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Statistics

STAB57H3

Mahinda Samarakoon

Winter

Description

UNIVERSITY OF TORONTO SCARBOROUGH
Department of Computer and Mathematical Sciences
Midterm Test, March 2012
STAB57H3 Introduction to Statistics
Duration: One hour and ▯fty minutes
Last Name: First Name:
Student number:
Aids allowed:
- The textbook (Probability and Statistics, Evans, M.J and Rosenthal, J. S)
- Class notes and any notes from tutorials
- A calculator (No phone calculators are allowed)
No other aids are allowed. For example you are not allowed to have any other textbook or
past exams.
All your work must be presented clearly in order to get credit. Answer alone (even though
correct) will only qualify for ZERO credit. Please show your work in the space provided;
you may use the back of the pages, if necessary, but you MUST remain organized. Show your
work and answer in the space provided, in ink. Pencil may be used, but then any re-grading
will NOT be allowed.
There are 11 pages including this page. Please check to see you have all the pages.
Circle your tutorial: TUT001(Mon 11:00-12:00) TUT002 (Mon 10:00-11:00) TUT003 (Thu
9:00-10:00)
Good Luck!
Question: 1 2 3 4 5 6 7 8 9 10 Total
Points: 10 5 10 6 11 6 8 11 4 9 80
Score: Page 2 of 11
1. Suppose that a reponse X follows the following probability distribution.
x 1 2 3 4
P(X = x) 0.2 0.4 0.3 0.1
(a) (3 points) If we use the expected value of the response as a predictor, then what is
the prediction of a future response X?
Solution: predicted value =1 ▯ 0:2 + 2 ▯ 0:4 + 3 ▯ 0:3 + 4 ▯ 0:1 = 2:3
(b) (4 points) Calculate the mean-squared error of the prediction in part (a) above.
2 2 2
Solution: MSE = E(X ▯ 2:3) = (1 ▯ 2:3) ▯ 0:2 + (2 ▯ 2:3) ▯ 0:4 + (2 ▯
2:3) ▯ 0:3 + (4 ▯ 2:3) ▯ 0:1 = 0:81
(c) (3 points) The mode of the distribution of the response X is 2. If we use the mode
of the response X as a predictor of X, then what is the mean-squared error of this
prediction?
Solution: MSE = E(X ▯ 2) = (1 ▯ 2) ▯ 0:2 + (2 ▯ 2) ▯ 0:4 + (2 ▯ 2) ▯ 2
2
0:3 + (4 ▯ 2) ▯ 0:1 = 0:9 Page 3 of 11
2. (5 points) Suppose that the life length (X, in years) of a machine is known to have an
Exponential(2) distribution. Determine the shortest interval that contains the value of
a future X with probability 0.99. [Recall that the pdf of exponential(▯ ) distribution is:
f(x) = ▯e ▯▯x for x ▯ 0 and zero otherwise.]
Solution: This is very similar to example 5.2.1 p 244 text.
Ru ▯2x
0 2e dx = 0:99
i.e. 1 ▯ e2u = 0:99
ln0:01
u = ▯2 = 2:3
The shortest interval is (0;2:3) Page 4 of 11
3. Suppose that we have a population ▯ = f▯ ;▯ ;▯ ;▯ ;▯ g and a quantitative measure-
1 2 3 4 5
ment X given by:
i 1 2 3 4 5
X(▯ )i 1.5 1.0 2.0 1.5 3.5
(a) (4 points) Calculate F X1:7) and f (X:5).
3 2
Solution: F (X:7) = 5 and f X1:5) = .5
(b) (3 points) Suppose you are going to to select a simple random sample of size 2 (i.e.
n = 2) from this population. Find the probability that f▯ ;▯ 2 w5ll become your
simple random sample.
Solution: 5 = 1▯2 = 2 = 0:1
(2) 5▯4 20
(c) (3 points) Suppose you are going to to select an i.i.d. sample (i.e. sampling with
replacement) of size 2 (i.e. n = 2) from this population. Find the probability that
f▯ 2▯ 5 will become your sample.
1 1 1
Solution: 5▯ 5 = 25 Page 5 of 11
4. The R code and the output below are intended to identify outliers in the grades of a
statistics class (not STAB57).
grades=scan("C:/Users/Admin/Desktop/grades.txt")
a=quantile(grades, 0.25)
b=quantile(grades, 0.75)
a
b
c = grades < (a - 1.5 * (b -a))
d = grades > (b + 1.5 * (b -a))
e = sum(c) + sum(d)
e
R output
> grades=scan("C:/Users/Admin/Desktop/grades.txt")
Read 40 items
> a=quantile(grades, 0.25)
> b=quantile(grades, 0.75)
> a
25%
60
> b
75%
77
> c = grades < (a - 1.5 * (b -a))
> d = grades > (b + 1.5 * (b -a))
> e = sum(c) + sum(d)
> e
[1] 3
(a) (2 points) Based on 1:5▯IQR rule, how many outliers are there in this data set?
Solution: 3 (This is e in the given R output)
(b) (4 points) One student in this class had 39 for this course (i.e. his grade in this
course is 39). Test whether this grade is an outlier based on 1:5 ▯ IQR rule. Show
all related calculations clearly.
Solution: Q1 ▯ 1:5 ▯ IQR = a ▯ 1:5 ▯ (b ▯ a) = 34:5 < 39 and so 39 is NOT
an outlier. Page 6 of 11
5. Let (X ;1 ) 2e a random sample (i.e i.i.d.) form the distribution with p.d.f.
( 1 ▯x=▯
▯e if 0 < x < ▯,
f▯(x) =
0 otherwise.
where ▯ > 0 is unknown. Let T = (X + X )=51 2
2
Note: This distribution has mean ▯ and variance ▯
(a) (5 points) Show that T is a biased estimator of ▯ and compute its bias when used
as an estimator of ▯. (Hint: Show that Bias (T) is a▯function of ▯ of the form
Bias (T) = k▯ and calculate the value of k.)
▯
2
Solution: EX = EX1= ▯ =) 2T = ▯ and so Bias (T5 = ET▯▯ = ▯0:6▯▯
T is biased since bias is not equal to 0.
(b) (6 points) Calculate the mean squared error of T when it is used an estimator of
2
▯. (Hint Show that MSE (T) is▯a function of ▯ of the form MSE (T) = c▯ and ▯
calculate the value of c.)
2 2 2 2 2

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