This document provides an overview of engineering statistics and probability concepts taught in an EN505 Engineering Statistics course. It defines key terms like random variables, sample spaces, events, probability, and distributions. Random variables can be either discrete or continuous, depending on whether they take countable or uncountable values. Probability is used to quantify the likelihood of events and is governed by rules like the addition rule and multiplication rule. Conditional probability and Bayes' theorem are also introduced to relate the probabilities of events given other information. Important discrete and continuous probability distributions are discussed.

1. EN505 Engineering Statistics
Fernando Tovia, Ph.D.
1 RANDOM VARIABLES AND PROBABILITY
1.1 Random Variables
Definition 1.1 A random experiment is an experiment such that the outcome cannot be
predicted in advance with absolute precision.
Definition 1.2 The set of all possible outcomes of a random experiment is called the
sample space. The sample space is denoted by Ω. An element of the sample space is
denoted by ω.
Example 1.1 Construct the sample space for each of the following random experiments:
1. flip a coin
2. toss a die
3. flip a coin twice
Definition 1.3 A subset of Ω is called an event. Events are denoted by italicized, capital
letters.
Example 1.2 Consider the random experiment consisting of tossing a die. Describe the
following events.
1. A = the event that 2 appears
2. B = the event that an even number appears
3. C = the event that an odd number appears
4. D = the event that a number appears
5. E = the event that no number appears
The particular set that we are interested in depends on the problem being considered.
However, a good thing to do when beginning any probability modeling problem is to
clearly define all the events of interest.
One graphical method of describing events defined on a sample space is the Venn
diagram. The representation of an event using a Venn diagram is given in Figure 1.1 Note
that the rectangle corresponds to the sample space, and the shaded region corresponds to
the event of interest.
Figure 1.1 Venn Diagram for Event A
1

2. Definition 1.4 Let A and B be two event defined on a sample space Ω. A is a subset of B,
denoted by A ⊂ B, if an only if (iff), ∀ ω ∈ A, ω ∈ B. (Figure 1.2)
Figure 1.2 Venn Diagram for A ⊂ B
Definition 1.5 Let A be an event defined on a sample space Ω. ω ∈ Ac iff ω ∉ A. Ac is
called the complement of A. (Figure 1.3)
Figure 1.3 Venn Diagram for Ac
Definition 1.6 Let A and B be two events defined on the sample space Ω. ω ∈ A ∪ B iff
ω ∈ A or ω ∈ B (or both). A ∪ B is called the union of A and B (see Figure 1.4).
Figure 1.4 Venn Diagram for A ∪ B
Let {A1, A2, …} be a collection of events defined on a sample space.
∞
ω ∈ U Aj
j =1
iff ∃ some j = 1, 2, … ∋ ω ∈ Aj
∞
UA
j =1
j is called the union of {A1, A2, …}
Definition 1.7 Let A and B be two events defined on the sample space Ω. ω ∈ A ∩ B iff
ω ∈ A and ω ∈ B. A ∩ B is called the intersection of A and B (see Figure 1.5).
Figure 1.5 Venn Diagram for
Let {A1, A2, …} be a collection of events defined on a sample space.
∞
ω ∈ I Aj
j =1
iff ω ∈ A ∀ j = 1, 2, …
∞
IA
j =1
j is called the intersection of {A1, A2, …}
Example 1.3 (example 1.2 continued)
1. Bc = C
2

3. 2. B ∪ C = D
3. A ∩ B = B
Theorem 1.1 Properties of Complements
Let A be an event defined on a sample space Ω. Then
(a)
(b)
Theorem 1.2 Properties of the Unions
Let A, B, C be events defined on a sample space Ω. Then
(a)
(b)
(c)
(d)
(e)
Example Prove Theorem 1.2 (c)
Theorem 1.3 Properties of the Intersection
Let A, B, and C be events defined on the sample space Ω. Then
(a)
(b)
(c)
(d)
(e)
Example 1.6 Prove theorem 1.3 (b)
3

4. Theorem 1.4 Distribution of Union and Intersection
Let A, B and C be events defined in the sample space Ω. Then
(a) A ∩ (B ∪ C) = (A ∩ B) ∪ (A∩C)
(b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Theorem 1.5 DeMorgan’s Law
Let A, B and C be events defined in the sample space Ω. Then
(a) (A ∪ B)c = Ac ∩ Bc
(b) (A ∩ B) c = Ac ∪ Bc
Definition 1.8 Let A and B be two events defined in the sample space Ω. A and B are said
to be mutually exclusive or disjoint iff A ∩ B = Ø (Figure 1.6). A collection of events
{A1, A2, … }, defined on a sample space Ω, is said to be disjoint iff every pair of events
in the collection is mutually exclusive.
Figure 1.6 Venn Diagram for Mutually Exclusive Events
Definition 1.9 A collection of events {A1, A2, …, An} defined on a sample space Ω, is
said to be a partition (Figure 1.7) of Ω iff
(a) the collection is disjoint
n
(b) UA j =Ω
j =1
Figure 1.7 Venn Diagram for a Partition
Example 1.7 (Example 1.2 continued) Using the defined event, identify:
(a) a set of mutually exclusive events
(b) a partition of the sample space
4

5. Definition 1.10 A collection of events, F, defined on a sample space Ω, is said to be a
field iff
(a) Ω ∈ F,
(b) if A Ω ∈ F, then Ac ∈ F, then
n
U A ∈ F,
j =1
j
We use fields to represent all the events that we are interested in study. To construct a
field:
1. we start with Ω
2. Ø is inserted by implication (Definition 1.10 (b)
3. we then add the events of interest
4. we then add complements and unions
Example 1.8 Suppose we perform a random experiment which consists of observing the
type of shirt worn by the next person entering a room. Suppose we are interested in the
following events.
L = the shirt that has long sleeves
S = the shirt has short sleeves
N = the shirt has no sleeves
Assuming that {L, S, N} is a partition of Ω, construct an appropriate field
Theorem 1.6 Intersection are in Fields
Let F be a field of events defined in the sample space Ω. Then if A1, A2, … , An ∈ F,
then
n
IA ∈ F
j =1
j
Example 1.9 Prove that if A, B ∈ F, then A ∩ B ∈ F.
5

6. Any meaningful expression containing events of interest, ∪ , ∩, and c can be shown to be
in the field.
Definition 1.11 Consider a set of elements, such as S = {a, b, c}. A permutation of the
elements is an ordered sequence of elements. The number of permutations of n different
elements is n! where
n! = n x (n-1) x (n-2) x …x 2 x 1
Example 1.10 List all the permutations of the elements S
Definition 1.12 The number of permutations of subsets of r elements selected from a set
of n different elements is
Another counting problem of interest is the number of subsets of r elements that can be
selected from a set of n elements. Here the order is not important, and are called
combinations.
Definition 1.13 The number of combinations, subsets of size r that can be selected from a
set of n elements, is denoted as
Example 1.11 The EN505 class has 13 students. If teams of 2 students can be selected,
how many different teams are possible?
1.2 Probability
Probability is used to quantify the likelihood, or chance, that an outcome of a random
experiment will occur.
Definition 1.14 A random variable is a real-valued function defined on a sample space.
Random variables are typically denoted by italicized capital letters. Specific values taken
on by a random variable are typically denoted by italicized, lower-case letters.
Definition 1.15 A random variable that can take on a countable number of values is said
to be a discrete random variable.
Definition 1.16 A random variable that can take on an uncountable number of values is
said to be a continuous random variable.
6

7. Definition 1.17 The set of possible values for a random variable is referred as a range of
a random variable.
Example 1.12 Consider the following experiments of random variables, define the
random variable, identify the range for each random variable, and classify it as discrete or
continuous.
1. flip a coin
2. toss a die until a 6 appears
3. quality inspection of a shipment of manufactured items.
4. arrival of customer to a bank
Definition 1.18 Let Ω be the random space for some random experiment. For any event
defined on Ω, Pr(·) is a function which assigns a number to the event. Pr(A) is called the
probability of event A provided the following conditions hold:
(a)
(b)
(c)
Probability is used to quantify the likelihood, or chance, that an event will occur within
the sample space.
7

8. Whenever a sample consists of N possible outcomes , the probability of each outcome is
1/N
Theorem 1.7 Probability Computational Rules
Let A and B events defined on a sample space Ω, and let {A1, A2, …, An} be a collection
of events defined on Ω. Then
(a)
(b)
(c)
(d)
(e)
(f)
Corollary 1.1 Union of Three or More Events
Let A, B, C and D be events defined on a sample space Ω. Then,
Pr( A ∪ B ∪ C ) = Pr( A) + Pr( B ) + Pr(C ) − Pr( A ∩ B ) − Pr( A ∩ C ) − Pr( B ∩ C ) + Pr( A ∩ B ∩ C )
and
Pr( A ∪ B ∪ C ∪ D ) = Pr( A) + Pr( B) + Pr(C ) + Pr( D) − Pr( A ∩ B) − Pr( A ∩ C ) − Pr( A ∩ D ) −
− Pr( B ∩ C ) − Pr( B ∩ D) − Pr(C ∩ D) + Pr( A ∩ B ∩ C ) + Pr( A ∩ B ∩ D) + Pr( B ∩ C ∩ D) +
− Pr( A ∩ B ∩ C ∩ D )
Example 1.11 Let A, B and C be events defined on a sample space Ω ∋
Pr(A) = 0.30
Pr(Bc) = 0.60
Pr(C) = 0.20
Pr(A ∪ B) = 0.50
Pr( B ∩ C ) = 0.05
A and C are mutually exclusive
Compute the following probabilities
(a) Pr(B)
8

9. (b) Pr( B ∪ C ) =
(c) Pr( A ∩ B )
(d) Pr( A ∪ C )
(e) Pr( A ∩ C )
(f) Pr( B ∩ C c )
(g) Pr( A ∪ B ∪ C ) =
1.3 Independence
Two events are independent if any one of the following equivalent statements is true
9

10. (1) P(A|B) = P(A)
(2) P(B|A) = P(B)
(3) P ( A ∩ B ) = P ( A) P ( B )
Example 2.29 (book work in class)
1.4 Conditional Probability
Definition 1.19. Let A and B events define on a sample space Ω ∋ B ≠ Ø. We refer to
Pr(A|B) as the conditional probability of event given the occurrence of event B, where
Pr( AC | B) =
probability of not A given B
10

11. note that Pr(A|Bc) ≠ 1-Pr(A|B)
Example 1.12
A semiconductor manufacturing facility is controlled in a manner such that 2% of
manufactured chips are subjected to high levels of contamination. If a chip is subjected to
high levels of contamination, there is a 12% chance that it will fail testing. What is the
probability that a chip is subjected to high levels of contamination and fails upon tests?
c=
f=
Pr(High c level) =
Pr(Fail | high c level) =
Pr(F∩C) =
Example 1.13
An air quality test is designed to detect the presence of two molecules (molecule 1 and
molecule 2). 17% of all samples contain both molecules, and 48% of all samples contain
molecule 1. If a sample contains molecule 1, what is the probability that it also contains
molecule 2?
M1 = molecule 1
M2 = molecule 2
Pr(M1∩M2) =
Pr(M1) =
Pr(M2|M1) =
Theorem 1.8 Properties of Conditional Probability
Let A and B be non-empty event defined on a sample space Ω. Then
(a) If A and B are mutually exclusive then Pr(A|B) = 0
(b) If A ⊂ B then Pr(A|B)>=Pr(A)
11

12. (c) If B ⊂ A then Pr(A|B) =1
Theorem 1.9 Law of Total Probability – Part 1
Let A and B be events defined on a sample space Ω ∋ A ≠ Ø, B ≠ Ø, Bc ≠ Ø. Then
Example 1.15
A certain machine’s performance can be characterized by the quality of a key component.
94% of machines with a defective key component will fail. Whereas only 1% of
machines with a non-defective key component will fail. 4% of machines have a defective
key component. What is the probability that the machine will fail?
F = fail
D = defective
Pr(D) =
Pr(F|D) =
Pr(F|Dc) =
Pr(F) =
Theorem 1.11 Bayes’ Theorem – Part 1
Let A and B be events defined on a sample space Ω ∋ A ≠ Ø, B ≠ Ø, Bc ≠ Ø. Then
Example 1.15 (Example 1.14 continues)
Suppose the machine fails. What is the probability that the key component was
defective?
Pr(D|F) =
12

13. Theorem 1.12 Law of Total Probability – Part 2
Let A be a non-empty event defined on a sample space Ω, and let {B1, B2, …, Bn}be a
partition of Ω ∋ Bj ≠ Ø ∀ j =1, 2, …, n. Then
Pr( A) =
Theorem 1.13 Bayes’ Theorem – Part 2
Let A be a non-empty event defined on a sample space Ω, and let {B1, B2, …, Bn}be a
partition of Ω ∋ Bj ≠ Ø ∀ j =1, 2, …, n. Then
Pr( A | B j ) Pr( B j ) Pr( A | B j ) Pr( B j )
Pr( B j | A) = = n
Pr( A)
∑ Pr( A | B ) Pr( B )
i =1
i i
13

14. 14

15. 2 DISCRETE RANDOM VARIABLES AND PROBABILITY
DISTRIBUTIONS
A discrete random is a random variable that can take on at most countable number of
values.
Definition 2.1 Let X be a discrete random variable having cumulative distribution
function F. Let x1, x2 , ... denote the possible values of X. Then f(x) is the probability
mass function (pmf) of X, if
a) f(x) = P(X = x)
b) f(xj) > 0, j = 1, 2, …
c) f(x) = 0, if x = xj, j = 1, 2,
∞
d) ∑ f (x ) = 1
j =1
j
Definition 2.2 The cumulative distribution function of a discrete random variable X, is
denoted by F(X), and is given by
F(X) = ∑ f ( x)
xj ≤x
And satisfies the following properties
a) F(X) = P(X ≤ x) = ∑ f (x )
xj ≤x
j
b) 0 ≤ F( X ) ≤ 1
c) if x ≤ y then F( X ) ≤ F( Y )
Example 2.1 Suppose X is a discrete random variable having pmf f and cmf F, where
f(1) = 0.1, f(2) = 0.4, f(3) = 0.2, f(4) = 0.3.
1. Construct the cumulative distribution function of X.
15

16. 2. Compute Pr(X ≤ 2).
3. Compute Pr(X < 4).
4. Compute Pr(X ≥ 2).
Definition 2.3 The mean or expected value of X, denoted as µ or E(X), is
µ = E ( X ) = ∑ xf ( x)
x
The variance of X, denoted by Var (X) and given by
σ 2 = V ( X ) = E ( x − µ )2 = ∑ ( x − µ )2 f ( x) = ∑ x 2 f ( x) − µ 2
x x
The standard deviation of X is σ = σ 2
Definition 2.4 Let X be a discrete random variable with a probability mass function f(x).
The expected value of X is denoted by E(X) and given by
∞
E( X ) = ∑ x j f ( x j )
j =1
2.1 Discrete Distributions
2.1.1 Discrete Uniform Distribution
Suppose a random experiment has a finite set of equally likely outcomes. If X is a random
variable such that there is one-to-one correspondence between the outcomes and the set
of integers {a, a + 1, … , b}, then X is a discrete uniform random variable having
parameters a and b.
Notation
Range
Probability Mass Function
16

17. Parameters
Mean
Variance.
Example 2.2 Let X ~ DU(1, 6).
1. Compute Pr(X = 2).
2. Compute Pr(X > 4)
2.1.2 The Bernoulli Random Variable
Consider a random experiment that either “succeeds” or “fails”. If the probability of
success is p, and we let X = 0 if the experiment fails and X = 1 if it succeeds, then X is a
Bernoulli random variable with probability p. Such a random experiment is referred to
as a Bernoulli trial.
Notation
Range
Probability Mass Function
Parameter
17

18. Mean
Variance
2.1.3 The Binomial Distribution
The binomial distribution denotes the number of success in n independent Bernoulli
trials with probability p of success on each trial.
Notation
Range
Probability Mass Function
Cumulative Distribution Function
Parameters
Mean
Variance
Comments if n = 1, then X ~ Bernoulli(p)
18

19. Example 2.3 Each sample of air has a 10% chance of containing a particular rare
molecule.
1. Find the probability that in the next 18 samples, exactly 2 contains the rare molecule.
2. Determine the probability that at least four samples contain the rare molecule
3. Determine the probability that there will be at least one sample with the rare molecule
but less than four?
2.1.4 The Negative Binomial Random Variable
The negative binomial random variable denotes the number of trials
until the kth success in a sequence of independent Bernoulli trials with
probability p of success on each trial.
Notation
Range
Probability Mass Function
19

20. Cumulative Distribution Function
Parameters
Mean
Variance
Example 2.4 A high-performance aircraft contains three identical computers. Only one is
used to operate the aircraft; the other two are spares that can be activated in case the
primary system fails. During one hour of operation, the probability of a failure in the
primary computer is 0.0005.
1. Assuming that each hour represents an independent trial, what is the mean time
to failure of all the three components?
3. What is the probability that all three computers fail within a 5-hour flight?
20

21. Comments: If k = 1, then X ~ geom(p), i.e. X is a geometric random variable having a
probability of success p
2.1.5 The Geometric Distribution
In a series of independent Bernoulli trials, with constant probability p of a success, let the
random variable X denote the number of trials until the first success. Then X has a
geometric distribution.
Notation
Range
Probability Mass Function
Cumulative Distribution Function
Parameters
Mean
Variance
Example 2.3 Consider a sequence of independent Bernoulli trials with a probability of
success p = 0.2 for each trial.
(a) What is the expected number of trials to obtain the first success?
21

22. (b) After the eight successes occurs, what is the expected number of trials to obtain
the ninth success?
2.1.6 The Hypergeometric Random Variable
Consider a population consisting of N members, K of which are denoted as successes.
Consider a random experiment during which n members are selected random from the
population, and let X denote the number of successes in the random sample. If the
members in the sample are selected from the population without replacement, then X is
a hypergeometric random variable having parameters N, k and n.
Notation
Range
Probability mass function
Parameters
22

23. Comments If the sample is taken from the population with replacement, then
X~bin(n,K/N). Therefore, if n<<N, we can use the approximation bin(n,K/
N) ≈ HG(N, K, n).
Example 2.4 Suppose a shipment of 5000 batteries is received, 150 of them being
defective. A sample of 100 is taken from the shipment without replacement. Let X denote
the number of defective batteries in the sample.
1. What kind of random variable is X, and what is the range of X?
2. Compute Pr(X = 5).
3. Approximate Pr(X = 5) using the binomial approximation to the hypergeometric.
2.1.7 The Poisson Random Variable
The Poisson random variable denotes the number of events that occurs in an interval of
length t when events occur at a constant average rate λ.
Notation
23

24. Probability Mass Function
Cumulative Distribution Function
Parameters
Comments
The Poisson random variable X equals the number of counts in the time interval t. The
counts in each subinterval is independent of other subinterval.
If np>5 or (1-p) > 5, then we can use the approximation, Poisson ≈ bin (n, p).
Mean
Variance
It is important to use consistent units in the calculations of probabilities, means, and
variances involving Poisson random variable.
Example 2.5 Contamination is a problem in the manufacture of optical storage disks. The
number of particles of contamination that occur on an optical disk has a Poisson
24

25. distribution, and the average number of particles per centimeter squared of media surface
is 0.1. The area of a disk under study is 100 squared centimeters.
a) Find the probability that 12 particles occur in the area of a disk under study.
b) Find the probability that zero particles occur in the area of the disk under the study.
c) Find the probability that 12 or fewer particles occur in the area of a disk under study.
2.1.8 Poisson Process
Up to this time in the course, we have discussed the assignment of probabilities to events
and random variables, and by manipulating these probabilities we can analyze “snap-
shots” of systems behavior at certain point in the time, or under certain conditions. Now,
we are going to study one of the most commonly recognized continuous-time stochastic
processes that allow us to study important aspects of systems behavior over a time
interval t.
Definition 2.5 Let {N(t), t≥ 0} be a counting process. Then, {N(t), t≥ 0}is said to be a
Poisson process having rate λ, λ > 0, iff
a.
start counting from zero
25

26. b.
The number of outcomes occurring in one time or (specific region) is
independent of the number that occurs in any other disjoint time
interval or region space, which can be interpreted that the Poisson
process has no memory.
c. the number of events in any interval (s, s + t) is a Poisson random
variable with mean λt.
d. The probability that more that more than one outcome will occur at the
same time is negligible.
This is denoted by N(t) ~ PP(λ), which λ refers to the average rate at which events occur.
The part (c) of the definition implies that
1)
2)
4)
Note that in order to be a Poisson proves the average event occurrence MUST BE
CONSTANT over time, otherwise the Poisson process would be an inappropriate model.
Also note, that t can be interpreted as the specific “time”, “distance”, “area”, “volume” of
interest.
Example 2.6 Customers arrive to a facility according to a Poisson process with rate λ =
120 customer per hour. Suppose we begin observing the facility at some point in time.
a) What is the probability that 8 customers arrive during a 5-minute interval?
b) On average, how many customers will arrive during a 3.2-minute interval?
26

27. c) What is the probability that more than 2 customers arrive during a 1-minute
interval?
d) What is the probability that 4 customers arrive during the interval that begins 3.3
minutes after we start observing and ends 6.7 minutes after we start observing?
e) On average, how many customers will arrive during the interval that begins 16
minutes after we start observing and ends 17.8 minutes after we start observing?
f) What is the probability that 7 customers arrive during the first 12.2 minutes we
observe, given that 5 customers arrive during the first 8 minutes?
27

28. g) If 3 customers arrive during the first 1.2 minutes of our observation period, on
average, how many customers will arrive during the first 3.7 minutes?
h) If 1 customer arrives during the first 6 seconds or our observations, what is the
probability that 2 customers arrive during the interval that begins 12 seconds after
we observing and ends 30 seconds after we start observing?
i) If 5 customers arrive during the first 30 seconds of our observations, on average,
how many customers will arrive during the interval that begins 1 minute after we
start observing and ends 3 minutes after we start observing?
j) If 3 customers arrive during the interval that starts 1 minute after we start
observing and ends 2.2 minutes after we start observing, on average, how many
customers will arrive during the first 3.7 minutes?
28

29. Example 2.7 (Binomial approximation)
In a manufacturing process where glass products are produced, defects or bubbles occur,
occasionally rendering the piece undesirable for marketing. It is known that, on average,
1 in every 1000 of these items produced has one or more bubbles. What is the probability
that a random sample of 8000 will yield fewer than 7 items possessing bubbles?
29

30. 3 CONTINUOUS RANDOM VARIABLES AND PROBABILITY
DISTRIBUTIONS
As stated earlier, a continuous random variable is a random variable that can take on an
uncountable number of values.
Definition 3.1 The probability density function of a continuous random variable X is a
nonnegative function f(x) defined ∀ real x ∋ for any set A of real numbers
Theorem 3.1 Integral of a Density Function
The function f is a density function iff
All probability computations for a continuous random variable can be answered using the
density function.
Theorem3.2 Probability Computational Rules for Continuous Random Variables
Let X be a continuous random variable having cumulative distribution function F and
probability density function f. Then
(a)
(b)
(c)
30

31. (d)
(e)
The mean or expected value of X, denoted as µ or E(X), is
The variance of X, denoted by Var (X) and given by
Example 3.1 Consider a continuous random variable X having the following density
function where c is a constant
 c(1 − x 2 ) 0 ≤ x ≤ 1
f ( x) = 
0 otherwise
1. What is the value of c?
2. Construct the cumulative distribution function of X.
31

32. 3. Compute Pr(0.2< X ≤ 0.8) =
4. Compute Pr(X >0.5) =
Part (d) of Theorem 1.3.2 states that the probability density function is the derivative of
the cumulative distribution function. Altough this is true, it does not provide adequate
intuition as to the interpretation of the density function. For a discrete random variable,
the probability mass function actually assigned probabilities to the possible values of the
random variables. Theorem 1.3.2 (b) states that the probability of any specific value for a
continuous random variable is 0. The probability density function is not the probability of
a specific value. It is, however, the relative likelihood ((as compared to other possible
values) that the random variable will be near a certain value.
Continuous random variables are typically specified in terms of the form of their
probability density functions. In addition, some continuous random variables have been
widely-used in probability modeling. We will consider some of these more commonly-
used random variables, including:
1. the uniform random variable,
2. the exponential random variable,
3. the gamma random variable,
4. the Weibull random variable,
5. the normal random variable,
6. the lognormal random variable,
7. the beta random variable,
3.1 The Uniform Continuous Random Variable
Notation
Range
Probability Density Function
32

33. Cumulative Distribution Function
Parameters
Mean
Variance
Comments As its name implies, the uniform random variable is used to represent
quantities that occur randomly over some interval of the real line.
An observation of a U(0,1) random variable is referred to as a random number.
Example 3.2 Verify that the equation for the cumulative distribution of the uniform
random variable is correct.
Example 3.3 The magnitude (measured in N) of a load applied to a steel beam is
believed to be a U(2000, 5000) random variable. What is the probability that the load
exceeds 4200 N?
33

34. 3.2 The exponential Random
The random variable X that equal the distance (time) between successive counts of a
Poisson process with mean λ (rate, events per time unit, i.e. arrivals per hour, failures per
day, etc.) has an exponential distribution with parameter λ.
Notation
Range
Probability Density Function
Cumulative Distribution Function
Parameters
Mean
Variance
Comments λ is called the rate of the exponential distribution.
34

35. Example 3.4 In a large computer network, user log-ons to the system can be modeled as
a Poisson process with a mean of 25 logs-on per hour. What is the probability that there
are no log-ons in an interval of 6 minutes?
What is the probability that the time until the next log-on is between 2 and 3 minutes?
Upon converting all units to hours,
Determine the interval of time that the probability that no log-on occurs in the interval is
0.90. Te question asks for the length of time x such that Pr(X>x) = 0.90.
What is the mean time until the next log-on?
What is the standard deviation of the time until the next log-on?
35

36. Theorem 3.3 The Memoryless Property of the Exponential Distribution
Let X be a continuous random variable. The X is an exponential random variable iff
Theorem 3.4 The Conditional Form of the Memoryless Prperty
Let X be a continuous random variable. Then X is an exponential random variable iff
Furthermore, no other continuous random variable possesses this property.There are
several implications of the memoryless property of the exponential random variable.
 First, if the exponential random variable is used to model the lifetime of a
device, then at every point in time until it fails, the device is as good as
new (from a probabilistic standpoint).
 If the exponential random variable is used to model an arrival time, then at
every point in time until the arrival occurs, it is as just began “waiting” for
the arrival.
Example 3.5 Suppose that the life length of a component is an exponential random
variable with rate 0.0001. Note that time units are hours. Determine the following.
a) What is the probability that the component lasts more than 2000 hours?
b) Given that the component lasts at least 1000 hours, what is the probability that
it lasts more than 2000 hours?
36

37. Theorem 3.5 Expectation under the Memoryless Property
Let X be an exponential random variable. Then
Example 3.6 (Example 3.5 continued)
a) Given that the component lasts at least 1000 hours, what is the expected value
of its life length?
b) Given that the component has survived 1000 hours, on average, how much
longer will it survive?
3.3 The Normal Distribution
Notation
Range
Probability Density Function
Cumulative Distribution Function no closed form expression
Parameters
37

38. Mean
Variance
Comments
Standard Normal Random Variable
If µ = 0 and σ = 1, then X is referred as the standard normal random variable. The
standard random normal variable is often denoted by Z.
The cumulative distribution of the standard normal random variable is denoted as
Φ ( z ) = Pr( Z ≤ z )
Appendix A Table I provides cumulative probabilities for a standard random variable.
For example, assume that Z is a standard normal random variable. Appendix A Table I
provides probabilities of the form Pr(Z≤ 1.53). Find in the column z 1.5 and find in the
row 0.03, then Pr(Z≤ 1.53) = 0.93699.
The same value can be obtained in Excel, type function icon (fx), statistical,
NORMSDIST(z), enter 1.53, and Excel will give you the result in the cell
=NORMSDIST(1.53) = 0.936992
The function
Is denoted a probability from Appendix A Table I. It is the cumulative distribution
function of a standard normal random variable. (see figure 4-13 page 124 from the
Montgomery book).
Example 3.7 (Example 4-12Montgomery)
Some useful results concerning a normal distribution are summarized in Fig 4-1413.
(textbook). For any random variable
38

39. 1)
2)
3)
4)
If X ~ N(µ,σ2), then (X -µ)/ σ ~N(0,1), which is known as the z-transformation. That is, Z
is a standard normal random variable.
Suppose X is a normal random variable with mean µ and standard deviation σ. Then,
39

40. Example 3.8 One key characteristic of a certain type of drive shaft is its diameter, and
the diameter is a normal distributed random variable having µ = 5 cm and σ = 0.08 cm.
a) What is the probability that the diameter of a given drive shaft is between 4.9
and 5.05 cm?
b) What diameter is exceeded by 90% of drive shafts?
c) Provide tolerances, symmetric about the mean, that capture 99% of drive shafts.
40

41. Example 3.9
The diameter of a shaft in an optical storage drive is normally distributed with mean
0.2508 inch and standard deviation 0.0005 inch. The specifications on the shaft are
±0.0015 inch. What proportion of shafts conforms to specifications?
3.3.1 Normal Approximation to the Binomial and Poisson Distributions
Binomial Approximation
If X is a binomial random variable with parameter n and p
Is approximately a standard random variable. To approximate a binomial probability with
a normal distribution a correction (continuity) factor is apllied.
41

42. The approximation is good for np > 5 and n(1-p) > 5.
Poisson Approximation
If X is a Poisson random variable with E(X) = λ and V(X) = λ,
is approximately a standard random variable. The approximation is good for λ > 5.
Example 3.10
The manufacturing of semiconductor chips produces 2% defective chips. Assume that
chips are independent and that a lot contains 1000 chips.
a) Approximate the probability that more than 25 chips are defective.
b) Approximate the probability that between 20 and 30 chips are defective
42

43. 3.4 Lognormal Distribution
Variables in the system sometimes follow an exponential relationship, where the
exponent is a random variable, say W, X = exp(W). If W has a normal distribution then
the distribution of X is called a lognormal distribution.
Notation
Range
Probability density function
Cumulative Distribution Function no closed form expression
Parameters
Comments If Y ~ N(µ, σ2) and X = eY, then X ~ LN((µ, σ2)
The lognormal random variable is often used to represent
elapsed times, especially equipment repair times, and
material properties.
Mean
Variance
43

44. Example 3.11 A wood floor system can be evaluated in one way by measuring its
modulus of elasticity (MOE) measured in 106 psi. One particular type of system is such
that its MOE is a lognormal random variable having µ = 0.375 and σ = 0.25.
1. What is the probability that a system’s MOE is less than 2?
2. Find the value of MOE that is exceeded by only 1% of the systems?
3.5 The Weibull Distribution
The Weibull distribution is often used to model the time until failure of many different
physical systems. It is used in Reliability time-dependent failures models, where the
failure distribution may be used to model both increasing and decreasing failure rates.
Notation
44

45. Range
Probability Density Function
Cumulative Distribution Function
Parameters
Mean
Variance
Comments If β = 1, then X ~ expon(1/η)
The Weibull random variable is most often used to
represent elapsed time, especially time to failure of a unit
of equipment.
Example 3.12 The time to failure of a power supply is a Weibull random variable having
β = 2.0 and η = 1000.0 hours. The manufacturer sells a warranty such that only 5% of the
power supplies fail before the warranty expires. What is the time period of the warranty?
45

46. 4 JOINT PROBABILITY DISTRIBUTIONS
Up to this point we have considered issues related to a single random variable. Now we
are going to consider situations in which we have two or more random variables that we
are interested in studying.
4.1 Two or more discrete random variables
Definition 4.1 The function f(x, y) is a joint probability distribution or probability
mass function of discrete random variables X and Y if
1.
2.
3.
Example 4.1 Let X denote the number of times a certain numerical control machine will
malfunction: 1, 2 or 3 times on a given day. Let Y denote the number of times a
technician is called on an emergency call. Their joint probability distribution is given as
f (x, y ) x
1 2 3
1 0.05 0.05 0.1
y 2 0.05 0.1 0.35
3 0 0.2 0.1
a) Find P(X<3, Y = 1)
b) Find the probability that the technician is called at least 2 times and the machine fails
no more than 1 time.
46

47. c) Find P(X>Y)
When studying joint probability distribution we are also interested in the probability
distributions of each variable individually, which is referred as the marginal probability
distribution.
Theorem 4.1 Let X and Y be discrete random variables having joint probability mass
functions f(x, y). Let x1 , x2 ,... denote the possible values of X, and let y1 , y2 ,... denote the
possible values of Y. Let f x ( x) denote the marginal probability mass function of X, and
let f y ( y ) denote the (marginal) probability mass function of Y. Then,
47

48. Example 4.2 Let X and Y be discrete random variables such that
f(1, 1) = 1/9 f(1, 2) = 1/6 f(1, 3) = 1/8
f(2, 1) = 1/18 f(2, 2) = 1/9 f(2, 3) = 1/9
f(3, 1) = 1/9 f(3, 2) = 1/9 f(3, 3) = 1/6
Find the marginal probability mass function of X and Y
48

49. Definition 4.2 The function f(x, y) is a joint probability density function of continuous
random variables X and Y if
1.
2.
3.
Example 4.3 A candy company distributes boxes of chocolates with a mixture of creams,
toffees, and nuts coated in both light and dark chocolates. For a randomly selected box,
let X and Y, respectively, be the proportions of the light and dark chocolates that are
creams and suppose that joint density function is
2 
 (2 x + 3 y ), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
f ( x, y ) =  5 
0,
 elsewhere  
a) verify condition 2
49

50. c) Find P[( X , Y ∈ A], where A={(x,y),|0 ≤ x ≤ 1/2, 1/4 ≤ y ≤ 1/2}
Theorem 4.2 Marginal Probability Density Function
Let X and Y be continuous random variables having probability density function f(x, y).
le. Let f x ( x) denote the marginal probability density function of X, and let f y ( y ) denote
the (marginal) probability density function of Y. Then,
50

51. Example 4.4 Let X and Y be continuous random variables such that
f(x, y) = 0.75 e-0.3y
Find the marginal probability density function of X and Y.
Theorem 4.3 The Law of the Unconscious Statistician
Let X and Y be discrete (continuous) random variables having joint probability mass
(density) function f(x,y). Let x1, x2, … denote the possible values of X, and let y1, y2, …
denote the possible values of Y. Let g(X,Y) be a real-valued function. Then
51

52. Example 4.5 Suppose X and Y are discrete random variables having joint probability
mass function f(x,y). Let x1, x2, … denote the possible values of X, and let y1, y2, … denote
the possible values of Y. What is E(X+Y)?
52

53. Theorem 4.4 Expectation of a Sum of Random Variables
Let X1, X2, …, Xn be random variables, and let a1, a2, …an be constants. Then
Example 4.6 What is the E(3X – 2Y + 4)?
Theorem 4.5 Independent Discrete Random Variables Let X and Y be random
variables having joint probability mass function f(x, y). Let fx(x) denote the marginal
probability mass function of X, and let fy(y) denote the marginal probability mass
function of Y. Then X and Y are said to be independent iff
53

54. Theorem 4.6 Independent Continuous Random Variables Let X and Y be random
variables having joint probability density function f(x, y). Let fx(x) denote the marginal
probability density function of X, and let fy(y) denote the marginal probability density
function of Y. Then X and Y are said to be independent iff
Example 4.6 Consider example 4.2. Are X and Y independents?
Example 4.7 Consider example 4.4. Are X and Y independent?
Definition 4.3 Let X and Y be random variables. The covariance of X and Y is denoted as
Cov(X,Y) and given by
54

55. A positive covariance indicates that X tends to increase (decrease) as Y increases
(decreases). A negative covariance indicates that X tends to decrease (increase) as Y
increases (decreases).
Example 4.8 Example 4.2 continued. Find the covariance of X and Y.
Theorem 4.7 Covariance of Independent Random Variables
Let X and Y be random variables. If X and Y are independent, then
Cov(X, Y) = 0.
Theorem 4.8 Variance of the Sum of Random Variables
Let X1, X2, … , XN be random variables. Then
Theorem 4.9 Variance of the Sum of Independent Random Variables
Let X1, X2, … , XN be independent random variables. Then
55

56. Definition 4.4 Let X and Y be two random variables. The correlation between X and Y is
denoted by ρxy and given by
Note that correlation and covariance have the same interpretation regarding the
relationship between the two variables. However, correlation does not have units and is
restricted to the range (-1, 1). Therefore, the magnitude of the correlation provides some
idea of the strength of the relationship between the two random variables.
56

57. 5 RANDOM SAMPLES, STATISTICS AND THE CENTRAL
LIMIT THEOREM
Definition 5.1 Independent random variables X1, X2, … ,Xn are called a random sample.
A randomly selected sample means that if a sample of n objects is selected, each subset
of size n is equally likely to be selected. If the number of objects in the population is
much larger than n, the random variables X1, X2, … ,Xn that represents the observations
from the sample can be shown to be approximately independent random variables with
the same distribution.
Definition 5.2 A statistic is a function of the random variables in a random sample.
Given the data, we calculate statistics all the time, such as the sample mean X and the
sample standard deviation S. Each statistic has a distribution and it is the distribution that
determines how well it estimates a quantity such as μ.
We begin our discussions by focusing on a single random variable, X. To perform any
meaningful statistical analysis regarding X, we must have data.
Let X be some random variable of interest. A random sample on X consists of n
observations on X: x1, x2, … , xn. We assume that these observations are independent and
identically distributed. The value of n is referred to as the sample size.
Definition 5.3 Descriptive statistics refers to the process of collecting data on a random
variable and computing meaningful quantities (statistics) that characterize the underlying
probability distribution of the random variable.
There are three points of interest regarding this definition.
• Performing any type of statistical analysis requires that we collect data on one or
more random variables.
• A statistic is nothing more than a numerical quantity computed using collected
data.
• If we knew the probability distribution which governed the random variable of
interest, collecting data would be unnecessary.
Types of Descriptive Statistics
1. measures of central tendency
• sample mean (sample average)
• sample median
• sample mode (discrete random variables only)
57

58. 2. measures of variability
• sample range
• sample variance
• sample standard deviation
• sample quartiles
Microsoft Excel has a Descriptive Statistics tool within its Data Analysis
ToolPak.
Computing the Sample Mean
• Most of your calculators have a built-in method for entering data and computing
the sample mean.
• Note the sample mean is a point estimate of the true mean of X. In other
words,
Computing the Sample Median
To compute the sample median, we first rank the data in ascending order and re-
number it: x(1), x(2), …. , x(n).
The sample median corresponds to the value that has 50% of the data above it and
50% of the data below it.
Computing the Sample Mode
The sample mode is the most frequently occurring value in the sample. It is
typically only of interest in sample data from a discrete random variable, because
sample data on a continuous random variable often does not have any repeated
values.
58

59. Compute the Sample Range
Computing the Sample Variance
• Why do we divide by n − 1? We divide by n − 1 because we have n − 1 degrees
of freedom. This refers to the fact that if we know the sample mean and n − 1
of the data values, we can compute the remaining data point.
• Note that the sample variance is a point estimate of the true variance. In
other words,
Computing the Sample Standard Deviation
• Note that the sample standard deviation is a point estimate of the true standard
deviation.
Theorem 5.1. If X1, X2, … ,Xn is a random sample of size n taken from a population with
mean μ and variance σ2, μ and variance σ2, and if X is the sample mean, the limiting form
of the distribution of
59

60. as n→∞, is the standard normal distribution.
5.1 Populations and Random Samples
The field of statistical inference consists of those methods used to draw
conclusions about a population. These methods utilize the information contained in a
random sample of observations from the population.
Statistical inference may be divided into two major areas:
• parameter estimation
• hypothesis testing
Both of these areas require a random sample of observations from one or more
populations, therefore, we will begin our discussions by addressing the concepts of
random sampling.
Definition 5.4 A population consists of the totality of the observations with which
we are concerned.
• We almost always use a random variable/probability distribution to model the
behavior of a population.
Definition 5.5 The number of observations in the population is called the size of the
population.
• Populations may be finite or infinite. However, we can typically assume the
population is infinite.
• In some cases, a population is conceptual. For example, the population of items
to be manufactured is a conceptual population.
Definition 5.6 A sample is a subset of observations selected from a population.
• We model these observations using random variables.
• If our inferences are to be statistically valid, then the sample must be
representative of the entire population. In other words, we want to ensure that we
take a random sample.
60

61. Definition 5.7 The random variables X1, X2, … , Xn are a random sample of size n
if X1, X2, … , Xn are independent and identically distributed.
• After the data has been collected, the numerical values of the observations are
denoted as x1, x2, … , xn.
• The next step in statistical inference is to use the collected data to compute one or
more statistics of interest.
5.2 Point Estimates
ˆ
Definition 5.8 A statistic, Θ , is any function of the observations in a random sample.
• In parameter estimation, statistics are used to estimate quantities of interest.
•
• The measures of central tendency and variability we considered in “Descriptive
Statistics” are all statistics.
•
Definition 5.9 A point estimate of some population parameter θ is a single
ˆ ˆ
numerical value θ of a statistic Θ .
•
Estimation problems occur frequently in engineering. The quantities that we will focus
on are:
• the mean µ of a population
• the standard deviation σ of a population
• the proportion p of items in a population that belong to a class of interest – p is the
probability of success for a Bernoulli trial
The point estimates that we use are:
•
•
•
61

62. 5.3 Sampling Distributions
A statistic is a function of the observations in the random sample. These observations
are random variables, therefore, the statistic itself is a random variable. All random
variables have probability distributions.
Definition 5.10 The probability distribution of a statistic is called a sampling
distribution.
• The sampling distribution of a statistic depends on the probability distribution
which governs the entire population, the size of the random sample, and the
method of sample selection.
Theorem 5.3 The Sampling Distribution of the Mean
If X1, X2, … , Xn are IID N(µ,σ2) random variables, then the sample mean is a
normal random variable having mean and variance
.
Thus, if we are sampling from a normal population then the sampling distribution of the
mean is normal. But what if we are not sampling from a normal population?
Theorem 5.4 The Central Limit Theorem
If X1, X2, … , Xn is a random sample of size n taken from a population with mean
µ and variance σ2, then as n → ∞,
is a standard normal random variable.
• The quality of the normal approximation depends on the true probability
distribution governing the population and the sample size.
• For most cases of practical interest, n ≥ 30 ensures a relatively good
approximation.
• If n < 30, then the underlying probability distribution must not be severely
non-normal.
Example 5.1 A plastics company produces cylindrical tubes for various industrial
applications. One of their production processes is such that the diameter of a tube is
normally distributed with a mean of 1 inch and a standard deviation of 0.02 inch.
(a) What is the probability that a single tube has a diameter of more than
1.015 inches?
62

63. X = diameter of a tube (measured in inches) ~ N( )
(b) What is the probability that the average diameter of five tubes is more than
1.015 inches?
n= X = average diameter ~ N( )
(c) What is the probability that the average diameter of 25 tubes is more than
1.015 inches?
n= X = average diameter ~ N( )
Example 5.2 The life length of an electronic component, T, is exponentially
distributed with a mean of 10,000 hours.
(a) What is the probability that a single component lasts more than 7500
hours?
(b) What is the probability that the average life length for 200 components is
more than 9500 hours?
E(T) = hours
σT = hours
63

64. Note that .
(c) What is the probability that the average life length for 10 components is
more than 9500 hours?
n is too small to use the CLT approximation
S 10
Note that T = .
10
If we had tried to use the CLT:
Now consider the case in which we are interested in studying two independent
populations. Let the first population have mean µ1 and standard deviation σ1, and let the
second population have mean µ2 and standard deviation σ2.
If we are interested in comparing the two means, then the obvious point estimate of
interest is
µ1 − µ 2 = X 1 − X 2 .
ˆ
What is the sampling distribution of this statistic?
64

65. Theorem 5.4 The Sampling Distribution of the Difference in Two Means
If we have two independent populations with means µ1 and µ2 and standard
deviations σ1 and σ2, and if a random sample of size n1 is taken from the first population
and a random sample of size n2 is taken from the second population, then the sampling
distribution of
is standard normal as n1 and n2 → ∞. If the two populations are normal, then the
sampling distribution of Z is exactly standard normal.
• Again, the approximation is relatively accurate if n1 ≥ 30 and n2 ≥ 30.
Example 5.3 The life length of batteries produced by Battery Manufacturer A is a
continuous random variable having a mean of 1500 hours and a standard deviation of 100
hours. The life length of batteries produced by Battery Manufacturer B is a continuous
random variable having a mean of 1400 hours and a standard deviation of 200 hours.
(a) Suppose 50 batteries of each type are tested. What is the probability that Battery
Manufacturer A’s sample average life length exceeds Battery Manufacturer B’s
by more than 75 hours?
(b) How would your answer change if only 12 batteries of each type were tested?
There is not enough information to answer the question. If we assume normality,
then we could proceed.
65

66. 5.4 Confidence Intervals
A point estimate provides only a single number for drawing conclusions about a
parameter. And if another random sample were selected, this point estimate would
almost certainly be different. In fact, this difference could be drastic.
For this reason, a point estimate typically does not supply adequate information to
an engineer. In such cases, it may be possible and useful to construct a confidence
interval which expresses the degree of uncertainty associated with a point estimate.
Definition 5.11 If θ is the parameter of interest, then the point estimate and
sampling distribution of θ can be used to identify a 100(1 − α )% confidence interval on
θ. This interval is of the form:
L and U are called the lower-confidence limit and upper-confidence limit. If L and U
are constructed properly, then
.
The quantity (1 − α) is called the confidence coefficient.
• The confidence coefficient is a measure of the accuracy of the confidence
interval. For example, if a 90% confidence interval is constructed, then the
probability that the true value of θ is contained in the interval is 0.9.
• The length of the confidence interval is a measure of the precision of the point
estimate. A general rule of thumb is that increasing the sample size improves the
precision of a point estimate.
Confidence intervals are closely related to hypothesis testing. Therefore, we will address
confidence intervals within the context of hypothesis testing.
66

67. 6 FORMULATING STATISTICAL HYPOTHESES
For many engineering problems, a decision must be made as to whether a particular
statement about a population parameter is true or false. In other words, we must either
accept the statement as being true or reject the statement as being false.
Example 6.1 Consider the following statements regarding the population of engineering
students at the Philadelphia University.
1. The average GPA is 3.0.
2. The standard deviation of age is 5 years.
3. 30% are afraid to fly
4. The average age of mothers is the same as the average age of fathers.
Definition 6.1 A statistical hypothesis is a statement about the parameters of one or
more populations.
• It is worthwhile to note that a statistical hypothesis is a statement about the
underlying probability distributions, not the sample data.
Example 6.2 (Ex. 6,1 continued) Convert each of the statements into a statistical
hypothesis.
1.
2.
3.
4.
To perform a test of hypotheses, we must have a contradictory statement about the
parameters of interest.
Example 6.3 Consider the following contradictory statements.
1. No, it’s more than that.
2. No, it’s not.
3. No, it’s less than that.
67

68. 4. No, fathers are older.
The result of our original statement and our contradictory statement is a set of two
hypotheses.
Example 6.4 (Ex. 6.1 continued) Combine the two statements for each of the examples.
1.
2.
3.
4.
Our original statement is referred to as the null hypothesis (H0).
• The value specified in the null hypothesis may be a previously established value
(in which case we are trying to detect changes to that value), a theoretical value
(in which case we are trying to verify the theory), or a design specification (in
which case we are trying to determine if the specification has been met).
The contradictory statement (H1) is referred to as the alternative hypothesis.
• Note that an alternative hypothesis can be one-sided (1, 3, 4) or two-sided (2).
• The decision as to whether the alternative hypothesis should be one-sided or two-
sided depends on the problem of interest.
Type I Error
Rejecting the null hypothesis H0 when it is true
68

69. For example, the true mean of example 2.4.1 is 3.0. However, for the randomly selected
sample we could observe that the test statistic x falls into the critical region. Therefore,
we could reject the null hypothesis in favor of the alternative hypothesis H1.
Type II Error
Failing to reject the null hypothesis when it is false.
6.1 Performing a Hypothesis Test
Definition 6.2 A procedure leading to a decision about a particular null and
alternative hypothesis is called a hypothesis test.
• Hypothesis testing involves the use of sample data on the population(s) of
interest.
• If the sample data is consistent with a hypothesis, then we “accept” that
hypothesis and conclude that the corresponding statement about the population is
true.
• We “reject” the other hypothesis, and conclude that the corresponding statement
is false. However, the truth or falsity of the statements can never be known with
certainty, so we need to define our procedure so that we limit the probability of
making an erroneous decision.
• The burden of proof is placed upon the alternative hypothesis.
Basic Hypothesis Testing Procedure
A random sample is collected on the population(s) of interest, a test statistic is computed
based on the sample data, and the test statistic is used to make the decision to either
accept (some people say “fail to reject”) or reject the null hypothesis.
Example 6.5 A manufactured product is used in such a way that its most important
dimension is its width. Let X denote the width of a manufactured product. Suppose
historical data suggests that X is a normal random variable having σ = 4 cm. However,
the mean can change due to fluctuations in the manufacturing process. Therefore, we
wish to perform the following hypothesis test.
H0:
H1:
69

70. The following procedure has been proposed.
Inspect a random sample of 25 products. Measure the width of each product. If the
sample mean is less than 188 cm or more than 192 cm, reject H0.
For the proposed procedure, identify the following:
(a) sample size
(b) test statistic
(c) critical region
(d) acceptance region
Is the procedure defined in Ex. 6.5 a good procedure? Since we are only taking a random
sample, we cannot guarantee that the results of the hypothesis test will lead to us making
the correct decision. Therefore, the question “Is this a good procedure?” can be broken
down into two additional questions.
1. If the null hypothesis is true, what is the probability that we accept H0?
2. If the null hypothesis is not true, what is the probability that we accept H0?
Example 6.6 (Ex. 6.5 continued) If the null hypothesis is true, what is the probability
that we accept H0?
70

71. note assumptions
Therefore, if the null hypothesis is true, then there is a 98.76% chance that we will make
the correct decision. However, that also means that there is a 1.24% chance that we will
make the incorrect decision (reject H0 when H0 is true).
• Such a mistake is called a Type I error, or a false positive.
• α = P(Type I error) = level of significance
• In our example, α = 0.0124. When constructing a hypothesis test, we get to
specify α.
If the null hypothesis is not true (i.e. the alternative hypothesis is true), then accepting H0
would be a mistake.
• Accepting H0 when H0 is false is called a Type II error, or a false negative.
•
•
• Unfortunately, we can’t answer this question (find a value for β) in general. Since
the alternative hypothesis is µ ≠ 190 cm, there are an uncountable number of
situations in which the alternative hypothesis is true.
• We must identify specific situations of interest and analyze each one individually.
Example 6.7 (Ex. 6.5 continued) Find the probability of a Type II error when µ = 189 cm
and µ = 193 cm.
For µ = 189 cm:
71

72. For µ = 193 cm:
Note that as µ moves away from the hypothesized value (190 cm), β decreases.
If we experiment with other sample sizes and critical/acceptance regions, we will see that
the values of α and β can change significantly. However, there are some general “truths”
for hypothesis testing.
1. We can explicitly control α (given that the underlying assumptions are true).
2. Type I and Type II error are inversely related.
3. Increasing the sample size is the only way to simultaneously reduce α and β.
4. We can only control β for one specific situation.
Since we can explicitly control α, the probability of a Type I error, rejecting H0 is a
strong conclusion. However, we can only control Type II errors in a very limited
fashion. Therefore, accepting H0 is a weak conclusion. In fact, many statisticians use
the terminology, “fail to reject H0” as opposed to “accept H0.”
• Since “reject H0” is a strong conclusion, we should put the statement about which
it is important to make a strong conclusion in the alternative hypothesis.
Example 6.8 How would the procedure change if we wished to perform the following
hypothesis test?
H0: µ ≥ 190 cm
72

73. H1: µ < 190 cm
Proposed hypothesis testing procedure: Inspect a random sample of 25 observations on
the width of a product. If the sample mean is less than 188 cm, reject H0.
6.1.1 Generic Hypothesis Testing Procedure
All hypothesis have a common procedure. The textbook identifies eight steps in this
procedure.
1. From the problem context and assumptions, identify the parameter of interest.
2. State the null hypothesis, H0.
3. Specify an appropriate alternative hypothesis, H1.
4. Choose a significance level α.
5. State an appropriate test statistic.
6. State the critical region for that statistic.
7. Collect a random sample of observations on the random variable (or from the
population) of interest, and compute the test statistic.
8. Compare the test statistic value to the critical region and decided whether or not to
reject H0.
6.2 Performing Hypothesis Tests on µ when σ is Known
In this section, we consider making inferences about the mean µ of a single population
where the population standard deviation σ is known.
• We will assume that a random sample X1, X2, … , Xn has been taken from the
population.
• We will also assume that either the population is normal or the conditions of the
Central Limit Theorem apply.
Suppose we wish to perform the following hypothesis test.
73

74. It is somewhat obvious that inferences regarding µ would be based on the value of the
sample mean. However, it is usually more convenient to standardize the sample mean.
Using what we know about the sampling distribution of the mean, it is reasonable to
conclude that the test statistic will be
If the null hypothesis is true, then the test statistic is a standard normal random variable.
Therefore, we only reject the null hypothesis if the value of Z0 is unusual for an
observation on a standard normal random variable.
Specifically, we reject H0 if:
where α is the specified level of significance. The acceptance region is therefore
Obviously, the acceptance and critical regions can be converted to expressions in terms of
the sample mean.
Reject H0 if X > a or X < b where
74

75. Example 6.9 Let X denote the GPA of an engineering student at the Philadelphia
University. It is widely known that, for this population, σ = 0.5. The population mean is
not widely known, however, it is commonly believed that the average GPA is 3.0. We
wish to test this hypothesis using a sample of size 25 and a level of significance of 0.05.
(a) Identify the null and alternative hypotheses.
(b) List any required assumptions.
(c) Identify the test statistic and the critical region.
Reject H0 if
(d) Suppose 25 students are sampled and the sample average GPA is 3.18. State and
interpret the conclusion of the test.
Z0 =
(e) What is the probability of a Type I error for this test?
(f) How would the results change if we had used α = 0.10?
75

76. Critical region changes.
We may also modify this procedure if the test is one-sided. This modification only
requires a change in the critical/acceptance regions. If the alternative hypothesis is
then a negative value of Z0 would not indicate a need to reject H0. Therefore, we
only reject H0 if
Likewise, if the alternative hypothesis is
Example 6.10 The Glass Bottle Company (GBC) manufactures brown glass beverage
containers that are sold to breweries. One of the key characteristics of these bottles is
their volume. GBC knows that the standard deviation of volume is 0.08 oz. They wish to
ensure that the mean volume is not more than 12.2 oz using a sample size of 30 and a
level of significance of 0.01.
(a) Identify the null and alternative hypotheses.
76

77. (b) Identify the test statistic and the critical region.
(c) Suppose 30 bottles are measured and the sample mean is 12.23. State and
interpret the conclusion of the test.
6.2.1 Computing P-Values
We have already seen that the choice of the value for the level of significance can
impact the conclusions derived form a test of hypotheses. As a result, we may be
interested in answering the question: How close did we come to making the opposite
conclusion? We answer this question using an equivalent decision approach that can be
used as an alternative to the critical/acceptance regions. This approach is called the P-
value approach.
Definition 6.3 The P-value for a hypothesis test is the smallest level of significance that
would lead to rejection of the null hypothesis.
How we compute the P-value depends on the form of the alternative hypothesis.
We reject H0 if
77

78. Example 6.11 (Ex. 6.9 continued) Compute the P-value for the test.
Note when α = 0.05,
But, when α = 0.10,
Example 612 (Ex. 6.10 continued) Compute the P-value for the test.
Since α = 0.01, P > α
6.2.2 Type II Error
In hypothesis testing, we get to specify the probability of a Type I error ( α). However,
the probability of a Type II error (β) depends on the choice of sample size (n).
Consider first the case in which the alternative hypothesis is H1: µ ≠ µ0.
Before we can proceed, we must be more specific about “H0 is false”. We will
accomplish this by saying:
where δ ≠ 0.
78

79.  X − µ0 
β = P − Z α / 2 ≤

≤ Zα / 2 µ = µ 0 + δ 

 σ n 
 σ σ 
β = P µ 0 − Z α / 2
 ≤ X ≤ µ 0 + Zα / 2 µ = µ0 + δ 

 n n 
 σ σ 
 µ 0 − Zα / 2 − ( µ0 + δ ) µ 0 + Zα / 2 − (µ0 + δ ) 
β = P 
n n
≤Z≤
 σ σ 
 
 n n 
If the alternative hypothesis is H1: µ > µ0, then
.
If the alternative hypothesis is H1: µ < µ0.
.
Example 6.13 (Ex. 6.10 continued) Let X denote the GPA of an engineering student at
the Philadelphia University. It is widely known that, for this population, σ = 0.5. The
population mean is not widely known, however, it is commonly believed that the average
GPA is 3.0. We wish to test this hypothesis using a sample of size 25 and a level of
significance of 0.05. In Example 16.5, we formulated this hypothesis test as
79

80. The corresponding test statistic and critical region are given by
X − 3.0
Z0 =
0.5 25
(a) If µ = 3.2, what is the Type II error probability for this test?
δ = µ − µ0 =
β = P( − 3.96 ≤ Z ≤ −0.04 ) = 0.4840
(b) If µ = 2.68, what is the Type II error probability for this test?
δ = µ − µ0 =
β = P(1.24 ≤ Z ≤ 5.16 ) = 0.1075
(c) If µ = 2.68, what is the power of the test?
power =
80

81. (d) If µ = 3.32, what is the power of the test?
power = 0.8925
Example 6.14 (Ex. 6.11 continued) The Glass Bottle Company (GBC) manufactures
brown glass beverage containers that are sold to breweries. One of the key characteristics
of these bottles is their volume. GBC knows that the standard deviation of volume is
0.08 oz. They wish to ensure that the mean volume is not more than 12.2 oz using a
sample size of 30 and a level of significance of 0.01. Example 16.6, we formulated this
hypothesis test as
H0: µ ≤ 12.2
H1: µ > 12.2
The corresponding test statistic and critical region are given by
Reject H0 if
(a) If µ = 12.27 oz, what is the Type II error probability for this test?
δ = µ − µ0 = 0.07
 0.07 30 
β = P Z ≤ 2.3263 −  = P( Z ≤ −2.47 ) = 0.0068
 0.08 
 
(b) If µ = 12.15 oz, what is the Type II error probability for this test?
This is a poor question. If µ = 12.15 oz, then “technically” the null hypothesis is
true. If we are truly concerned with detecting this, we should have used a two-
sided alternative hypothesis.
81

82. 6.2.3 Choosing the Sample Size
The expressions for β allow the determination of an appropriate sample size. To choose
the proper sample size for our test, we must specify a value of β for a specified value of
δ.
For the case in which H1: µ ≠µ0, the symmetry of the test allows us to always specify a
positive value of δ. If we specify a relatively small value of β (≤ 0.1), then the lower side
of the equation becomes negligible. So, the equation for β reduces to:
This yields:
δ n
Z 1− β = − Z β = Z α / 2 −
σ
δ n
= Zα / 2 + Z β
σ
For both cases in which the alternative hypothesis is one-sided:
82

83. Example 6.14 (Ex. 6.10 continued) Let X denote the GPA of an engineering student at
the Philadelphia University. It is widely known that, for this population, σ = 0.5. The
population mean is not widely known, however, it is commonly believed that the average
GPA is 3.0. We wish to test this hypothesis using a sample of size n and a level of
significance of 0.05. In Example 16.5, we formulated this hypothesis test as
H0: µ = 3.0
H1: µ ≠ 3.0
The corresponding test statistic and critical region are given by
(a) If we want β = 0.10 at µ = 3.2, what sample size should we use?
δ = 0.2
n = 66
(b) If we want β = 0.10 at µ = 3.25, what sample size should we use?
δ = 0.25
n=
( Z 0.025 + Z 0.10 ) 2 0.5 2 = (1.96 + 1.282) 2 0.5 2 = 42.04
0.25 2 0.25 2
n = 43
(c) If we want β = 0.05 at µ = 3.2, what sample size should we use?
δ = 0.2
n=
( Z 0.025 + Z 0.05 ) 2 0.5 2 = (1.96 + 1.645) 2 0.52 = 81.2
0.2 2 0.2 2
83

84. n = 82
Example 6.15 (Ex. 6.11 continued) The Glass Bottle Company (GBC) manufactures
brown glass beverage containers that are sold to breweries. One of the key characteristics
of these bottles is their volume. GBC knows that the standard deviation of volume is
0.08 oz. They wish to ensure that the mean volume is not more than 12.2 oz using a
sample size of n and a level of significance of 0.01. Example 16.6, we formulated this
hypothesis test as
The corresponding test statistic and critical region are given by
If we wish to have a test power of 0.95 at µ = 12.25 oz, what is the required sample size
for this test?
6.2.4 Choosing the Sample Size
The expressions for β allow the determination of an appropriate sample size. To choose
the proper sample size for our test, we must specify a value of β for a specified value of
δ.
For the case in which H1: µ ≠µ0, the symmetry of the test allows us to always specify a
positive value of δ. If we specify a relatively small value of β (≤ 0.1), then the lower side
of the equation becomes negligible. So, the equation for β reduces to:
 δ n
β = P Z ≤ Z α / 2 −


 σ 

84

85. This yields:
δ n
Z 1− β = − Z β = Z α / 2 −
σ
δ n
= Zα / 2 + Z β
σ
(Z α/2 + Zβ ) σ 2
2
n=
δ2
For both cases in which the alternative hypothesis is one-sided:
(Z α + Zβ ) σ 2
2
n=
δ2
Example 6.14 (Ex. 6.10 continued) Let X denote the GPA of an engineering student at
the Philadelphia University. It is widely known that, for this population, σ = 0.5. The
population mean is not widely known, however, it is commonly believed that the average
GPA is 3.0. We wish to test this hypothesis using a sample of size n and a level of
significance of 0.05. In Example 16.5, we formulated this hypothesis test as
H0: µ = 3.0
H1: µ ≠ 3.0
The corresponding test statistic and critical region are given by
X − 3.0
Z0 =
0.5 n
Reject H0 if Z0 < −Zα/2 = −Z0.025 = −1.96 or if Z0 > Zα/2 = 1.96
(a) If we want β = 0.10 at µ = 3.2, what sample size should we use?
δ = 0.2
n=
( Z 0.025 + Z 0.10 ) 2 0.5 2 = (1.96 + 1.282) 2 0.5 2 = 65.7
0.2 2 0.2 2
n = 66
(b) If we want β = 0.10 at µ = 3.25, what sample size should we use?
85

86. δ = 0.25
n=
( Z 0.025 + Z 0.10 ) 2 0.5 2 = (1.96 + 1.282) 2 0.5 2 = 42.04
0.25 2 0.25 2
n = 43
(c) If we want β = 0.05 at µ = 3.2, what sample size should we use?
δ = 0.2
n=
( Z 0.025 + Z 0.05 ) 2 0.5 2 = (1.96 + 1.645) 2 0.52 = 81.2
0.2 2 0.2 2
n = 82
Example 6.15 (Ex. 6.11 continued) The Glass Bottle Company (GBC) manufactures
brown glass beverage containers that are sold to breweries. One of the key characteristics
of these bottles is their volume. GBC knows that the standard deviation of volume is
0.08 oz. They wish to ensure that the mean volume is not more than 12.2 oz using a
sample size of n and a level of significance of 0.01. Example 16.6, we formulated this
hypothesis test as
H0: µ ≤ 12.2
H1: µ > 12.2
The corresponding test statistic and critical region are given by
X − 12.2
Z0 =
0.08 n
Reject H0 if Z0 > Zα = Z0.01 = 2.3263
If we wish to have a test power of 0.95 at µ = 12.25 oz, what is the required sample size
for this test?
δ = 0.05
β = 0.05
n=
( Z 0.01 + Z 0.05 ) 2 0.082 = ( 2.326 + 1.645) 2 0.082 = 40.4
0.05 2 0.05 2
86

87. n = 41
6.3 Statistical Significance
A hypothesis test is a test for statistical significance. When we reject H 0, we are
stating that the data indicates a statistically significant difference between the true mean
and the hypothesized value of the mean. When we accept H 0, then we are stating that
there is not a statistically significant difference. Statistical difference and practical
significance are not the same. This is especially important to recognize when the sample
size is large.
6.3.1 Introduction to Confidence Intervals
As we have previously discussed, the sample mean is the most often used point
estimate for the population mean. However, we also pointed out that two different
samples would most likely result in two different sample means. Therefore, we define
confidence intervals as a means of quantifying the uncertainty in our point estimate.
If θ is the parameter of interest, then the point estimate and sampling distribution of θ can
be used to identify a 100(1 − α )% confidence interval on θ. This interval is of the
form:
L ≤ θ ≤ U.
L and U are called the lower-confidence limit and upper-confidence limit.
If L and U are constructed properly, then
P(L ≤ θ ≤ U) = 1 − α.
The quantity (1 − α) is called the confidence coefficient. The confidence coefficient is a
measure of the accuracy of the confidence interval. For example, if a 90% confidence
interval is constructed, then the probability that the true value of θ is contained in the
interval is 0.9.
87

88. The length of the confidence interval is a measure of the precision of the point
estimate. A general rule of thumb is that increasing the sample size improves the
precision of a point estimate.
6.3.2 Confidence Interval on µ when σ is Known
We can use what we have learned to construct a 100(1 − α )% confidence
interval on the mean, assuming that (a) the population standard deviation is known, and
(b) the population is normally distributed (or the conditions of the Central Limit Theorem
apply).
P( − Z α / 2 ≤ Z ≤ Z α / 2 ) = 1 − α
 X −µ 
P − Z α / 2 ≤ ≤ Zα / 2  = 1−α
 σ n 
 
 σ σ 
P X − Z α / 2
 ≤ µ ≤ X + Zα / 2  = 1−α

 n n
Such a confidence interval is called a two-sided confidence interval. We can also
construct one-sided confidence intervals for the same set of assumptions (σ known,
normal population or Central Limit Theorem conditions apply).
The 100(1 − α)% upper-confidence interval is given by
 σ 
P µ ≤ X + Z α
  = 1−α

 n
and the 100(1 − α)% lower-confidence interval is given by
 σ 
P µ ≥ X − Z α
  = 1−α .

 n
Example 6.16 Let X denote the GPA of an engineering student at the Philadelphia
University. It is widely known that, for this population, σ = 0.5. The population mean is
not widely known, however, we have a collected a sample of size 25 from the population.
The resulting sample mean was 3.18.
88

89. (a) What assumptions, if any, are required to use this data to construct a confidence
interval on the mean GPA?
GPA is normally distributed.
(b) Construct a 95% confidence interval on µ and interpret its meaning.
σ 0.5
X ± Z 0.025 = 3.18 ± 1.96
n 25
2.984 ≤ µ ≤ 3.376
P( 2.984 ≤ µ ≤ 3.376 ) = 0.95
(c) Construct a 99% confidence interval on µ and compare it to the confidence
interval obtained in part (b).
σ 0.5
X ± Z 0.005 = 3.18 ± 2.58
n 25
2.922 ≤ µ ≤ 3.438
more accurate, but less precise
(d) Construct a 95% upper-confidence interval on µ and interpret its meaning.
σ 0.5
X + Z 0.05 = 3.18 + 1.645
n 25
µ ≤ 3.3445
P( µ ≤ 3.3445) = 0.95
(e) Construct a 95% lower-confidence interval on µ and interpret its meaning.
σ 0.5
X − Z 0.05 = 3.18 − 1.645
n 25
µ ≥ 3.0155
P( µ ≥ 3.0155) = 0.95
(f) Combine the two confidence intervals obtained in parts (d) and (e). Is this
confidence interval superior to the one constructed in part (b)?
89

90. 3.0155 ≤ µ ≤ 3.3445
No, it is only a 90% confidence interval.
6.3.3 Choosing the Sample Size for a Confidence Interval on µ when σ is Known
The percentage of a confidence interval is a measure of the accuracy of the
confidence interval.
The half-width of the confidence interval, E, is a measure of the precision of the
confidence interval. For a two-sided confidence interval, E = (U – L)/2. For an upper-
confidence interval, E = U − θ and for a lower-confidence interval, E = θ − L.
For a given level of accuracy (α), we can control the precision of the confidence
interval using the sample size. For the two-sided confidence interval on µ, we specify a
value of E and note that:
σ
E = Zα / 2 .
n
Then, we can solve for n.
2
Z σ 
n =  α/2 
 E 
For the one-sided confidence intervals:
2
Z σ 
n= α  .
 E 
Example 6.17 (Ex. 6.16 continued)
(a) If we wish to construct a 95% confidence interval on µ that has a half-width of
0.1, how many students should we survey?
2 2
 Z σ   1.96 ⋅ 0.5 
n =  0.025  =   = 96.04
 E   0.1 
n = 97
90

91. (b) If we wish to construct a 95% upper-confidence interval on µ that has a half-
width of 0.1, how many students should we survey?
2 2
 Z σ   1.645 ⋅ 0.5 
n =  0.05  =   = 67.65
 E   0.1 
n = 68
(c) If we wish to construct a 90% confidence interval on µ that has a half-width of
0.1, how many students should we survey?
2 2
 Z σ   1.645 ⋅ 0.5 
n =  0.05  =   = 67.65
 E   0.1 
n = 68
6.3.4 Using Confidence Intervals to Perform Hypothesis Tests on µ when σ is
Known
Thus far, we have considered two methods of evaluating hypothesis tests: critical
regions and P-values. A third, equivalent method is to use a confidence interval.
1. Specify: µo, α, n
2. If H1: µ ≠ µo, construct a 100(1 − α)% confidence interval on µ.
If H1: µ > µo, construct a 100(1 − α)% lower-confidence interval on µ.
If H1: µ < µo, construct a 100(1 − α)% upper-confidence interval on µ.
3. Reject H0 is µo is not contained in that confidence interval.
Example 6.17 (Ex. 6.10 continued) Let X denote the GPA of an engineering student at
the Philadelphia University. It is widely known that, for this population, σ = 0.5. The
population mean is not widely known, however, it is commonly believed that the average
GPA is 3.0. We wish to test this hypothesis using a sample of size 25 and a level of
significance of 0.05.
From Ex. 6.10:
H0: µ = 3.0
H1: µ ≠ 3.0
91

92. Suppose the sample mean is 3.18. Use a confidence interval to evaluate the hypothesis
test.
α = 0.05, H1: ≠
95% confidence interval
From Ex. 6.16:
2.984 ≤ µ ≤ 3.376
3.0 is in the confidence interval
fail to reject H0
Example 6.18 (Ex. 6.11 continued) The Glass Bottle Company (GBC) manufactures
brown glass beverage containers that are sold to breweries. One of the key characteristics
of these bottles is their volume. GBC knows that the standard deviation of volume is
0.08 oz. They wish to ensure that the mean volume is not more than 12.2 oz using a
sample size of 30 and a level of significance of 0.01.
From Ex. 3.2.2:
H0: µ ≤ 12.2
H1: µ > 12.2
Suppose the sample mean is 12.23. Use a confidence interval to evaluate the hypothesis
test.
α = 0.01, H1: >
99% lower-confidence interval
σ 0.08
X − Z 0.01 = 12.23 − 2.3263
n 30
µ ≥ 12.1960
12.2 is in the confidence interval
fail to reject H0
92

93. 6.4 Hypothesis Test on μ and σ unkown
What if σ is Unknown?
Suppose we are interested in studying the mean of a population, but we do not
know the value of the population standard deviation?
• We can use the procedures defined in section 2.3 and replace σ with S, provided
that the sample size is large (n ≥ 30).
• When the sample size is small and σ is unknown, then we must assume that the
population is normally distributed.
The t Distribution
Suppose we wish to perform the following hypothesis test.
H0: µ = µ0
H1: µ ≠ µ0
Suppose we have collected a random sample of size n and that we have used this
sample data to compute the sample mean X and the sample standard deviation S.
If σ were known then we would compute the test statistic:
X − µ0
Z0 = .
σ n
Therefore, a logical approach is to replace σ with S. The resulting test statistic is:
X − µ0
T0 = .
S n
Before we can proceed, we should analyze the sampling distribution of this test statistic.
Theorem 6.1 The t Distribution
Let X1, X2, … , Xn be a random sample from a normal population having mean µ.
The quantity
X −µ
T=
S
n
93

94. has a t distribution with n – 1 degrees of freedom.
While we won’t discuss the details of the t distribution, it is important to recognize
two points regarding the t probability density function.
• First, it is symmetric about 0.
• Second, as the number of degrees of freedom increases, the t distribution
approaches the standard normal distribution. This explains why it is OK to use
the procedures from section 2.3 when n ≥ 30 (at 29 degrees of freedom there is
little difference between t and Z).
Example 6.19 Suppose T has a t distribution with 7 degrees of freedom. Find the
following:
(a) P(T > 2.365)
Excel function TDIST(x, degrees of freedom, 1 or 2 tails)=P(2.3625, 7, 1) = 0.025
Note Excel gives you the value P(X>x)
(b) P(T > 1.415)
0.10
(c) P(T < −3.499)
P(T > 3.499) = 0.005
(d) P(T > −2.8) = 1 – Pt (T<2.88) = 0.9867
(e) the value a such that P(T >a) = 0.05
a = t0.05,7 = 1.895
(f) the value of a such that P(T > a) = 0.01
a = t0.01,7 = 2.998
(g) the value of a such that P(T < a) = 0.9975
a = t0.0025,7 = 4.029
94

95. The Critical Region
If the null hypothesis is true, then T0 has a t distribution with n − 1 degrees of
freedom. Therefore, we reject H0 if
T0 > tα/2,n-1 or T0 < −tα/2,n-1.
• If the alternative hypothesis is H1: µ > µ0, then we reject H0 if T0 > tα,n-1.
• If the alternative hypothesis is H1: µ < µ0, then we reject H0 if T0 < −tα,n-1.
Example 6.20 Let X denote the age of an Engineering Statistics student at the
Philadelphia University. Suppose we wish to test the hypothesis that the mean age is
21.0 years using a sample of size 14 and a level of significance of 0.05.
(a) State the null and alternative hypothesis.
H0: µ = 21
H1: µ ≠ 21
(b) What assumptions are required before we can proceed with this test of
hypotheses?
Age is normally distributed.
(c) How could we test the validity of this assumption?
Probability plot of the data.
(d) What would have to be done to avoid making this assumption?
Survey at least 16 more students (n ≥ 30).
(e) Identify the test statistic and acceptance region for this test.
X − µ0 X − 21
T0 = =
S n S 14
Fail to reject H0 if −tα/2,n-1 ≤ T0 ≤ tα/2,n-1
Fail to reject H0 if −2.160 ≤ T0 ≤ 2.160
NOTE: In Excel ® the function is TINV(probability, degrees of freedom)
Probability: is the associated probability with a two-sided t-distribution
95

96. (f) Suppose the sample mean is 22.67 years and the sample standard deviation is 4.08
years. State and interpret the conclusion for this test.
22.67 − 21
T0 = = 1.532
4.08 14
Fail to reject H0 and conclude that µ = 21.
(g) What would be required to solve this problem using a Z-test instead of a t-test?
n ≥ 30 or σ known
Example 6.21 A construction company purchases shear pins for use in several of their
projects. These pins are supposed to have a mean tensile strength of at least 500 psi, but
destructive testing is required to measure the tensile strength of a pin. Therefore, a
sample of 12 pins are evaluated and used to verify that the tensile strength is acceptable
using a level of significance of 0.10.
(a) State the null and alternative hypothesis.
H0:
H1:
(b) Identify the test statistic and critical region for this test.
Reject H0 if.
Reject H0 if.
Note in Excel use TINV(2*(0.1),11) because it is one-sided test and excel return
two-sided test.
(c) Suppose the sample mean is 491.3 psi and the sample standard deviation is 13.2
psi. State and interpret the conclusion for this test.
491.3 − 500
T0 = = −2.2832
13.2 12
96

97. P-Values
As before, we can use P-values to draw conclusions regarding a t-test.
H1: µ ≠ µ0 P = 2 P ( t n −1 > T0 )
H1: µ > µ0 P = P(tn-1 > T0)
H1: µ < µ0 P = P(tn-1 < T0)
We reject H0 if P < α.
Example 6.22 (Ex. 6.20 continued) Perform the hypothesis test using a P-value.
Example 6.23 (Ex. 6.21 continued) Perform the hypothesis test using a P-value.
Type II Errors and Sample Size
Unfortunately, we cannot construct an expression for β for these tests, and as a result, we
cannot construct expressions for computing an appropriate sample size. Some cases have
been analyzed numerically, and the corresponding OC curves are in your text – Charts
VII(e), VII(f), VII(g) and VII(h).
The value on the horizontal axis is
µ − µ0
d= .
σ
Note that the value in the denominator must be our best guess for σ.
97

98. Example 6.24 (Ex. 6.20 continued)
(a) If the true population mean is 22.0 years and the true population standard
deviation is 1.5 years, then what is the probability of a Type II error?
Chart V (a)
(b) Suppose the true population standard deviation is 1.25 years and we wish to have
a power of 0.9 when the true population mean is 22.5 years. What sample size
should we use?
Confidence Intervals
As before, we can also use the sample data to construct a 100(1−α)% confidence
interval on the population mean.
 S S 
P X − t α / 2, n −1
 ≤ µ ≤ X + t α / 2, n −1  = 1−α

 n n
The 100(1 − α)% upper-confidence interval is given by
 S 
P µ ≤ X + t α , n −1
  = 1−α

 n
and the 100(1 − α)% lower-confidence interval is given by
 S 
P µ ≥ X − t α , n −1
  = 1−α .

 n
We can then use these confidence intervals to perform the hypothesis test.
Specify: µ0, α, n
98

99. If H1: µ ≠ µ0, construct a 100(1 − α)% confidence interval on µ.
If H1: µ > µ0, construct a 100(1 − α)% lower-confidence interval on µ.
If H1: µ < µ0, construct a 100(1 − α)% upper-confidence interval on µ.
Reject H0 is µ 0 is not contained in that confidence interval.
Example 6.25 (Ex. 6.20 continued) Perform the hypothesis test using an appropriate
confidence interval.
α = 0.05, H1: ≠
95% confidence interval
Example 6.26 (Ex. 6.21 continued) Perform the hypothesis test using an appropriate
confidence interval.
α = 0.10, H1: <
Therefore, with a 90% confidence interval (90% of the times ) the mean
will fall µ ≤ 496.49, therefore 500 is not in the confidence interval
Z-Tests versus t-Tests
Even if we know the population standard deviation, we can use the t-test instead
of the Z-test. The reason we don’t do this is:
• A Z-test is more powerful than the corresponding t-test.
• A confidence interval based on Z is tighter than the corresponding confidence
interval based on t.
99

100. 6.5 Inference on the Variance of a Normal Population
The Chi-Square Distribution
Sometimes it is of interest to perform a hypothesis test on the population standard
deviation. For example, we may wish to test the value of σ prior to testing µ.
Assumptions
• population is normally distributed (cannot make use of the Central Limit
Theorem)
Potential Tests
Test Statistic
Theorem 3.5.1 The Chi-Square Distribution
Let X1, X2, … , Xn be a random sample from a normal population. The quantity
χ 2
=
( n − 1) S 2
σ2
has a chi-square distribution with n – 1 degrees of freedom.
• A table of probabilities for the chi-square distribution is provided in the
Appendix of your text.
• Note that the chi-square distribution is not symmetric. In fact, the range of the
chi-square random variable is (0,∞).
Example 6.27 Suppose Y has a chi-square distribution with 13 degrees of freedom.
Find the following:
(a)
(b) P(Y < 5.89) =
100

101. (c) P(Y > 0)
(d) the value a such that P(Y >a) = 0.005
In Excel CHIINV( probability, degrees of freedom )
(e) the value of a such that P(Y < a) = 0.01
Performing the Test
Critical Regions
H1: σ ≠ σ 0
2 2
H1: σ > σ 0
2 2
H1: σ < σ 0
2 2
P-Values
H1: σ ≠ σ 0
2 2
H1: σ > σ 0
2 2
H1: σ < σ 0
2 2
OC Function and Sample Size
σ
must use Charts VII(i, j, k, l, m, n) with λ =
σ0
Confidence Intervals
101

102.  ( n − 1) S 2 ( n − 1) S 2  = 1 − α
two-sided P 2 ≤σ2 ≤ 2 
 χ χ1−α / 2, n −1 
 α / 2, n −1 
 ( n − 1) S 2
P σ 2 ≤ 2

 = 1−α
upper  
 χ 1−α , n −1 
 2 ( n − 1) S 2 
lower P σ ≥  =1−α
 χ α , n −1 
2
 
Example 6.28 In an effort to improve their quality, the Glass Bottle Company
(GBC) has completed a variance reduction initiative. As a result of this initiative, GBC
now believes that the standard deviation of the volume of their brown glass bottles is 0.05
oz. They wish to test this hypothesis using a sample of size 20 and a level of significance
of 0.05.
(a) State any required assumptions.
Volume is normally distributed.
(b) What if further testing revealed that this assumption was invalid?
You would be stuck.
(c) State the null and alternative hypotheses, the test statistic and critical region.
H0:
H1:
(d) Suppose the sample standard deviation is 0.065 oz. State and interpret the
conclusion of the test.
102

103. (e) Construct a 95% confidence interval on the population standard deviation.
Example 6.29 Let X denote the age of an Engineering Statistics student at the
Philadelphia University and consider the following hypothesis test.
H0:
H1:
We wish to perform this test using a level of significance of 0.05.
(a) Suppose we wish to detect a true standard deviation of 6.0 years with probability
0.95. What sample size should we use?
power of the test = 0.95
Chart VII(k) ⇒ n = 15
(b) If this sample size is used, what is the probability of a Type II error if σ = 4.5
years?
(c) Suppose this sample size is used and S = 4.4025 years. Use a P-value to
determine the conclusion resulting from the test.
103

104. 6.6 Population Proportions
Sometimes we wish to perform a hypothesis test on a population proportion p.
Examples of population proportions are:
• proportion of registered voters that favor term limits
• proportion of manufactured items that are defective
• probability of winning a game
• probability of successfully completing a mission
The general definition of p is the proportion of members of the population that
belong to the class of interest. In some cases, p corresponds to the probability of success
for a Bernoulli trial.
Testing Population Proportions
Assumptions
• population is very large (or infinite)
Potential Tests
Test Statistic
For a random sample of size n, we compute X = the number of members
sampled that belong to the class of interest or X = the number of successes.
Note that we are going to take advantage of the normal approximation to the
binomial. Therefore, all of our results are “approximate”, and the larger n is, the better
the approximation is.
Critical Regions
H1: p ≠ p0
H1: p > p0
104

105. H1: p < p0
P-Values
H1: p ≠ p0
H1: p > p0
H1: p < p0
OC Function
H1: p ≠ p0
 p 0 (1 − p 0 ) p 0 (1 − p 0 ) 
 p0 − p − Zα / 2 p0 − p + Zα / 2 
 n n 
β = P ≤Z≤ 
 p (1 − p ) p (1 − p ) 
 
 n n 
 p 0 (1 − p 0 ) 
 p0 − p + Zα 
 n 
H1: p > p0 β = P Z ≤ 
 p (1 − p ) 
 
 n 
 p 0 (1 − p 0 ) 
 p0 − p − Zα 
 n 
H1: p < p0 β = P Z ≥ 
 p (1 − p ) 
 
 n 
Choice of Sample Size
 Z α / 2 p 0 (1 − p 0 ) + Z β p (1 − p ) 
2
two-sided test n= 
 p − p0 
 
p 0 (1 − p 0 ) + Z β p (1 − p ) 
2
 Zα
one-sided test n= 
 p − p0 
 
Confidence Intervals
X
Recall p =
ˆ .
n
 p (1 − p )
ˆ ˆ p (1 − p ) 
ˆ ˆ
two-sided P p − Z α / 2

ˆ ≤ p ≤ p + Zα / 2
ˆ  = 1−α

 n n 
105

106.  p (1 − p ) 
ˆ ˆ
upper P p ≤ p + Z α

ˆ  = 1−α

 n 
 p (1 − p ) 
ˆ ˆ
lower P p ≥ p − Z α

ˆ  = 1−α

 n 
Reject H0 if p0 is not in the confidence interval.
Choice of Sample Size for Confidence Intervals
2 2
Z  Z 
two-sided n =  α / 2  p (1 − p ) n =  α / 2  0.25
 E   E 
2
Z 
one-sided n =  α  0.25
 E 
Example 6.30 A beverage producer purchases their brown glass bottles from GBC.
However, they purchase their caps from another supplier. Prior to using a shipment of
these caps in their bottling process, they inspect 500 of them. Each inspected cap is
classified as defective or non-defective. The producer has agreed with their supplier that
the true proportion defective should not exceed 0.04. They wish to use the inspection
data to perform an appropriate test using a level of significance of 0.10.
(a) State the null and alternative hypotheses, the test statistic and the critical region.
H0: p = 0.04
H1: p > 0.04
(b) For a particular shipment, suppose that 27 of the inspected caps are defective.
State and interpret the conclusion of the test.
Z0 = 1.5975
.
(c) Compute the P-value for the test.
106

107. (d) What is the power of the test when p = 0.08?
 0.04(1 − 0.04 ) 
 0.04 − 0.08 + 1.282 
β = P Z ≤  = P ( Z ≤ −2.3709 ) = 0.0089
500
 0.08(1 − 0.08) 

 

 500 
(e) If the producer wishes to have a Type II error probability of 0.05 when p = 0.1,
then what sample size should they use?
Example 6.31 Let p denote the proportion of engineering students at the Philadelphia
University that are afraid to fly.
(a) Suppose we wish to construct a 95% confidence interval on p such that the error
in our point estimate is 15% or less. What sample size should we use?
(b) Suppose this sample size is used and 9 of the surveyed students are afraid to fly.
Construct a point estimate of p.
(c) Construct a 95% confidence interval on p.
(d) Use this confidence interval to evaluate the following hypothesis test.
H0: p = 0.3
H1: p ≠ 0.3
0.3 is in the confidence interval ⇒ fail to reject H0
(e) What is the level of significance for the test evaluated in part (d)?
107

108. 0.05 (95% confidence interval)
108

109. 7 DECISION MAKING FOR TWO SAMPLES
Comparing Two Populations
In many cases, we wish to use hypothesis testing to compare two populations. We denote
these as Population 1 and Population 2, and we use the population numbers as subscripts
on parameters, statistics, and sample sizes.
Population 1
• parameter of interest: µ1, σ1 or p1
• sample size: n1
ˆ
• statistic of interest: X 1 , S1 or p1
Population 2
• parameter of interest: µ2, σ2 or p2
• sample size: n2
ˆ
• statistic of interest: X 2 , S2 or p 2
7.1 Comparing Two Population Means, Variance Known
Assumptions
• the two populations are independent
• σ1 and σ2 are known
• the populations are normally distributed (or the conditions of the Central Limit
Theorem hold)
Parameter of Interest
µ1 − µ2
Point Estimate
Potential Tests
H0: µ1 − µ2 = ∆0 H0: µ1 − µ2 ≤ ∆0 H0: µ1 − µ2 ≥ ∆0
109

110. Test Statistic
Critical Region
H1: µ1 − µ2 ≠ ∆0 Reject H0 if Z0 > Zα/2 or if Z0 < −Zα/2
H1: µ1 − µ2 > ∆0 Reject H0 if Z0 > Zα
H1: µ1 − µ2 < ∆0 Reject H0 if Z0 < −Zα
P-Values
H1: µ1 − µ2 ≠ ∆0
H1: µ1 − µ2 > ∆0
H1: µ1 − µ2 < ∆0
OC Function
∆ = µ1 − µ2
H1:µ1− µ2 ≠ ∆0
 
 
 ∆ − ∆0 ∆ − ∆0 
β = P − Z α / 2 − ≤ Z ≤ Zα / 2 − 
 σ 12 σ 2
2
σ 12 σ 2 
2
 + + 
 n1 n 2 n1 n 2 
 
 
 ∆ − ∆0 
β = P Z ≤ Z α − 
 σ 12 σ 2 
2
 + 
 n1 n 2 
H1: µ1 − µ2 > ∆0
 
 
 ∆ − ∆0 
β = P Z ≥ − Z α − 
 σ 12 σ 2 
2
 + 
 n1 n 2 
H1: µ1 − µ2 < ∆0
110

111. Choice of Sample Size
n = n1 = n2
two-sided test
one-sided test
Using OC Curves for Computing β and Sample Size
H1: µ1 − µ2 ≠ ∆0 use Charts VII(a) and VII(b)
H1: µ1 − µ2 > ∆0 use Charts VII(c) and VII(d)
H1: µ1 − µ2 < ∆0 use Charts VII(c) and VII(d)
Confidence Intervals
two-sided
 σ 12 σ 2
2
σ 12 σ 2 
2
P ( X − X ) − Z + ≤ µ1 − µ 2 ≤ ( X 1 − X 2 ) + Z α / 2 +  = 1−α
 1 2 α/2
n1 n 2 n1 n 2 
 
upper
 σ 12 σ 2 
2
P µ 1 − µ 2 ≤ ( X 1 − X 2 ) + Z α +  = 1−α
 n1 n 2 
 
lower
 σ 12 σ 2 
2
P µ 1 − µ 2 ≥ ( X 1 − X 2 ) − Z α +  = 1−α
 n1 n 2 
 
Note: Fail to reject if zero (0) is within the confidence interval
Choice of Sample Size for Confidence Intervals
111

112. two-sided
one-sided
Examples
Example 7.1 An automobile manufacturer wishes to purchase its batteries from an
outside supplier. The key quality characteristic of these batteries is voltage. Two
suppliers are competing for the manufacturer’s business (Supplier 1 and Supplier 2).
Both suppliers have normally distributed voltage and a standard deviation in voltage of
0.15 v. Therefore, the manufacturer only needs to compare their means. The QC director
has decided to make this comparison using a two-sample hypothesis test and a level of
significance of 0.05.
(a) State the null and alternative hypothesis.
H0: µ1 − µ2 = 0
H1: µ1 − µ2 ≠ 0
(b) Identify the test statistic.
(c) What sample size should be used if the manufacturer wishes to detect a difference
of 0.25 volts with probability 0.9?
(d) Identify the critical region.
Reject H0 if Z0 > 1.96 or if Z0 < −1.96
(e) If the sample size prescribed in part (c) is used and the resulting sample means are
X 1 = 12.06 and X 2 = 11.98 , then what is the conclusion resulting from the test?
112

113. Fail to reject H0 and conclude that µ1 − µ2 = 0.
(f) Find the P-value for this test.
P = 2 P ( Z > 1.0 6 ) = 0.2861
(g) What is the Type II error probability for this test if µ1 − µ2 = 0.15?
β = P(−3.96 ≤ Z ≤ −0.04) = 0.4840
(h) Construct a 95% confidence interval on µ1 − µ2.
0.15 2 0.15 2
(12.06 − 11.98) − 1.96 + ≤ µ1 − µ 2
8 8
0.15 2 0.15 2
≤ (12.06 − 11.98) + 1.96 +
8 8
−0.067 ≤ µ1 − µ2 ≤ 0.227
(i) If we wanted the confidence interval in part (h) to have a half-width of 0.05 volts
or less, how many additional batteries would have to be tested?
113

114. Example 7.2 Let X1 denote the GPA of a female engineering student at the Philadelphia
University, and let X2 denote the GPA of a male engineering student at the Philadelphia
University. It is common knowledge that σ1 = σ2 = 0.5. A female student has made the
dangerous claim that females are better engineering students than males, and she wishes
to support this claim with GPA data on 8 female and 12 male engineering students. The
results of this data collection are X 1 = 3.44 and X 2 = 3.27 .
(a) Formulate an appropriate hypothesis test.
H0: µ1 − µ2 = 0
H1: µ1 − µ2 > 0
(b) What assumption(s), if any, do we need to make to proceed with this hypothesis
test?
GPA is normally distributed. The two populations are independent.
(c) Does the data support the female student’s claim? Demonstrate that all three
hypothesis testing approaches yield the same result. Use a level of significance of
0.10.
Critical Region
Reject H0 if Z0 > Zα = 1.282
P-Value (smallest level of significance)
Confidence Interval
90% lower-confidence interval
µ1 − µ 2 ≥ −0.1226
114

115. 7.2 Inference on the Means of Two Populations, Variances
Unknown
If the population variances are unknown but the sample sizes exceed 30, then we
can use the two-sample Z-test (lecture 20), replacing σ1 with S1 and σ2 with S2.
Otherwise, we must use one of three two-sample t-tests.
7.2.1 Variances Unknown but Equal
Assumptions
• the two populations are independent
• σ1 and σ2 are unknown, but σ = σ1 = σ2
• the populations are normally distributed
Parameter of Interest
µ1 − µ2
Point Estimate
X1 − X 2
Potential Tests
H0: µ1 − µ2 = ∆0 H0: µ1 − µ2 = ∆0 H0: µ1 − µ2 = ∆0
H1: µ1 − µ2 ≠ ∆0 H1: µ1 − µ2 > ∆0 H1: µ1 − µ2 < ∆0
Test Statistic
2
The pooled estimator of σ2, denoted by S p is defined by
115

116. Critical Region
ν = n1 + n2 − 2
H1: µ1 − µ2 ≠ ∆0 Reject H0 if T0 > tα/2,ν or if T0 < −tα/2,ν
H1: µ1 − µ2 > ∆0 Reject H0 if T0 > tα,ν
H1: µ1 − µ2 < ∆0 Reject H0 if T0 < −tα,ν
P-Values
H1: µ1 − µ2 ≠ ∆0 P = 2 P ( t ν > T0 )
H1: µ1 − µ2 > ∆0 P = P(tν > T0)
H1: µ1 − µ2 < ∆0 P = P(tν < T0)
Using OC Curves for Computing β and Sample Size
∆ = µ1 − µ2
n = n1 = n2
∆ − ∆0
d=
2σ
sample sizes on the OC curves are actually 2n − 1
two-sided test use Charts VII(e) and VII(f)
one-sided test use Charts VII(g) and VII(h)
n* + 1
n=
2
Confidence Intervals
two-sided
 1 
P ( X 1 − X 2 ) − t α / 2,ν S p ≤ µ 1 − µ 2 ≤ ( X 1 − X 2 ) + t α / 2,ν S p
1 1 1  = 1−α
 + +
 n1 n 2 n1 n 2 

upper
 1 1 
P µ1 − µ 2 ≤ ( X 1 − X 2 ) + tα ,ν S p
 +  = 1−α
 n1 n2 

lower
116

117.  1 
P µ 1 − µ 2 ≥ ( X 1 − X 2 ) − t α ,ν S p
1  = 1−α
 +
 n1 n 2 

Example 7.3 A bridge construction firm purchases their shear pins from an
outside supplier. The key characteristic of these pins is their tensile strength. Currently,
the firm purchases their pins from Supplier 1. Recently, a sales representative from
Supplier 2 visited the firm. Supplier 2 claimed that, although their pins have the same
standard deviation as Supplier 1, their mean tensile strength is larger. As a result of this
visit, the construction firm has asked you to test this claim using 15 of each supplier’s
pins and a level of significance of 0.01.
(a) State the null and alternative hypotheses.
(b) State any assumptions that are required to perform this test.
Tensile strength is normally distributed.
(c) Identify the test statistic and the critical region.
The results of the test were:
X 1 = 22351.5 psi S1 = 98.8 psi
X 2 = 22461.5 psi S2 = 103.4 psi
(d) Estimate the two suppliers’ standard deviation.
14( 98.8) + 14(103.4)
2 2
S =
2
p = 10226.5
28
117

118. Sp = 101.1 psi
(e) State the conclusion resulting from the test and find the P-value for the test.
T0 =
( 22351.5 − 22461.5) = −2.9797
1 1
101.1 +
15 15
(f) What is the power of the test if Supplier 2’s mean tensile strength is 100 psi more
than Supplier 1?
(g) If we wish to detect a superiority (for Supplier 2) of 150 psi with probability 0.90,
then what sample size should we have used?
(h) Construct a 99% confidence interval that is relevant to this problem.
7.2.2 Variances Unknown and Unequal
Assumptions
118

119. • the two populations are independent
• σ1 and σ2 are unknown and not believed to be equal
• the populations are normally distributed
Parameter of Interest
µ1 − µ2
Point Estimate
X1 − X 2
Potential Tests
Test Statistic
T0 =
(X 1 − X 2 ) − ∆0
S 12 S 2
2
+
n1 n 2
When the null hypothesis is true, this statistic has approximately a t
distribution. Therefore, all the methods in this section are approximate.
Critical Region
2
 S 12 S 2 
2
 + 
n n2 
 1 
ν= −2 (round down)
(S n1
1
2
S2 n
+ 2 2
) 2
( ) 2
n1 + 1 n2 + 1
H1: µ1 − µ2 ≠ ∆0 Reject H0 if T0 > tα/2,ν or if T0 < −tα/2,ν
H1: µ1 − µ2 > ∆0 Reject H0 if T0 > tα,ν
H1: µ1 − µ2 < ∆0 Reject H0 if T0 < −tα,ν
P-Values
H1: µ1 − µ2 ≠ ∆0 P = 2 P ( t ν > T0 )
H1: µ1 − µ2 > ∆0 P = P(tν > T0)
H1: µ1 − µ2 < ∆0 P = P(tν < T0)
119

120. Confidence Intervals
two-sided
 S 12 S 2
2
S 12 S 2 
2
P ( X 1 − X 2 ) − t α / 2,ν + ≤ µ 1 − µ 2 ≤ ( X 1 − X 2 ) + t α / 2,ν +  = 1−α
 n1 n 2 n1 n 2 
 
upper
 S2 S2 
P µ1 − µ 2 ≤ ( X 1 − X 2 ) + tα ,ν 1 + 2  = 1 − α
 n1 n2 
 
lower
 S 12 S 22 
P µ 1 − µ 2 ≥ ( X 1 − X 2 ) − t α ,ν +  = 1−α
 n1 n 2 
 
Example 7.4 The bridge construction firm purchases another type of shear pin from an
outside supplier. Currently, the firm purchases their pins from Supplier A. Recently, a
sales representative from Supplier B visited the firm. Supplier B claimed that their mean
tensile strength is larger. As a result of this visit, the construction firm has asked you to
test this claim using 23 of each supplier’s pins and a level of significance of 0.05.
(a) State the null and alternative hypotheses.
H0: µA − µB = 0
H1: µA − µB < 0
(b) Identify the test statistic and the critical region.
Reject H0 if T0 < −t0.05,ν
The results of the test were:
X A = 5263.8 psi SA = 37.3 psi
X B = 5512.2 psi SB = 52.8 psi
120

121. (c) State the conclusion resulting from the test and find the P-value for the test.
T0 =
( 5263.8 − 5512.2) = −18.43
37.3 2 52.8 2
+
23 23
2
 37.3 2 52.8 2 

 23 + 23  
 
ν= − 2 = 41.2 = 41
(37.3 2
23 ) 2
+
(52.8 2
23 ) 2
23 + 1 23 + 1
(d) Construct a 95% confidence interval that is relevant to this problem.
7.3 Pairs of Observations, the Paired t- Test
A special case of the two-sample t-tests occurs when the observations of the two
populations of interest are collected in pairs.
An automobile manufacturer wishes to compare the mean tread wear for front
axle tires as compared to rear axle tires. Tires are installed on 10 cars and they are driven
under defined experimental conditions. The resulting observations would be:
(X11, X21) (total wear on front tires of car 1, total wear on rear tires of car 1)
(X12, X22) (total wear on front tires of car 2, total wear on rear tires of car 2)
We only use this approach when we believe that pairing the observations results
in a large positive correlation between the observations within each pair. If we don’t
believe this to be true, then we are better off using the tests from the previous two
sections.
7.3.1 The Paired t-Test
Let (X11, X21), (X12, X22), … , (X1n, X2n) be a set of n paired observations. Define
the differences between each pair of observations as
Dj = X1j − X2j j = 1, 2, … n.
121

122. Having computed the difference for each pair, we then compute the average of the
differences, D , and the standard deviation of the differences SD.
Assumptions
• the Dj’s are iid normal random variables
Parameter of Interest
µD = µ1 − µ2
Point Estimate
D
Potential Tests
H0: µ1 − µ2 = ∆0 H0: µ1 − µ2 = ∆0 H0: µ1 − µ2 = ∆0
H1: µ1 − µ2 ≠ ∆0 H1: µ1 − µ2 > ∆0 H1: µ1 − µ2 < ∆0
Test Statistic
D − ∆0
T0 =
SD
n
Critical Region
H1: µ1 − µ2 ≠ ∆0 Reject H0 if T0 > tα/2,n-1 or if T0 < −tα/2,n-1
H1: µ1 − µ2 > ∆0 Reject H0 if T0 > tα,n-1
H1: µ1 − µ2 < ∆0 Reject H0 if T0 < −tα,n-1
P-Values
H1: µ1 − µ2 ≠ ∆0 P = 2 P ( t n −1 > T0 )
H1: µ1 − µ2 > ∆0 P = P(tn-1 > T0)
H1: µ1 − µ2 < ∆0 P = P(tn-1 < T0)
Confidence Intervals
two-sided
122

123.  S S 
P D − t α / 2, n −1 D ≤ µ 1 − µ 2 ≤ D + t α / 2, n −1 D  = 1 − α
 
 n n
upper
 S 
P µ 1 − µ 2 ≤ D + t α , n −1 D  = 1 − α
 
 n
lower
 S 
P µ 1 − µ 2 ≥ D − t α , n −1 D  = 1 − α
 
 n
Example 7.5 A consumer ratings magazine has developed a test to compare the
mean life length of two brands of batteries. Five electronic toys were used to make the
comparison. Two of each type of toy were purchased, and one toy of each type was
powered by each type of battery (Type 1 and Type 2). The resulting durations of toy
usage are given below.
Toy Battery Type 1 Battery Type 2
1 52.6 hr 61.4 hr
2 103.4 hr 112.8 hr
3 68.2 hr 67.1 hr
4 88.4 hr 92.3 hr
5 111.6 hr 121.5 hr
(a) State the null and alternative hypotheses.
H0: µ1 − µ2 = 0
H1: µ1 − µ2 ≠ 0
(b) State any assumptions that you are making.
Dj = X1j − X2j, j = 1, 2, … , 5
D1, D2, … , D5 are iid normal random variables.
(c) Identify the test statistic and the critical region (use α = 0.05).
123

124. Reject H0 if T0 > tα/2,n-1 = 2.776 or if T0 < −tα/2,n-1 = −2.776
(d) State the conclusion resulting from the test and find the P-value for the test.
Reject H0 and conclude that µ1 − µ2 ≠ 0.
(e) Construct a 95% confidence interval that is relevant to this problem.
124

125. 7.4 Inference on the Ratio of Variances of Two Normal
Populations
7.4.1 The F Distribution
One of the most useful distributions in statistics is the F distribution. Let W and Y
be two independent random each one with u and v number of degrees of freedom
respectively. Then the ratio of two
W
F= u
Y
v
has the probability density function
u
 u + v  u  2 ( u 2 )
−1
Γ   x
f ( x) =  2  v  , 0<x < ∞
( u +v) / 2
 u   v   u  
Γ   Γ     x + 1
 2   2   v  
Then, the ratio is said to follow the F distribution with u degrees of freedom in
the numerator and v number of degrees of freedom in the denominator.
Theorem 7.1 The F Distribution
Consider two independent normal populations having standard deviations σ1 and σ2
respectively. Suppose a random sample of size n1 is taken from the first population, and
let S1 denote the sample standard deviation. Suppose a random sample of size n2 is taken
from the second population, and let S2 denote the sample standard deviation. Then:
S 12 σ 12
F=
S 22 σ 2
2
has an F distribution with n1 − 1 numerator degrees of freedom and n2 − 1 denominator
degrees of freedom.
Example 7.6 Suppose W has an F distribution with 8 numerator degrees of freedom and
13 denominator degrees of freedom.
(a) P(W > 1.49) =
Note that Excel and the tables give you the P(W > x)
125

126. (b) P(W < 2.77) =
(c) P(W > 0) = 1
(e) Find w such that P(W > w) = 0.025
(f) Find w such that P(W < w) = 0.99 ⇒ P(W >w) = 1-0.99 = 0.01
1 1
w = f0.01,8,13 = FINV(0.01, 8, 13) = 4.30208 = =
FINV(0.99,13,8) 0.232447
Note that
1
f 1−α , n1 −1, n2 −1 =
f α , n2 −1, n1 −1
7.4.2 Hypothesis Testing on the Ratio of Two Variances
Suppose we wish to compare the standard deviations of two populations.
Assumptions
• the two populations are independent
• the populations are normally distributed
Parameter of Interest
Point Estimate
Potential Tests
H0: σ 12 = σ 2
2
H0: σ 12 = σ 2
2
H0: σ 12 = σ 2
2
H1: σ 12 ≠ σ 2
2
H1: σ 12 > σ 2
2
H1: σ 12 < σ 2
2
126

127. Test Statistic
S 12
F0 = 2
S2
• If the null hypothesis is true, then the test statistics has an F distribution.
Performing the Test
Critical Region
H1: σ 12 ≠ σ 2 Reject H0 if F0 > f α / 2, n1 −1, n2 −1 or if F0 < f 1−α / 2, n1 −1,n2 −1
2
H1: σ 12 > σ 2 Reject H0 if F0 > f α , n1 −1, n2 −1
2
H1: σ 12 < σ 2
2
Reject H0 if F0 < f 1−α , n1 −1,n2 −1
P-Values
H1: σ 12 ≠ σ 2
2
{ ( ) (
P = 2 min P f n1 −1, n2 −1 > F0 , P f n1 −1,n2 −1 < F0 )}
H1: σ 12 > σ 2
2
(
P = P f n1 −1, n2 −1 > F0 )
H1: σ 12 < σ 2
2
P = P( f n1 −1, n2 −1 <F )
0
Using OC Curves for Computing β and Sample Size
σ1
λ= n = n1 = n2
σ2
two-sided test use Charts VII (o) and VII (p)
one-sided test use Charts VII (q) and VII (r)
σ2
Use λ = for H1: σ 12 < σ 2 .
2
σ1
Confidence Intervals
S2 σ 2 S2 
two-sided P 12 f 1−α / 2, n2 −1, n1 −1 ≤ 12 ≤ 12 f α / 2, n2 −1,n1 −1  = 1 − α
S 
 2 σ 2 S2 
σ 2 S 2 
upper P 12 ≤ 12 f α , n2 −1, n1 −1  = 1 − α
σ 
 2 S2 
σ 2 S 2 
lower P 12 ≥ 12 f 1−α , n2 −1, n1 −1  = 1 − α
σ 
 2 S2 
127

128. Reject H0 if 1 is not in the confidence interval.
Example 7.7 A company which manufactures aircraft is studying the purchase of the
glass used to make the windshields for one of its aircraft. Two suppliers, Supplier 1 and
Supplier 2, are competing for this business. The characteristic of the glass that is of
particular interest is its thickness. It has already been determined that both suppliers have
the same mean thickness. However, Supplier 1 claims to have superior performance with
respect to variability. The manufacturer wishes to test this claim using a sample of size
10 (for each supplier) and a level of significance of 0.01.
(a) State the null and alternative hypothesis, the test statistic, and the critical region.
S 12
F0 = 2
S2
(b) State any required assumptions.
Glass thickness is normally distributed, and the two suppliers are independent.
(c) Suppose S1 = 0.023 in and S2 = 0.028 in. State and interpret the conclusion of this
test.
(d) What is the P-value for this test?
Because of the nature of the F distribution tables, I do not recommend using P-
values to perform this hypothesis test.
128

129. (e) Construct a confidence interval on σ1 / σ2 that is relevant to this problem.
99% upper confidence interval
(f) What is the power of this test if σ1 = 0.25σ2?
(g) What sample size should we have used if we want the Type II error probability to
be 0.1 when σ1 = 0.2σ2?
Example 7.8 Let X1 denote the GPA of a female engineering student at the
Philadelphia University, and let X2 denote the GPA of a male engineering student at the
Philadelphia University. Consider the following hypothesis test.
129

130. Suppose we wish to perform this test using a level of significance of 0.05.
(a) If we want the power of the test to be 0.9 when σ1 = 3.2σ2, then what sample size
should we use?
(b) If this sample size is used, what is the probability of a Type II error when σ1 =
2.4σ2?
(c) If this sample size is used, what is the probability of a Type I error?
α = 0.05
(d) Identify the test statistic and the critical region for this test.
(e) Suppose the test is implemented and S1 = 0.5315 and S2 = 0.5476. State and
interpret the conclusion resulting from this test.
(f) Verify that a confidence interval would have lead to the same conclusion.
95% confidence interval
130

131. 7.5 Performing the Test on Two Population Proportions
Assumptions
• the two populations are independent and large
Note that all procedures are approximate based on the normal approximation to the
binomial.
Parameter of Interest
p1 − p2
Point Estimate
p1 − p 2
ˆ ˆ
Potential Tests
H0: p1 = p2 H0: p1 = p2 H0: p1 = p2
H1: p1 ≠ p2 H1: p1 > p2 H1: p1 < p2
Test Statistic
p1 − p 2
ˆ ˆ
Z0 =
1 1 
p (1 − p )  + 
ˆ ˆ 
 n1 n 2 
X1 + X 2
where p =
ˆ
n1 + n 2
Critical Region
H1: p1 ≠ p2 Reject H0 if Z0 > Zα/2 or if Z0 < −Zα/2
H1: p1 > p2 Reject H0 if Z0 > Zα
H1: p1 < p2 Reject H0 if Z0 < −Zα
P-Values
H1: p1 ≠ p2 P = 2 P( Z > Z 0 )
H1: p1 > p2 P = P(Z > Z0)
H1: p1 < p2 P = P(Z < Z0)
OC Function
q1 = 1 − p1
q2 = 1 − p2
n p + n2 p 2 n1q1 + n2 q 2
p= 1 1 q=
n1 + n 2 n1 + n2
131

132. p1q1 p 2 q 2
σ= +
n1 n2
H1: p1 ≠ p2
 − Z α / 2 pq (1 n1 + 1 n 2 ) − ( p1 − p 2 ) Z α / 2 pq (1 n1 + 1 n 2 ) − ( p1 − p 2 ) 
β = P ≤Z≤ 
 σ σ 
 
 Z pq (1 n1 + 1 n 2 ) − ( p1 − p 2 ) 
H1: p1 > p2 β = P Z ≤ α 
 σ 
 
 − Zα pq (1 n1 + 1 n 2 ) − ( p1 − p 2 ) 
H1: p1 < p2 β = P Z ≥ 
 σ 
 
Choice of Sample Size
n = n1 = n2
two-sided test n=
(Z α/2 ( p1 + p 2 )( q1 + q 2 ) / 2 + Z β p1 q1 + p 2 q 2 )
2
( p1 − p 2 ) 2
one-sided test n=
(Z α ( p1 + p 2 )( q1 + q 2 ) / 2 + Z β p1 q1 + p 2 q 2 )
2
( p1 − p 2 ) 2
Confidence Intervals
q1 = 1 − p1
ˆ ˆ
q2 = 1 − p2
ˆ ˆ
ˆ ˆ ˆ ˆ
p1q1 p 2 q 2
σ=
ˆ +
n1 n2
two-sided P( p1 − p 2 − Z α / 2σ ≤ p1 − p 2 ≤ p1 − p 2 + Z α / 2σ ) = 1 − α
ˆ ˆ ˆ ˆ ˆ ˆ
upper P( p1 − p 2 ≤ p1 − p 2 + Z α σ ) = 1 − α
ˆ ˆ ˆ
lower P( p1 − p 2 ≥ p1 − p 2 − Z α σ ) = 1 − α
ˆ ˆ ˆ
Reject H0 if 0 is not in the confidence interval.
Example 7.9 The California legislature is considering a new bill which will affect the
citizens of Los Angeles and San Francisco. Newspaper reports have indicated that 80%
132

133. of the citizens of these cities oppose the provisions of the bill. The legislature has hired
an independent firm to determine if, in fact, the proportion is the same for citizens of both
cities. Other sources have indicated that the proportion is actually larger for citizens of
Los Angeles. The firm intends to use a level of significance of 0.05.
(a) State the null and alternative hypotheses, the test statistic and the critical region.
p1 = proportion of Los Angeles citizens who oppose the bill
p2 = proportion of San Francisco citizens who oppose the bill
(b) If p1 = 0.85 and p2 = 0.75, the firm wishes to have a Type II error probability of
0.10. How many citizens of each city should they survey?
(c) Suppose the sample size identified in part (b) is used and the results are
X1 = 228 and X2 = 211. Evaluate the results of the test using a P-value.
133

134. .
(d) What is the power of this test if p1 = 0.82 and p2 = 0.78?
(e) Construct a confidence interval that is relevant to this problem.
134