Statistics 270
Sample Final Exam
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO!
Instructions:
1.
2.
3.
4.
5.
Read all questions carefully.
Define all variables/events used in your solutions.
Show all of your work.
Cross out any material you do not wish to have considered.
Correct answers with insufficient justification or accompanied by additional
incorrect statements will not receive full credit.
6. Write all answers on the test paper
Name:
Student Number:
Signature:
1. A random sample of students from a
large class was selected. Among the
variables recorded were height (in
inches) and gender. At the right are
the box-plots that compare males and
females with respect to height.
\
a. From the box-plot, what is the median height of the males (about)?
b. From the box-plot, what percent of females were taller than 66 inches?
c. From the box-plot, about how tall is the tallest woman?
2. Short answers.
a. State the central limit theorem.
b. A research paper includes a 95% confidence interval. In terms of repeated
sampling, interpret the 95% confidence level.
3. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is
selected, and the alcohol content of each bottle is determined. Let μ denote the average
alcohol content for the population of all bottles of the brand under study. Suppose that the
resulting 95% confidence interval is (8.0, 9.6).
a. Would a 90% confidence interval calculated from this same sample have been
narrower or wider than the given interval? Explain your reasoning.
b. Consider the following statement: There is a 95% chance that μ is between 8 and
9.6. Is this statement correct? Why or why not?
c. Consider the following statement: If the process of selecting a sample of size 50
and then computing the corresponding 95% interval is repeated 100 times, 95 of
the resulting intervals will include μ. Is this statement correct? Why or why not?
4. Students in a statistics class were asked whether or not they generally felt sleep
deprived and also were asked how many hours they usually slept per night. The
researcher wished to assess if the mean number of hours of sleep per night was
significantly lower for the sleep deprived group as compared to the not sleep
deprived group. The following information was reported:
Group
Not sleep deprived
Sleep deprived
Sample size (n)
35
51
Sample mean
7.10
5.90
Sample standard deviation
1.16
1.35
a. The researcher plans to use a two sample t-test to see if there is evidence that the
mean number of hours of sleep per night was significantly lower for the sleep
deprived group as compared to the not sleep deprived group. What are the
assumptions underlying the use of this test?
b. The researcher is going use a two sample t-test to see if there is evidence that the
mean number of hours of sleep per night was significantly lower for the sleep
deprived group as compared to the not sleep deprived group. What are the
appropriate hypotheses for assessing the evidence of this study?
H0:__________________________________________________
H1:__________________________________________________
c. Compute the pooled estimate of the common population standard deviation.
d. Report the value of the test statistic for testing the hypotheses in part c.
e. Report the value of the p-value for testing the hypotheses in part (c).
f. Based on the results in parts d and e, what do you conclude about that hypotheses
in part b?
5. Government regulations indicate that the total weight of cargo in a certain kind of
airplane cannot exceed 330 kg. On a particular day a plane is loaded with 100 boxes
of a particular item only. Historically, the weight distribution for the individual
boxes of this variety has a mean 3.2 kg and standard deviation 0.4 kg.
a.
What is the distribution of the sample mean weight for the boxes (give the pdf
and appropriate parameter values)?
b. What is the probability that the observed sample mean is larger than 3.5 kg?
c. What is the probability that the government regulation is met?
6. Let X be a gamma random variable with pdf
 1
x  1e  x /  for x  0

f ( x)     ( )

0 otherwise.

Derive the expected value of X.
7. Let X, Y and Z be random variables with joint pdf
8xyz for 0  x  1,0  y  1,0  z  1
f ( x)  
0 otherwise.

Compute the covariance between X and Y.
8. Light bulbs of a certain type are advertised as having an average lifetime 800
hours. The price of these bulbs is very favourable, so a potential customer has
decided to go ahead with a purchase arrangement unless it can be conclusively
demonstrated that the true average lifetime (in hours) is smaller than what is
advertised. A random sample of 21 bulbs was selected, the lifetime of each bulb
determined, and the appropriate hypotheses were tested using MINITAB,
resulting in the accompanying output.
Lifetime
Sample Mean
738.44
Sample Standard Deviation
38.30
Test the appropriate hypothesis for this study. What conclusion would be
appropriate for a significance level of .05? A significance level of .01?
Formula Sheet
Descriptive Statistics
n
Sample Mean: x 
x
i 1
i
n
n
Sample Variance: s 2 
 x
i 1
i
 x
2
n 1
Sample median:
 n  1
If n is odd then the sample median is the 
 ordered value
 2 
th
If n is even then the sample median is the mean of the
n 
  1
2 
th
ordered values.
Range: Max-Min
Empirical rule for bell shaped distributions:
 68% of the data lie in the interval x  s
 95% of the data lie in the interval x  2s
 99% of the data lie in the interval x  3s
n
 
2
th
and the
Useful Probability Formulas
Addition Rules:
 P( A  B)  P( A)  P( B)  P( A  B)
 P( A  B  C )  P( A)  P( B)  P(C )  P( A  B)  P( A  C )  P( B  C )  P( A  B  C )
Complement Rule: P( A)  1  P( A' )
Multiplication Rule: P( A  B)  P( A) P( B | A)  P( B) P( A | B)
Law of Total Probability: P( A)  P( A | B) P( B)  P( A | B' ) P( B' )
General Version of the Law of Total Probability:
Suppose (A1, A2, …, Ak) form a partition of the sample space,
k
then P( B)   P( B | Ai ) P( Ai ) .
i 1
Bayes Theorem: P( A | B) 
P( B | A) P( A)
P( B | A) P( A)  P( B | A' ) P( A' )
General Version of Bayes Theorem:
Suppose (A1, A2, …, Ak) form a partition of the sample space, then
P( B | A j ) P( A j )
P( A j | B)  k
.
 P( B | Ai ) P( Ai )
i 1
Mutually Independent:
Events A1, A2, …, An are mutually independent if for every k (k=2, 3, …, n) and
every index set i1, …, ik, P( Ai1  Ai2  ...  Aik )  P( Ai1 ) P( Ai2 )...P( Aik ) ,
Discrete Random Variables
Expected value:
k
E ( X )   p ( xi ) xi   X
i 1
E (aX  b)  aE ( X )  b
Variance:
k
 2  V ( X )   ( xi   ) 2 p ( xi )
i 1
V (aX  b)  a 2V ( X )
Covariance:
Cov( X ,Y )  E((Y   X )(Y  Y ))
Common Distributions:
Bernoulli:
pmf: p( x)  p x (1  p)1 x ;   p ;  2  p(1  p)
Binomial:
n
pmf: p( x)    p x (1  p) n  x ;   np ;  2  np(1  p)
 x
Hyper-geometric:
 M  N  M 
 

x  n  x 
M
 N  n  M  M 

pmf: p( x) 
; n
;  2  n
 1  
N
N
N
 N  1  N 
 
n
Poisson:
e   x
pmf: p( x) 
;    ; 2  
x!
Continuous Random Variables
Expected value:
E ( X )   xf ( x)dx   X
E (aX  b)  aE ( X )  b
Variance:
 2  V ( X )   ( x   X ) 2 f ( x)dx
V (aX  b)  a 2V ( X )
Covariance:
Cov( X ,Y )  E((Y   X )(Y  Y ))
Useful Formulas for the Normal Distribution
observation  mean x  

standard deviation


z  score 

Percentile: x  z  

If X has the N ( ,  ) distribution, then the variable Z 
X 

has the N (0,1)
distribution.
Statistical Inference
One Sample Inference for the Population Mean
Unknown Population Standard Deviation
Confidence Interval
x  t*
s
Known Population Standard Deviation
Confidence Interval

x  z / 2
df = n – 1
n
n
Sample Size for Desired Width
 z 
n   2  / 2
w

One-Sample t-Test
t
x  0
s
n




2
One-Sample z-Test
z
df = n – 1
Confidence Level
Z
90%
1.645
x  0
 n
95%
1.95
99%
2.575
Two Sample Inference for the Population Mean
Pooled Standard Deviation
sp 
(n1  1)s12  (n 2  1) s 22
n1  n 2  2
Confidence Interval
x1  x2   t * s p
1
1

n1 n2
df = n1  n2  2
Pooled Two-Sample t-Test
x1  x 2
t
sp
1
1

n1 n 2
df = n1  n2  2