Math 295
November 11, 2002
Solutions #9
Problem A. (From text, question 5.3.11) In a much-hyped TV event, 100 people independently
chose between Schlitz and Bud in a blind taste test. In the event, 54 chose Bud. Let p
be the true fraction of randomly-chosen people that would have (in the long run) chosen
Bud.
a. Construct (by the usual method) a 95% confidence interval for p.
 1 
The confidence interval is pˆ  1.96  

 n

pˆ (1  pˆ )

with pˆ 
54
and
100
n = 100... hence, 0.54 + 0.098 or [ 0.442, 0.638 ]
b. Summarize this result in 5 words or less from the point of view of a Budweiser
marketing manager.
“More people chose Bud.”
c. Summarize this result in 5 words or less from the point of view of a Schlitz marketing
manager.
“It wasn’t statistically significant.”
d. In standard usage, what is the “margin of error” of this “poll” ?
Use ½ in place of

pˆ (1  pˆ )
…
M.O.E. is
 1 
 1/ 2  = 0.098.
 n
1.96  
Problem B. (From text, question 5.3.1) One measure of respiratory function is called
FEV1/VC. In the general population, FEV1/VC is normally distributed with mean 0.80
and standard deviation 0.09. For people who have been exposed to a certain enzyme, we
think the distribution is still normal and still has standard deviation  = 0.09, but we
think the mean  might be different. Here are the FEV1/VC values for 19 people who
were exposed to the enzyme:
.61
.72
.78
.85
.70
.64
.84
.73
.63
.82
.83
.85
.76
.88
.82
.87
.67
.82
.74
a. Find the obvious point estimate for  (the mean of exposed people generally).
x  0.7663
1
b. Construct a 95% confidence interval for .
 1 
Use x  (1.96) 
  . We are assuming that we know , so we can
 19 
use it in preference to sx (which is 0.0859). Since we are assuming
that the underlying population is normal, we can expect X-bar to be
normally distributed even if n < 30.
This formula gives
0.7663  0.0405 or
0.7258,0.8068 .
c. Is it plausible that  = .80; that is, that the exposed people have the same
distribution as the general population?
From the point of view of the 95% confidence interval, yes, since
=0.80 is within the interval.
Even if 0.80 were a bit outside the interval, it would be hard to say
that this value is “implausible” based on a 95% significance level.
Problem C. A certain random variable W is normally distributed with mean 15.0 and standard
deviation 4.0. What is P(W > 24.33), the probability that any particular draw from this
distribution is greater than 24.33 ?
W  15.0
. Then Z is the “standardized” version of W, and
4.0
Z has a standard normal distribution. Clearly, W > 24.33 if and only if
24.33  15.0
 2.33 . From a table or calculator, the standard normal
Z>
4.0
lower tail probability corresponding to 2.33 is 0.990. We want the
Define Z 
upper tail probability, P(W>24.33) = P(Z>2.33) = 1 – 0.990 = 0.010.
2
Problem D. Examination of the daily percentage returns from holding UAL stock during the
period 1987-2001 gives these summary values:
number of observations:
3785
average daily return:
0.000185 (that is, about (2/100)% per day)
sample standard deviation (sx):
0.021115 (meaning, typical days’ returns
were on the scale of +– 2%)
(Should I have listed all 3785 values? Ask for them if you want them.)
Suppose we have a theory that the returns are all drawn independently from the same
distribution, and that it has mean , standard deviation .
a. What are reasonable point estimates for  and ?
ˆ  x  0.000185, ˆ  sx  0.021115.
b. Find any reasonable confidence interval for .
95% confidence interval:
1.96  0.021115
or 0.000185  0.000673
0.000185 
3785
For a 90% confidence interval, use 1.65 in place of 1.96, etc.
to get 0.000185  0.000566 .
c. Suppose that this distribution is actually normal, and that  is exactly what your
point estimate says it is. From these observations only, what do you think is the
probability that  is actually positive?
Since 0.00 is well inside even a 90% confidence interval (or even a
one-sigma confidence interval) we can’t have much confidence in 
being positive. If we want to say that UAL is a good stock to invest
in, we should have some evidence beyond even this 15-year data set.
Problem E. Here are the amounts of monetary damage, in each of the 101 years 1900-2000,
from eruptions of Mt. St. Helens:
1900
1901
1902
etc.
1979
1980
1981
etc.
2000
$0
$0
$0
$0
$0
$ 4 billion
$0
$0
$0
Assume these are independent draws from some distribution with mean .
3
a. Do the best you can to construct a 95% confidence interval for .
Use n = 101, x 
$4 bil.
 $39.6 mil. , and
101
sx  $398 mil.
This gives a 95% confidence interval of $ 39.6 mil +– $77.6 mil.
or [ –$38 mil , +$117.2 mil. ].
b. Does it make any sense?
Not much sense. Negative numbers aren’t possible for this variable,
and even if they were possible, they have never been observed in 101
random draws. So including negative numbers in the confidence
interval is pretty silly.
What’s going on here is that the underlying distribution is nowhere
near normal, and despite the central limit theorem and the fact that
n ≥ 30, X isn’t close to a normal distribution either. So the logic
behind the confidence interval doesn’t hold up well.
This situation is a serious problem for insurance companies. There are
better methods, but no really good methods.
4