MATH220 Test 1 Solutions Fall 2014 1. (12 pts - it

MATH220
Test 1
Solutions Fall 2014
1. (12 pts) True or False: Put a T or F in front of each number.
T. 1. Both stratified sample and cluster sample divide the population
into groups.
F. 2. In a cluster sample, the population is divided into groups, and a
random sample from each group is drawn.
F. 3. A sample of convenience is never acceptable.
F. 4. A confounder makes it easier to draw conclusions from a study.
F. 5. In an observational study, subjects are assigned to treatment
groups at random.
F. 6. Observational studies are generally more reliable than experiments.
F.7. If every individual in a sample has of an equal chance of being
selected, then this must be a simple random sample.
T. 8. In a case-control study, the outcome has occurred before the
subjects are sampled.
T. 9. A nominal variable is a qualitative variable with no natural
ordering.
T. 10. A person’s height is an example of a continuous variable.
T. 11. A double blind experiment means neither the investigator nor
the subjects know who has been assigned to which treatment.
T. 12. In a randomize experiment, the treatment groups are not too
different from each other except that they receive different treatments.
2. (14 pts) Find Q1, Q2, Q3, IQR and use quartiles and IQR to check
potential outliers for this data set.
70, 10, 20, 30, 40, 50, 60, 70, 80, 90, 110, 205.
The ordered values are:
10, 20, 30, |
40, 50, 60,
1
||
70, 70, 80,
|
90, 110, 205.
Q1 = 35
Q2 = 65
Q3 = 85
IQR = Q3 − Q1 = 50
Lower outlier boundary = Q1 − 1.5 ∗ IQR = 35 − 1.5 ∗ 50 = −40
Upper outlier boundary = Q5 + 1.5 ∗ IQR = 85 + 1.5 ∗ 50 = 160.
Outliers are 205 since 205 > 160. No values are smaller than the lower
outlier boundary.
3. (16 pts) The table shows the frequency of primary food choices of 200
alligators captured in Florida lakes.
Primary food choice
frequency
Fish
Invertebrate
Other
Reptile
Bird
86
58
30
16
10
relative frequency
0.43
0.28
0.15
0.08
0.05
(a) Put a column of relative frequency (proportion or percent) in the
table.
(b) Make a bar chart for the variable Primary food choice. Put the
relative frequency on the vertical axis.
Omitted.
(c) What is the mode for primary food choice?
The mode is Fish which has the higher frequency among all the
categories.
(d) In addition to a bar chart, what chart can be used as well to
display this variable? Only name this chart. You do not need to
make the chart.
A pie chart can also be used to display a categorical variable.
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4. (20 pts) The table below classifying auto accidents by survival status
(S=survival, D=died) and seat belt status of the individuals involved
in the accident.
Belt
Yes
No
Outcome
S
412,368
162,527
D
510
1601
Total
412,878
164,128
Total
574,895
2111
577,006
(a) Estimate P (D), the probability that the individual died in the
auto accident, i.e., find the the proportion of the individuals who
died in the auto accident among all who had auto accidents.
2111
= 0.0037.
P (D) = 577,006
(b) Estimate the probability that the individual wore a seat belt given
that he/she survived the accident.
P (Y es|S) = 412,368
= 0.7173.
574,895
(c) Estimate the probability that the individual died, given that he/she
(1) wore, (2) did not wear a seat belt. Note you need to give two
probabilities here.
510
P (D|Y es) = 412878
= 0.0012
1601
P (D|N o) = 164218 = 0.0097.
5. (8 pts) Following are the prices (in dollars) of 4 randomly chosen undergraduate textbooks. Find the variance of this data set.
100, 135, 85, 180.
x¯ = 125, s2 =
(100−125)2 +(135−125)2 +(85−125)2 +(180−125)2
4−1
3
= 1783.33.
6. (13 pts) In Harrisonburg, 60% of the household subscribe to NY Times,
45% subscribe to Washington Post, and 75% subscribe to at least one
of the two newspapers. Let N indicate subscribing to NY Times, W
indicate subscribing to Washington Post.
(a) Find P (N or W).
P(N or W)= 0.75.
(b) What is the probability that a randomly selected household subscribe to neither of them?
P (Neither) = 1-0.75=0.25.
(c) Find P( N and W).
Note P(N or W)= P(N)+P(W) - P(N and W), so
P(N and W)= P(N)+P(W)- P(N or W)= 0.60+0.45-0.75=0.30.
(d) What is the probability that a randomly selected household subscribes to exactly one newspaper?
P(Exactly one) = P(N or W)- P(N and W)=0.75-30=0.45.
7. (16 pts) It is known that 10% of a population carry a certain disease.
A doctor is about to check 4 randomly selected people from this population.
(a) What is the probability all the 4 people carry the disease?
P (all) = 0.14 = 0.0001.
(b) What is the probability that none of the 4 people carry the disease?
P (none) = 0.94 = 0.6561.
(c) What is the probability that at least one person carry the disease?
P(at least one) = 1- P(none) =1-0.6561=0.3439.
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