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Math 2210
Fall 2015
Test 1
Name:
1. A grocery store manager is interested in the mean amount of money spent by the store’s
customers per visit to the store and collects data from a random sample of customers.
(a) (3 points) Identify the parameter.
Solution: Mean amount of money spent by ALL of the stores customers.
(b) (3 points) Identify the statistic.
Solution: Mean amount of meny spent by the sample of customers.
(c) (3 points) Which value would be known after data collection is complete: the parameter or the statistic?
Solution: Statistic
2. An online retailer is interested in the mean daily volume of traffic to her web site. For
each day in the sample, she records the total amount of time spent (in minutes) at the
site by all visitors.
(a) (3 points) Identify the data type:
A. categorical
B. quantitative discrete
C. quantitative continuous.
(b) (6 points) Name the sampling method used for each scenario. Choose from: simple
random, cluster, stratefied, convenience, and systematic.
i. The retailer uses the data from the last 35 days.
Solution: convenience
ii. The retailer randomly selects five Sundays, five Mondays, five Tuesdays, five
Wednesdays, five Thursdays, five Fridays, and five Saturdays from the last 365
days.
Solution: stratefied
iii. The retailer randomly selects 35 days out of the last 365 days, sampling with
replacement.
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Fall 2015
Solution: simple random sample
(c) (3 points) Which method described in part (b) is likely to produce the greatest
non-sampling error?
Solution: The convenience sample (i)
3. Carter wishes to test the effect of using Brand X commercial fertilizer for growing tomatoes. He purchases twelve “Better Boy” tomato plants from and randomly selects six to
plant in a garden plot on the shady north side of his house and plants the other six in
a garden plot on the sunny south side of the house. Carter applies the fertilizer to the
plants on the south side of the house. He applies no fertilizer to the plants on the north
side of the house. Carter waters and weeds both garden plots equally. For each plant,
Carter records the number of days until the first tomato is ripe.
(a) (2 points) Identify the expanatory variable.
Solution: Fertilizer/No fertilizer
(b) (2 points) Identify the response variable.
Solution: The number of days until the first tomato is ripe
(c) (2 points) Identify the control group.
Solution: The group of plants that received no fertilizer
(d) (4 points) Identify the single largest source of non-sampling error in Carter’s method.
What should Carter do to minimize this error? (Carter does not have space for any
more tomato plants, so do not consider sample size in your discussion.) Please use
complete sentences.
Solution: He puts all of the plants in the control group (no fertilizer) in the
shade. Now sun/shade is a confounding variable. Carter should put all of his
plants in the same growing conditions.
4. The “25-hydroxy vitamin D” level in the blood, measured in nanograms per milliliter
(ng/ml), represents the stores of vitamin D available to the body for use. The following
frequency table details the “25-hydroxy vitamin D” levels for a sample of 20 adolescent
boys.
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Relative Frequency
Fall 2015
x
Frequency
12
2
18
5
0.25
0.35
21
2
0.1
0.45
23
5
0.25
0.7
24
3
0.15
0.85
27
2
0.1
0.95
33
1
0.05
1
2
20
Cumulative Relative Frequency
= 0.1
0.1
(a) (6 points) Complete the table. Use fractions or three decimal places for proportions.
Solution: See table.
(b) (3 points) List the first five “25-hydroxy vitamin D” levels in order from lowest to
highest.
Solution: 12, 12, 18, 18, 18
(c) (6 points) Find the mean and standard deviation of the “25-hydroxy vitamin D”
measurements. Use the correct symbol for each. Round to one decimal place.
Solution: x̄ = 21.5, s = 4.9
(d) (4 points) Find the “25-hydroxy vitamin D” level that is 2 standard deviations
above the mean. Round to one decimal place.
Solution: 31.3
5. It is desireable to have consistency in product characteristics. Maintaining consistent
size is particularly difficult in agricultural products. The following table provides the
weights, in ounces, of a random sample of 18 potatoes.
3.7
4.3
4.7
4.7
5.2
6.0
6.0
6.3
6.7
7.0
7.3
7.3
7.6
7.6
7.8
7.9
7.9
7.9
(a) (3 points) Find the IQR of the potato data.
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Solution: Q3 − Q1 = 7.6 − 5.2 = 2.4 ounces
(b) (4 points) Find the boundary value for a high outlier using Q3 + (1.5)IQR.
Solution: 7.6 + (1.5)(2.4) = 11.2 ounces
(c) (3 points) 75% of potatoes in this sample weigh more than
ounces.
Solution: 5.2
(d) (4 points) Find the percentile of 7.3 ounces.
Solution: percentile =
x+0.5y
(100)
n
=
10+(0.5)(2)
(100)
18
= 61%, so P61 = 7.3
(e) (3 points) Interpret the percentile you found in part (e).
Solution: 61% of the potatoes in this data set weigh less than 7.3 ounces; 39%
weigh more than 7.3 ounces.
6. Of American adults, 64% own a smart phone and 70% have high-speed broadband
connections at home. 56% of American adults own a smart phone and have high-speed
broadband connections at home. Let P be the event that an American adult owns a
smart phone. Let B be the event that an American adult has a broadband connection
at home. Pew research, 2015.
(a) (4 points) Draw a Venn diagram of this scenario.
Solution:
(b) (3 points) Find the probability that an American adult has a smart phone or a
broadband connection at home.
Solution: 0.78
(c) (2 points) Find the probability that an American adult has NEITHER a smart
phone nor broadband at home.
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Fall 2015
Solution: 0.22
(d) (3 points) Are P and B independent events? Show your work.
Solution: Check: P(P AND B) = P(P) * P(B)?
P(P AND B) = 0.56
P(P) * P(B) = 0.64 * 0.70 = 0.448
No, they are not independent events.
7. The following table summarizes successes and failures of 692 subjects who were using different methods when trying to stop smoking. The smoking/not smoking determination
was made five months after treatment began. Use a fraction or round to three decimal
places for probabilities.
Nicotine Gum (G) Nicotine Patch (P) Nicotine Inhaler (I)
Smoking (S)
191
263
95
Not Smoking (N)
59
57
27
(a) (3 points) Find the probability that a subject was not smoking.
Solution:
number not smoking
total number
=
59+57+27
692
= 0.207
(b) (3 points) What is the probability that a subject was using a nicotine patch and
was not smoking.
Solution:
57
692
= 0.082
(c) (3 points) What is the probability that a person was smoking assuming that they
were using nicotine gum?
Solution: Restrict focus to the nicotine gum column:
191
250
= 0.764
(d) (2 points) Mark the circle next to each pair of mutually exclusive events. (There
may be more than one such pair.)
√
G and P
G and S
I and N
√
S and N
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Fall 2015
8. A box contains eight balls. Three of the balls are red. Five of the balls are yellow.
Suppose that you randomly draw two balls without replacement.
Let
Let
Let
Let
R1 = first ball is red.
Y1 = first ball is yellow.
R2 = second ball is red.
Y2 = second ball is yellow.
Use a fraction or round to three decimal places for probabilities.
(a) (4 points) Draw a tree diagram of the situation. Solution:
(b) (3 points) Find P (two red balls).
Solution: P(red on first AND red on second)
= P(red on first)P(red on second|red on first)
6
= 0.107
= 38 ∗ 27 = 56
(c) (3 points) Find P (at least one yellow ball).
Solution: P(at least one yellow) = 1-P(both red) = 1 −
6
56
=
50
56
= 0.893