Sample Exam #2 BioStats

Sample Exam #2
BioStats
Instructions. No books, notes, or calculators are allowed.
1. (15 points) Find the minimum sample size required to estimate a population
proportion, p, given that the margin of error is four percentage points for a 95%
confidence interval where p is estimated as pˆ = 12.5%.
Solution:
2
z0.05/2
pˆ(1 − pˆ)
E2
=
(1.96)2 (0.125)0.875
= 262.6094 so n = 263.
0.042
2. (20 points) The Common Murre population of birds became endangered in California due to an oil spill. Biologists want to know if their efforts to restore the
colony have been successful. Since the breeding colony of birds is site dependent, the mean number of breeding pairs of Common Murres were observed at
a specific rock known to be a breeding site. Find the p–value of a test that the
mean number of breeding pairs of Common Murre is greater than 400. Use the
following data collected from seven days of observations:
404 398 402 406 399 405 403
Assume the data comes from a normal distribution.
Solution: Let H0 : µ = 400 and H1 : µ > 400. One calculates x¯ = 402.4286
√ = 2.147487. The p–value
and s2 = 8.952381. The t(7 − 1) statistic is x¯s/−400
7
= 1 − 0.9623164 = 0.03768361 (between 0.025 and 0.05 using the Table A3).
3. (15 points) Medical researchers reviewing the risk of death for elderly patients
taking dementia drugs found the following data.
Number of Patients
Total Deaths
Dementia Drugs
500
23
Placebo
450
12
BioStats
Sample Exam #2
Find a 95% confidence interval for the difference in death rate proportions.
Solution:
s
23
477
23
12
lower bound =
−
− 1.96 500 500 +
500 450
500
s
23
477
23
12
upper bound =
−
+ 1.96 500 500 +
500 450
500
12
450
438
450
= −0.004304544
450
12
450
438
450
450
= 0.04297121
Thus the 95% confidence interval is (−0.0043, 0.043)
4. (15 points) Using the following data, use a significance level of 0.05 to determine
whether there is a significant linear correlation between the two variables.
x
y
Solution:
0
8
5
X
n = 5,
3
2
4
6
5
9
xj = 24,
j=1
5
X
j=1
x2j
= 194,
5
X
j=1
12
12
5
X
yj = 37,
j=1
yj2
= 329,
5
X
xj yj = 219.
j=1
Thus
5(219) − 24(37)
p
= 0.6277225.
r=p
5(194) − 242 5(329) − 372
r
The t(5 − 2) statistic is q
= 1.3967. For H0 : ρ = 0 versus H1 : ρ 6= 0,
1−r2
5−2
one has a p–value of 0.2569 (or greater than 0.20 using Table A3). There
is not sufficient evidence to support the conclusion of a significant linear
correlation.
BioStats
Sample Exam #2
5. (20 points) It is known that 41% of Americans have type A blood, 9% have type
B, 4% has type AB and 46% have type O. A random sample of 200 stomach
cancer patients yielded 92 with blood type A, 20 with type B, 4 with type AB
and 84 having type O. Is the data significant at the 5% level to enable us to reject
the null hypothesis that the blood type distribution of stomach cancer suffers is
the same as the general population?
Solution: One has
o1 = 92,
o2 = 20,
o3 = 4,
o4 = 84
and
e1
e2
e3
e4
=
=
=
=
np1
np2
np4
np4
= 200(0.41) = 82,
= 200(0.09) = 18,
= 200(0.04) = 8 and
= 200(0.46) = 92.
Thus the test statistic is
4
X
(92 − 82)2 (20 − 18)2 (4 − 8)2 (84 − 92)2
(oj − ej )2
=
+
+
+
= 4.1374.
ej
82
18
8
92
j=1
Using Table A-4, since χ20.05 (4 − 1) = 7.815 > 4.1374 we cannot reject at
5% significance H0 , that the distribution of blood types for stomach cancer
patients is the same as the general population.
6. (15 points) Consider the following Two-Way ANOVA table (missing some values).
Source
df
Iron-form
1
Dosage
2
Iron-form:Dosage 2
Error
—
Total
107
SS
MS
F
P
2.074 2.074 5.99 0.0161
15.588 —–
—– 0.0007
—–
0.405 1.17 0.3145
35.296 —–
53.768
BioStats
Sample Exam #2
Fill in the missing values. Is there an effect from interaction between Iron–form
and Dosage?
Solution:
df for Error = 107 − 1 − 2 − 2 = 102
SS for Iron-form:Dosage = 2(0.405) = 0.810
15.588
MS for Dosage =
= 7.794
2
25.296
MS for Error =
= 0.346
102
7.794
F for Dosage =
= 22.53
0.346
Source
df
SS
MS
F
Iron-form
1
2.074 2.074 5.99
Dosage
2 15.588 7.794 22.53
Iron-form:Dosage 2
0.810 0.405 1.17
Error
102 35.296 0.346
Total
107 53.768
P
0.0161
0.0007
0.3145
There is not interactive effect (p–value of 0.3145 for H0 : no effect).