Stat 503 Div
Solutions to Homework #6 (32 points)
Problem 6.42
In a study of human blood types in nonhuman primates, a sample of 71 orangutans were tested and 14
were found to be blood type B. Construct a 95% confidence interval for the relative frequency of
blood type B in the orangutan population.
(8)
n = 71 and y = 14.  = the relative frequency of blood type B in the orangutan population.
p~ = (14+2)/(71+4) = 16/75 = 0.2133
SE p~ 
~1  p
~
p
0.21330.7867 

 0.0473
n4
75
95% CI for  is
p~  1.96SE = 0.2133  (1.96)(0.0473) = 0.2133  0.0927 = [0.12, 0.31]
We are 95% certain that 0.12 <  < 0.31. That is, we are 95% confident that the population frequency of blood
type B in the orangutan population is between 0.12 and 0.31.
Problem 6.45
The ability to taste the compound phenylthiocarbamide (PTC) is a genetically controlled trait in
humans. In Europe and Asia, about 70% of people are tasters. Suppose a study is being planned to
estimate the relative frequency of tasters in a certain Asian population, and it is desired that the
standard error of the estimated relative frequency should be 0.01. How many people should be
included in the study?
(4)
Let  be the relative frequency of PTC tasters in the specific Asian population to be studied. From the earlier
~ . We wish SE to be 0.01.
studies in Europe and Asia, we use 70% as a rough estimate for , and therefore for p
~(1  p
~)
~1  p
~ (0.7)(0.3)
p
p

 2100
we can solve for n to obtain n  4 
n4
SE 2
0.012
~ of approximately 0.01.
A sample of size n = 2096 should result in a standard error for p
Since SE =
Problem 7.2
Compute the standard error of (y1 –y2) for the following data:
Sample 1 Sample 2
N
10
10
217
y 125
S
44.2
28.7
SE(y1 - y2) =
s12 s 22


n1 n 2
(3)
44.2 2 28.7 2

= 16.7
10
10
Problem 7.14
Prothrombin time is a measure of the clotting ability of blood. For ten rats treated with an antibiotic
and ten control rats, the prothrombin times (in seconds) were reported as follows:
(8)
Antibiotic Control
N
10
10
23
y 25
S
10
8
(a) Construct a 90% confidence interval for the difference in population means. (Assume that the two
populations from which the data came are normally distributed.) Note: Formula (7.1) yields 17.2
degrees of freedom for these data.
(6)
Let 1 represent the mean prothrombin time for rats with the antibiotic, and 2 represent the mean prothrombin
time for rats without the antibiotic (control).
s12 s22
10 2 8 2



= 4.050
n1 n2
10 10
SEy1-y2 =
t.05 = 1.740 for df = 17.
A 90% CI for 1 - 2 is y1-y2  t.05 SEy1-y2 = (25 – 23)  (1.740)(4.050) = 2  7.05
= [-5.05, 9.05]  [-5, 9]. -5 < 1 - 2 < 9.
(b) Interpret the confidence interval from part (a) in the context of this setting.
We are 90% confident that the true difference in mean prothrombin time between antibiotic and control rats is
between -5 and +9 seconds. The time for rats with antibiotic could be up to 9 seconds longer, or could be as
much as 5 seconds shorter than the time for the control rats.
Problem 7.16
In a field study of mating behaviour in the Mormon cricket (Anabrus simplex), a biologist noted that
some females mated successfully while others were rejected by the males before coupling was
complete. The question arose whether some aspect of body size might play a role in mating success.
The accompanying table summarizes measurements of head width (mm) in the two groups of females.
(2)
(9)
Successful Unsuccessful
n
22
17
8.500
8.441
y
s
0.289
0.262
(a) Construct a 95% confidence interval for the difference in population means. Note: Formula (7.1)
yields 36 degrees of freedom for these data.
Let 1 be the mean head width of successful females and 2 be the mean head width of unsuccessful females.
SE =
s12
n1

s 22
n2

2
(6)
2
0.289
0.262

= 0.0885
22
17
t(36)0.025 2.028 [Okay to use t for df = 30 (t=2.042)or 40 (t=2.021) or some sensible average.]
A 95% CI for 1 - 2 is
y1-y2  t.025 SEy1-y2 = (8.500 - 8.441)  (2.028)(0.0885) = 0.059  0.180 = [-0.121,0.239]
(b) Interpret the confidence interval from part (a) in the context of this setting.
We are 95% confident that the difference in mean head width of the two populations of females (successful
minus unsuccessful) is between -0.121 and +0.239. The mean head width for successful females could be larger
by as much as 0.239 mm, or could be smaller by up to 0.121 mm, when compared to the mean head width for
the population of unsuccessful females.
(3)