STAT 112 --- Quiz 8 --- Solutions
Problem 1. The biting rate of a particular species of fly was investigated in a study reported
in the Journal of the American Mosquito Control Association (Mar. 1995). Biting rate was defined
as the number of flies biting a volunteer during 15 minutes of exposure. This species of fly is
known to have a median biting rate of 5 bites per 15 minutes on Stanbury Island, Utah. However,
it is theorized that the median biting rate is higher in bright, sunny weather. To test this theory,
122 volunteers were exposed to the flies during a sunny day on Stanbury Island. Of these
volunteers, 95 experienced biting rates greater than 5.
a. (3 pts) Set up the null and alternative hypotheses for the test.
Let η denote the median biting rate of this particular species of fly.
H0: η = 5 versus H1: η > 5
b. (5 pts) Calculate the approximate p-value of the test.
Since the sample size by far exceeds 10, we’ll use the normal approximation to the
binomial to compute the p-value=P(X≥S) where S=number of volunteers experiencing
biting rates of more than 5 per minute. We are given that S=95, hence the z-test statistic
is given by
z
95  .5  .5(122)
.5 122
 6.066
The p-value of this test is the area under the standard normal curve to the right of the
observed test statistic value of 6.066. Thus, p-value ~ 0, providing very strong evidence
against the null and in favor of the alternative.
c. (2 pts) Make the appropriate conclusion at α=.01.
See answer in b. above.
Problem 2. (10 pts) Suppose you want to compare two treatments A and B. In particular, you
wish to determine whether the distribution for population B is shifted to the right of the
distribution for population A.
a. (3 pts) Specify the null and alternative hypotheses you would test.
Let DA and DB denote the distributions of populations A and B, respectively.
H0: DA = DB versus H1: DA is shifted to the left of DB
b.
Sample
A
rank
Sample
B
rank
(7 pts) Suppose you obtained the following independent random samples of observations
on experimental units subjected to the two treatments. Conduct the test of hypotheses
described in part a. using α = .05.
37
40
33
29
42
33
35
28
8
65
9
35
3.5
47
2
52
10
3.5 6.5 1
13
6.5
11
12
34
TA
5
48.5
TB
42.5
Since nB < nA, the test statistic is TB. The rejection region is TB ≥ 39 from table XV, with nA =
4, nB = 10 and α = .05.
Since the observed value of the test statistic 42.5 is greater than 39, we reject the null in
favor of the alternative to conclude that there is sufficient evidence to indicate the
distribution for population B is shifted to the right of the distribution for population A at α
= .05.