), where n
is the sample size, and "sqrt" stands for "square root of". You are
reading a journal paper and trying to find out what the original sample
size was (which the authors failed to state, and the editors did not catch
the omission). In a table, the authors state that "the standard error for
the data was 10.3 kg." Elsewhere in the paper, you find that the sample
variance was 1,697.44 kg^2. Now, solve for the original sample size, n.
5. We have a random sample of n = 5 weights (kg) of a particular kind of
animal: 3.1, 3.4, 3.6, 3.7, 4.0. The sample mean, X-bar, is 3.56 kg.
Now compute the sample variance of these data, using the standard
"machine formula" calculation for the sample variance, s^2:
s^2 = {Sum([Xi^2]) - [(Sum[Xi])^2]/n}/(n-1)
6. The following data (X) are known to come from a lognormal distribution;
that is, the natural logarithms of the data, Y = ln(X), will follow a
normal (bell-shaped) distribution. Here are the data in the original
units: 3.67, 4.01, 3.85, 3.92, 3.71, 3.88, 3.74, 3.82 ml.
a. Compute the sample mean of the log-transformed data.
b. Compute the sample variance of the log-transformed data. Take the sqrt
to get the sample standard deviation.
c. It turns out that the 95% confidence interval (CI) for the mean of the
log-transformed data is xbar +- 2.365*(std. deviation/sqrt(n)). Compute the 95%
CI for the log-transformed data; then transform each of the endpoints back
to get the 95% CI in the original units [ml]. Comment upon the symmetry of
the CI for the log-transformed units [ln(ml)], and the back-transformed
units [ml].