Download stat 190, exam 1 name: do not open the test until you are told to do so

STAT 190, EXAM 1
NAME:
DO NOT OPEN THE TEST UNTIL
YOU ARE TOLD TO DO SO
1. (2 points each) Determine if each of the underlined value is a parameter or a statistic.
(a) We want to learn the average score of this Stat 190 course and we see that this average
score is 88. Parameter
(b) In a survey, 66.4% of students in the survey use alcohol. Statistic
(c) 48.4% of all U.S. residents are men. Parameter
(d) We randomly pick 1000 people and see that 48.4 % of them are men. Statistic
(e) We want to check if a coin is fair by flipping it 100 times and we get a tail 48 times.
Statistic
2. (2 points each) Determine whether the variable is qualitative or quantitative.
(a) The number of students at Truman State University. Quantitative
(b) The temperature in this room. Quantitative
(c) Student ID number. Qualitative
(d) The cost of the Stat 190 textbook. Quantitative
(e) The number on a car license plate. Qualitative
3. We want to analyze the final grades of a particular Calculus class of 30 students and we obtain
the frequency distribution table
Grade
1
2
3
4
Number of students (frequency)
3
7
10
10
where 4.0 is an A, 3.0 is a B, 2.0 is a C, and 1.0 is a D.
(a) (2 points) Find the mean of the final grades. Show your work (using a statistical calculator to get a quick number is not accepted).
The mean is
4 × 10 + 3 × 10 + 2 × 7 + 1 × 3
= 2.9.
µ=
30
(b) (4 points) Find the variance and the standard deviation of the grades. Show your work
(using a statistical calculator to get a quick number is not accepted).
The variance is
(4 − 2.9)2 × 10 + (3 − 2.9)2 × 10 + (2 − 2.9)2 × 7 + (1 − 2.9)2 × 3
' 0.9567.
30
The standard deviation is
√
0.9567 ' 0.978.
2
(c) (4 points) Draw a frequency histogram for this problem. What is the shape of the
distribution.
It is a skewed left distribution.
4. Given the sample below
x 3 4 5 7
8
y 4 6 7 10 14
Given that x̄ = 5.4, sx ' 2.1, ȳ = 8.2, sy = 3.9. We treat x as a explanatory variable
and y as a response variable.
(a) (5 points) Compute the linear correlation coefficient r. Show how you use the formula
to get the answer.
The formula implies that r is
(3 − 5.4)(4 − 8.2) + (4 − 5.4)(6 − 8.2) + (5 − 5.4)(7 − 8.2) + (7 − 5.4)(10 − 8.2) + (8 − 5.4)(14 − 8.2)
= 0.97
(5 − 1) × 2.1 × 3.9
(b) (2 points) Could we have a linear relation between x and y? Are the two variable
positively associated or negatively associated? Explain how you have the answers for
these two questions.
We could have a linear relation between x and y because r is close to 1. It is a positive
association because r is positive.
(c) (3 points) Find the equation of the least-squares regression line ŷ = b1 x + b0 .
(d) The formulas give
b1 ' 0.97 ×
3.9
= 1.80
2.1
and
b0 ' 8.2 − 1.80 × 5.4 = −1.52.
Thus the equation is ŷ = 1.80x − 1.52.
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5. (3 points each) The following data represent the age of 44 U.S. presidents on their inauguration
days
42, 43, 46, 46, 47, 47, 48, 49, 49, 50,
50, 51, 51, 51, 51, 52, 52, 54, 54, 54,
54, 54, 55, 55, 55, 55, 56, 56, 56, 57,
57, 57, 57, 58, 60, 61, 61, 61, 62, 64,
64, 65, 68, 69
(a) Find the quartiles: Q1 , Q2 , Q3 .
The quartiles are
Q2
=
Q1
=
Q3
=
54 + 55
= 54.5,
2
50 + 51
= 50.5,
2
57 + 58
= 57.5.
2
(b) Find the interquartile range IQR.
The interquartile range is
Q3 − Q1 = 57.5 − 50.5 = 7.
(c) Find the lower and upper fences.
The lower fence is
50.5 − 1.5 × 7 = 40,
and the upper fence is
57.5 + 1.5 × 3 = 68.
(d) Find any outlier.
69 is an outlier.
(e) Draw a boxplot.
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6. (3 points each) We randomly select 2 random people from a group of 10 people including 7
men and 3 women.
(a) What is the probability that both of them are men? Hint: do you see that it is an “AND”
of two events?
The probability that the first person is a man AND the second person is a man is
7 6
7
. =
.
10 9
15
(b) What is the probability that we select at least a woman? Hint: what is the complement
of this case?
The complement of this event is the case where both of them are men. Thus the answer
is
7
8
1−
=
.
15
15
(c) What is the probability that the first person is a man and the second person is a woman?
The probability is
7
7 3
. =
.
10 9
30
(d) What is the probability that the first person is a woman and the second person is a man?
The probability is
3 7
7
. =
.
10 9
30
(e) What is the probability that we select a man and a woman where we do not care who is
the first and who is the second? Hint: Do you see that it is an “OR” of two events?
It is either part c OR part d. Since part c and part d are disjoint event, the addition
rule gives
7
7
7
+
=
.
30 30
15
7. The following data represent the number of different communication activities (e.g., cell phone,
text messaging, and email) used by a random sample of 400 teenagers in a given week
Activities
Male
Female
Total
0
21
21
42
1-2
81
52
133
3-4
60
56
116
5-7
38
71
109
Total
200
200
400
(a) (6 points) We want to check if the events “male” and “0 activities” are independent.
Follow the following steps: define one event as A and the other as B; find P (B|A) and
P (B); give a conclusion if the two events are independent with your computations.
We note that
21
P (male|0 activities) =
= 0.5,
42
200
P (male) =
= 0.5.
400
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Thus the two events are independent.
(b) (3 points) Are the events “male” and “0 activities” disjoint? Explain why (a yes or no
without any explanation will not be accepted).
No, because there are 21 men with 0 activities.
(c) (6 points) We want to check if the events “female” and “5-7 activities” are independent.
Follow the following steps: define one event as A and the other as B; find P (B|A) and
P (B); give a conclusion if the two events are independent with your computations.
Again, we need to check if
P (female|5-7 activities) = P (female).
We note that
P (female|5-7 activities)
=
P (female)
=
71
= 0.65,
109
200
= 0.5.
400
Thus the two events are not independent.
(d) (3 points) Are the events “female” and “5-7 activities” disjoint? Explain why (a yes or
no without any explanation will not be accepted).
No, because there are 71 women with 5-7 activities.
(e) (2 points) Are the events “1-2 activities” and “3-4 activities” disjoint? Explain why (a
yes or no without any explanation will not be accepted).
Yes, because we cannot have any person with both “1-2 activities” and “3-4 activities”.
8. Using the multiplication counting technique to answer the questions below.
(a) (5 points) Four students come into a classroom with 20 chairs. How many ways can they
seat?
The number of ways is 20.19.18.17 = 116280.
(b) (5 points) A social security number has the form xxx-xx-xxxx where x is a digit from 0
to 9. How many Social Security numbers can be formed?
The number of SSNs is 109 .
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