J.Opet- Senior Composition

The Central Limit Theorem andthe Theory of Hypothesis Testing,

with an Application to RentalIncome from Two Populations

A Senior Comprehensive Projectby

Johnathon M. OpetAllegheny College

Meadville, PA

December 8, 2015

Submitted to the Department of Mathematics in partial fulfillment ofthe requirements for the degree of Bachelor of Science.

Project Advisor: Dr. Anthony Lo BelloSecond Reader: Dr. Rachel Weir

I hereby recognize and pledge to fulfill my responsibilities, as definedin the Honor Code, and to maintain the integrity of both myself and

the College community as a whole.

Pledge:

—————————————————————–Johnathon M. Opet

Acknowledgements

I want to thank my parents, Dee and John Opet, for all the hard work they

have done for me and making this opportunity a reality. You two have taught

me that things are never handed but are deserved through hard work. You

have shown me how to become a man and I could never thank you both enough

for what you have taught me and what you have given me. I love you both,

thank you for giving me the world.

Abstract

This senior project presents the proof of the Central Limit Theorem

and a discussion of the theory of Hypothesis Testing, and concludes with

some applications to rental incomes of college and residential houses in

Northeast Ohio and Northwest and Southwest Pennsylvania.

Contents

1 Preliminary Definitions and Theorems 1

2 The Central Limit Theorem 9

3 The Theory of Hypothesis Testing 20

4 Application: Rental Income from Two

Populations 25

1 Preliminary Definitions and Theorems

Definition 1.1. A Sample Space, X, is a nonempty set of points, called

outcomes.

Definition 1.2. A Sigma Field of Events, A, is a non-empty collection of

subsets of X satisfying two conditions:

1. If A ∈ A then A′ ∈ A

2. If A1, A2, . . . , An, · · · ∈ A then∞⋃i=1

Ai ∈ A.

The elements of A are called events.

Theorem 1. If A is a sigma field of events of a sample space, X, then X ∈ A.

Proof. Since A is nonempty, there exists an event A ∈ A. Then we know that

A′ ∈ A. Therefore A ∪ A′ ∈ A, but A ∪ A′ = X, so X ∈ A.

Corollary 2. If A is a sigma field of events of a sample space, X, then ∅ ∈ A

Proof. Since X ∈ A, X′ ∈ A. But X′ = ∅, so ∅ ∈ A.

Definition 1.3. Probability, P, is a function with domain A and range in

the real numbers that satisfies the following conditions:

1. P (A) ≥ 0 for all events A.

2. P (X) = 1 where X is the sample space.

3. If A1, A2, ... is a mutually disjoint sequence of events, then

P

(∞⋃i=1

Ai

)=∞∑i=1

P (Ai)

.

1

Definition 1.4. A random variable X is a function from the sample space X

to the set of real numbers.

Definition 1.5. A Discrete Random Variable X is a random variable

whose range is finite or countably infinite. A Continuous Random Variable

X is a random variable whose range is an interval of length greater than 0.

Definition 1.6 ([3]). If X is a discrete random variable, the function given

by f(x) = P (X = x) for each x with the range of X is called the probability

function of X.

Definition 1.7 ([3]). A function, f(x), from the set of real numbers to the set

of real numbers is called a probability density function of the continuous

random variable X if and only if

P (a ≤ X ≤ b) =

∫ b

a

f(x)dx

for any real constants a and b with a ≤ b.

Definition 1.8 ([3]). The rth Moment about the origin of a random

variable X is

µ′r = E(Xr) =∑x

xr · f(x), r = 0, 1, 2, . . . ,

when X is discrete, and

µ′r = E(Xr) =

∫ ∞−∞

xr · f(x)dx, r = 0, 1, 2, . . . ,

2

when X is continuous.

Definition 1.9 ([3]). µ′1 is the mean or expected value of the random

variable X and is denoted by µ or E(X).

Definition 1.10. The rth moment about the mean of the random variable

X is

µr = E[(X− µ)r] =∑x

(x− µ)r · f(x), r = 0, 1, 2, . . . ,

when X is discrete, and

µr = E[(X− µ)r] =

∫ ∞−∞

(x− µ)r · f(x)dx, r = 0, 1, 2, . . . ,

when X is continuous.

Definition 1.11. µ2 is the variance of the random variable X and is denoted

by σ2 or var(X). The positive square root, σ, of the variance is the standard

deviation.

Theorem 3.

σ2 = µ′2 − µ2

for any random variable X with mean µ and variance σ2.

3

Proof.

σ2 =E[(X− µ)2

]=E(X2 − 2µx+ µ2)

=E(X2)− 2µE(X) + E(µ2)

=E(X2)− 2µ · µ+ µ2

=µ′2 − 2µ2 + µ2

=µ′2 − µ2.

Definition 1.12. If X1,X2, . . . ,Xn are discrete random variables, the func-

tion given by f(x1, x2, . . . , xn) = P (X1 = x1,X2 = x2, . . . ,Xn = xn) for each

point (x1, x2, . . . , xn), where xi is in the range of Xi, is the joint distribution

of X1,X2, . . . ,Xn.

Definition 1.13. If X1,X2, . . . ,Xn are continuous random variables, the

function F (x1, x2, . . . , xn), defined by F (x1, x2, . . . , xn) = P (X1 < x1, . . . ,Xn <

xn), is the joint cumulative distribution function of X1,X2, . . . ,Xn and

the function f(x1, . . . , xn) = ∂nF∂xn···∂x2∂x1 is the joint probability density function

of X1,X2, . . . ,Xn.

Definition 1.14 ([3]). If f(x1, x2, . . . , xn) is the value of the joint probability

distribution of the n random variables X1,X2, . . . ,Xn at (x1, x2, . . . , xn), and

fi(xi) is the value of the marginal distribution of Xi at xi for i = 1, 2, . . . , n,

4

then the n random variables are independent if and only if

f(x1, x2, . . . , xn) = f1(x1) · f2(x2) · · · fn(xn)

for all (x1, x2, . . . , xn) where xi is within the range of Xi.

Theorem 4 (Linear Property of Expectations).

E(aX + bY) = aE(X) + bE(Y)

Proof. We give the proof for the continuous case. Suppose the random vari-

ables X and Y have probability density functions g(x) and h(y) respectively,

and let f(x, y) be the joint probability density function of X and Y. Then

E(aX + bY) =

∫ ∞−∞

∫ ∞−∞

(ax+ by)f(x, y)dydx

=a

∫ ∞−∞

∫ ∞−∞

xf(x, y)dydx+ b

∫ ∞−∞

∫ ∞−∞

yf(x, y)dxdy

=a

∫ ∞−∞

x

(∫ ∞−∞

f(x, y)dy

)dx+ b

∫ ∞−∞

y

(∫ ∞−∞

f(x, y)dx

)dy

=a

∫ ∞−∞

xg(x)dx+ b

∫ ∞−∞

yh(y)dy

=aE(X) + bE(Y).

Definition 1.15. If X and Y are discrete random variables with joint prob-

5

ability function f(x, y), then the covariance of X and Y is

cov(X,Y) =∑y

∑x

(x− E(X))(y − E(Y))f(x, y).

If X and Y are continuous random variables with joint probability density

function f(x, y), then

cov(X,Y) =

∫ ∞−∞

∫ ∞−∞

(x− E(X))(y − E(Y))f(x, y)dydx.

Theorem 5.

V ar(aX + bY) = a2σ2X + 2ab cov(X,Y) + b2σ2

Y

Proof. We give the proof for the continuous case. Suppose the random vari-

ables X and Y have probability density functions g(x) and h(y) respectively,

and let f(x, y) be the joint probability density function of X and Y. Then

V ar(aX + bY) =

∫ ∞−∞

∫ ∞−∞

[(ax+ by)− (aE(X) + bE(Y))]2f(x, y)dydx

=

∫ ∞−∞

∫ ∞−∞

[ax− aE(X) + by − bE(Y)]2f(x, y)dydx

=

∫ ∞−∞

∫ ∞−∞

[a(x− E(X)) + b(y − E(Y))]2f(x, y)dydx

=a2∫ ∞−∞

∫ ∞−∞

(x− E(X))2f(x, y)dydx

+ 2ab

∫ ∞−∞

∫ ∞−∞

(x− E(X))(y − E(Y))f(x, y)dydx

+ b2∫ ∞−∞

∫ ∞−∞

(y − E(Y))2f(x, y)dydx

6

=a2∫ ∞−∞

(x− E(X))2[ ∫ ∞−∞

f(x, y)dy

]dx+ 2abcov(X,Y)

+ b2∫ ∞−∞

(y − E(Y))2[ ∫ ∞−∞

f(x, y)dx

]dy

=a2∫ ∞−∞

(x− E(X))2g(x)dx+ 2abcov(X,Y)

+ b2∫ ∞−∞

(y − E(Y))2h(y)dy

=a2var(X) + 2ab cov(X,Y) + b2var(Y)

Theorem 6. If X and Y are independent random variables, then cov(X,Y) =

0.

Proof. For the continuous case, if f(x, y), g(x), and h(y) are the same functions

as in the previous theorem, then

cov(X,Y) =

∫ ∞−∞

∫ ∞−∞

(x− E(X))(y − E(Y))f(x, y)dydx

=

∫ ∞−∞

∫ ∞−∞

(x− E(X))(y − E(Y))g(x)h(y)dydx

=

∫ ∞−∞

(x− E(X))g(x)dx

∫ ∞−∞

(y − E(Y))h(y)dy

=0 · 0 = 0.

Theorem 7. If X1,X2, . . .Xn is a sequence of independent random variables

with common mean, µ, and common variance, σ2, and if X = X1+X2+···+Xn

n,

7

then E(X) = µ and var(X) = σ2

n.

Proof. This theorem follows from Theorems 4, 5, and 6.

Definition 1.16. [1] X has Normal Distribution with parameters µ and σ

if it has probability density function given by

n(x;µ, σ) =1

σ√

2πe

12(x−µσ

)2

where −∞ < x <∞ and σ > 0.

Theorem 8.

ln(1 + x) = x− x2

2+x3

3− x4

4+− · · ·

for |x| < 1.

Proof. If |x| < 1, then 11+x

= 1− x+ x2 − x3 +− · · · .

Therefore, for |x| < 1

ln(1 + x) =

∫ x

0

1

1 + tdt

=

∫ x

0

(1− t+ t2 − t3 + · · · )dt

=x− x2

2+x3

3− x4

4+− · · · .

8

2 The Central Limit Theorem

Before we introduce the Central Limit Theorem, we must first understand

the background mathematics behind the proof of the theorem. First, we will

introduce the moment generating function.

Definition 2.1 ([3]). The moment generating function of a random vari-

able X, where it exists, is given by

MX(t) = E(etX) =∑x

etx · f(x)

when X is discrete and f is its probability function and

MX(t) = E(etX) =∫∞−∞ e

tx · f(x)dx

when X is continuous and f is its probability density function.

In addition, we will also need to introduce a property of moment generating

functions.

Theorem 9.

MX(t) = 1 + µt+ µ′2t2

2!+ µ′3

t3

3!+ · · · (1)

Proof. Let X be a continuous random variable with probability density func-

tion f(x) and let Y = eXt. Then

MX(t) = E(Y) =

∫ ∞−∞

yf(x)dx

=

∫ ∞−∞

extf(x)dx

9

Since we know the expansion of the Maclaurin Series for eu is eu =∑∞

n=0un

n!,

substituting u = xt we get

E(Y) =

∫ ∞−∞

∞∑n=0

(xt)n

n!f(x)dx

=∞∑n=0

tn

n!

∫ ∞−∞

xnf(x)dx.

Since we know∫∞−∞ x

nf(x)dx = µ′n we get

E(Y) =∞∑n=0

µ′ntn

n!.

Theorem 10. If a and b are constants; then

MaX+b(t) = ebtMX(at). (2)

Proof. We will give the proof for the case when X is a continuous random

variable. Let a and b be constants, then

10

MaX+b(t) =

∫ ∞−∞

e(ax+b)tf(x)dx

=

∫ ∞−∞

eaxtebtf(x)dx

= ebt∫ ∞−∞

exatf(x)dx

= ebtMX(at).

For example, if we wanted to find the moment generating function of the

random variable Y = 2X + 8, we could use this theorem to get

M2X+8 = e8tMX(2t).

Another tool we will need in order to prove the Central Limit Theorem is:

Theorem 11. If X1,X2, . . . , and Xn are independent random variables and

Y = X1 + X2 + · · ·+ Xn, then

MY(t) =n∏i=1

MXi(t) (3)

where MXi(t) is the value of the moment-generating function of Xi at t.

Proof. For the discrete case, let X1,X2, . . . , and Xn be independent random

variables and Y = X1 + X2 + · · ·+ Xn. Then we know

11

f(x1, x2, . . . , xn) = f1(x1) · f2(x2) · ... · fn(xn).

Then we can write

MY(t) = E(eYt)

= E[e(X1+X2+···+Xn)t]

=∑x1

· · ·∑xn

e(x1+x2+···+xn)tf(x1, x2, . . . , xn)

=∑x1

ex1tf1(x1) ·∑x2

ex2tf2(x2) · · ·∑xn

exntfn(xn)

=n∏i=1

MXi(t)

This theorem states that the moment-generating function of the sum of

n independent random variables is the product of their moment-generating

functions. As in the case of Theorem 10, this theorem allows us to manipulate

the moment generating function into a more simple form to work with.

Theorem 12. If two random variables have the same moment generating func-

tion, then they have the same probability distribution.

Proof. Omitted

12

Theorem 13. If X1,X2, . . . , are random variables such that

limn→∞

MXn(t) = MX(t)

then

limn→∞

P (Xn < a) = P (X < a)

for all real numbers a.

Proof. Omitted

We now proceed to discuss how we estimate an unknown population vari-

ance σ2 when we need such an estimate to do a hypothesis test.

Definition 2.2 ([3]). Let X1,X2, . . . ,Xn be a random sample from a distri-

bution having unknown finite variance σ2. Then

S2 =n∑i=1

(Xi −X)2

n− 1(4)

is called the sample variance of X1,X2, . . . ,Xn.

Theorem 14 ([3]). E(S2) = σ2

13

Proof. Using Equation (4) to calculate the expected value, we get

E(S2) =E

[ n∑i=1

(Xi −X)2

n− 1

]=

1

n− 1E

[ n∑i=1

(Xi −X)2]

=1

n− 1E

[ n∑i=1

(Xi2 − 2XiX + X

2)

]=

1

n− 1E

[ n∑i=1

Xi2 − 2X

n∑i=1

Xi + nX2]

=1

n− 1E

[ n∑i=1

Xi2 − 2nX

2+ nX

2]

=1

n− 1E

[ n∑i=1

Xi2 − nX

2]

=1

n− 1

[ n∑i=1

E(Xi2)− nE(X

2)

]

Since E(X2i ) = µ′2, and µ′2 = σ2 + µ2, and since E(X

2) = σ2

n+ µ2, we get

E

[ n∑i=1

(Xi −X)2

n− 1

]=

1

n− 1

[ n∑i=1

E(Xi2)− nE(X

2)

]=

1

n− 1[n(σ2 + µ2)− n(

σ2

n+ µ2)]

=1

n− 1[nσ2 + nµ2 − σ2 − nµ2)]

=1

n− 1[(n− 1)σ2]

=σ2.

14

Because of this theorem, we call S2 an unbiased estimator of σ2. Its

value s2 is called the estimate of σ2. We use the value of this unbiased

estimator, S2, to estimate σ2 from data when σ2 is unknown.

Theorem 15. The moment generating function of a random variable X with

normal distribution with parameters µ and σ is eµt+t2σ2

2 .

Proof.

MX(t) =

∫ ∞−∞

1√2πσ

e−12((x−µ)/σ)2extdx

=1√2πσ

∫ ∞−∞

ext−12((x−µ)/σ)2dx.

Looking at the exponent, xt− 12(x−µ

σ)2 we can see that,

xt− 1

2

(x− µσ

)2

=− 1

2σ2[−2xtσ2 + x2 − 2xµ+ µ2]

=− 1

2σ2[x2 − (2µ+ 2tσ2)x+ µ2]

=− 1

2σ2[x2 − (2µ+ 2tσ2)x+ (µ+ tσ2)2 − (µ+ tσ2)2 + µ2]

=− 1

2σ2[x− (µ+ tσ2)]2 +

1

2σ2[(µ+ tσ2)2 − µ2]

=− 1

2

[x− (µ+ tσ2)

σ

]2+

1

2σ2[µ2 + 2µtσ2 + t2σ4 − µ2]

=− 1

2

[x− (µ+ tσ2)

σ

]2+ µt+

t2σ2

2.

15

So,

MX(t) =

∫ ∞−∞

1√2πσ

ext−12((x−µ)/σ)2dx

=eµt+t2σ2

2

∫ ∞−∞

1√2πσ

e−12

[x−(µ+tσ2)

σ

]2dx

=eµt+t2σ2

2 .

If µ = 0 and σ = 1, then X has standard normal distribution and the

moment generating function is

MX(t) =et2

2 .

Theorem 16. If X and Y are independent random variables with normal

distribution with parameters µX, σ2X, and µY, σ2

Y respectively, then aX + bY

has a normal distribution with parameters µ = aµX + bµY and σ2 = a2σ2X +

b2σ2Y.

Proof. This result follows from Theorems 10, 11, 12, and 15.

The Central Limit Theorem is one of the fundamental theorems of probabil-

ity. What the Central Limit Theorem states is that the standardized average

value of a large sample of independent random variables will have approxi-

mately standard normal distribution regardless of any other aspects like the

underlying distribution [3].

16

Theorem 17 ([2], The Central Limit Theorem). Let X1,X2, . . . ,Xn, . . .

be independent and identically distributed random variables with mean µ and

standard deviation σ. Let X be the sample mean with sample size n. Then, as

n approaches infinity, the distribution of

Z =X− µ√σ2/n

, (5)

approaches that of the standard normal distribution.

Proof. If we refer back to Equation (10), we get

MZ(t) = M X−µσ/√n

(t) =M√nσ

X−√nµσ

(t)

=e−√nµt/σ ·MX

(√nt

σ

)=e−

√nµt/σ ·MnX

(t

σ√n

)

Since we know nX = X1 +X2 + · · ·+Xn, and since the Xi’s have the same

distribution, then it follows from Equation (3) that

MZ(t) = e−√nµt/σ ·

[MX

(t

σ√n

)]n

where X is a random variable with the same distribution as the Xi’s.

If we take the natural log of each side we obtain,

17

lnMZ(t) = −√nµt

σ+ n · ln

[MX

(t

σ√n

)]

by the properties of logarithms. If we expand MX

(t

σ√n

)as a power series in

t, we obtain

lnMZ(t) = −√nµt

σ+ n · ln

[1 + µ′1

t

σ√n

+ µ′2t2

2σ2n+ µ′3

t3

6σ3n√n

+ · · ·]

where µ′1, µ′2, µ

′3, . . . , are the moments about the origin of the population

distribution, that is, those of the original random variables Xi.

We can use the infinite series ln(1 + x) = x− 12x2 + 1

3x3−+ · · · for |x| < 1

to get

lnMZ(t) =−√nµt

σ+ n

{[µ′1

t

σ√n

+ µ′2t2

2σ2n+ µ′3

t3

6σ3n√n

+ · · ·]

− 1

2

[µ′1

t

σ√n

+ µ′2t2

2σ2n+ µ′3

t3

6σ3n√n

+ · · ·]2

+1

3

[µ′1

t

σ√n

+ µ′2t2

2σ2n+ µ′3

t3

6σ3n√n

+ · · ·]3 −+ · · ·

}.

Then, collecting the powers of t, we obtain

18

lnMZ(t) =

(−√nµ

σ+

√nµ′1σ

)t+

(µ′22σ2− µ′1

2

2σ2

)t2

+

(µ′3

6σ3√n− µ′1 · µ′2

2σ3√n

+µ′1

3

3σ3√n

)t3 + · · ·

and since µ′1 = µ and σ2 = µ′2 − µ2 so µ′2 = σ2 + µ2, we can reduce this to

lnMz(t) =1

2t2 +

(µ′36− µ′1µ

′2

2+µ′1

3

6

)t3

σ3√n

+ · · ·

Finally, observing that the coefficient of t3 is a constant times 1√n

and in

general, for r ≥ 2, the coefficient of tr is a constant times 1√nr−2

, we get

limn→∞

lnMZ(t) =1

2t2.

Thus

limn→∞

MZ(t) = et2/2,

which is the moment generating function of the random variable with standard

normal distribution.

19

3 The Theory of Hypothesis Testing

Definition 3.1. [3] A statistical hypothesis is an assertion or conjecture

about the distribution of one or more random variables. If a statistical hy-

pothesis completely specifies the distribution, it is referred to as a simple

hypothesis; if not, it is referred to as a composite hypothesis.

In order to be able to construct the correct criteria for testing statistic

hypotheses, we must first form our Null Hypothesis, H0, and our Alterna-

tive hypothesis, H1. The null hypothesis is the claim that will be tested and

assumed true until proven unlikely by the data. In contrast, the alternative

hypothesis is always a claim that denies the null hypothesis. After formulating

the two hypotheses for our test, there are two types of errors that can possibly

occur. The table below represents the relationship of our claim of H0 and the

truth or falsity of the hypothesis formed.

Do Not Reject H0 Reject H0

H0 is True Correct Type I ErrorH1 is True Type II Error Correct

Table 1: [2] Type I & II Error Table

When testing our hypothesis, we must be careful to avoid errors, which are

labeled in red. A Type I error is an error when we reject H0 when it is in

fact true, and a Type II error is an error when we do not reject H0 when it

is false.

20

Definition 3.2. α, the level of significance, is the probability of commit-

ting a Type I error, and β is the probability of committing a Type II error.

Since a Type I error is considered more serious than a Type II error, we

want to reduce that probability by selecting a small value for α. If rejecting the

null hypothesis would cause the closing down of a business, we would select a

very small value for α like .01 or .001 to reduce the risk of committing a Type

I error. If there is no terrible consequence associated with rejecting the null

hypothesis, then we would choose a slightly larger α like .1 or .05. Once we

have picked our α, we are able to generate our critical values for a two-tailed

test.

Definition 3.3. The decision whether to accept or reject the null hypothesis

is made based on the test statistic value, z, of a random variable Z. The

numbers −zα2

and zα2

are the critical values of Z. If −zα2< z < −zα

2, then we

accept the null hypothesis, whereas if zα/2 < z or z < −zα2, then we reject the

null hypothesis.

In our example, we will be concerned with the situation where our decision

regarding H0 depends on the value of z of a random variable Z with standard

normal distribution

z =x− µσ/√n

(6)

where x is the value of the sample mean of size n from a population with mean

µ and variance σ2. The test statistic, z, can be located in the non-rejection

region or in the rejection region. Graphically this is represented below:

21

Looking at the graph, if the test statistic is located in (−∞,−zα/2) or

(zα/2,∞), then it is in the rejection region, which means that we must not

accept H0. If the test statistic is located in the interval (−zα/2, zα/2), then it is

in the non-rejection region, which means that we must accept H0. If the test

statistic is either equal to, or relatively close to the critical values, then it is

up to the experimenter to decide whether to defer judgment.

Example 3.1. Suppose that it is known from experience that the standard

deviation of the weight of 8-ounce packages of cookies made by a certain bakery

is 0.16 ounces. To check whether its production is under control on a given

day, that is, to check whether the true average weight of the packages is 8

ounces, employees select a random sample of 25 packages and find that their

mean weight is x = 8.091 ounces. Is the bakery telling the truth when they

state that their package of cookies are 8-ounces? [3]

Solution: The null hypothesis is that the package of cookies does weigh

8-ounces, in other words, H0 : µ = 8. In contrast, the alternative hypothesis is

that the package of cookies does not weight 8-ounces, in other words, H1 : µ 6=

8. Since the application is associated with a lot of risk (if the null hypothesis

22

is rejected the bakery must stop production), we will choose a low level of

significance namely, α = 0.01.

Now that we have found our value for α, we must find the critical values.

When α = .005 and n = 25, the critical values are z.005 = 2.575 and −z.005 =

−2.575.

After finding the critical values, we must calculate the test statistic with

the given data. Substituting n = 25, σ = 0.16, x = 8.091, and µ = 8 into

Equation (6), we obtain

z =8.091− 8

0.16/√

25= 2.844.

This can be represented graphically below:

Looking at the graph, we know that the rejection region consists of the union

of the two intervals (−∞,−2.575) and (2.575,∞). Since our critical value

is greater than 2.575, the null hypothesis must be rejected. Therefore it is

highly unlikely that the 8-ounce packages of cookies actually weigh 8-ounces.

Therefore, the bakery must shut down production of their packages of cookies

and fix the problem immediately.

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Example 3.2. The owner of Tony’s Tires claims that his manufactured tires

last on average 22,000 miles. We wanted to test whether or not Tony was

telling the truth. Using the data collected, what can we say about his claim?

Solution: We must first produce our null hypothesis and alternative hy-

pothesis. Based on the information given, the null hypothesis H0 : µ = 22, 000

miles and the alternative hypothesis H1 : µ 6= 22, 000 miles. We choose

α = .01. After collecting and analyzing the data, we were able to obtain

the information that n = 100, x = 21, 800 miles, and s = 1, 295. Since we

know n ≥ 30, we can apply the Central Limit Theorem and the z-statistic test

to obtain

z =x− µs/√n

=21, 800− 22, 000

1295/√

100

≈ −1.54.

We know that the test statistic is -1.54. Since α = .01, we want critical

value for z.005 and −z.005. Looking Table IV in [3], we get a value of z.005 =

2.575 and −z.005 = −2.575 [3]. So we know that our rejection region is the

union of the intervals (−∞,−2.575) and (2.575,∞), and our non-rejection

region is (−2.575, 2.575). Since −2.575 < −1.54 < 2.575, we must accept our

null hypothesis, which means that the data in this case do not contradict the

owner’s claim.

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4 Application: Rental Income from Two

Populations

Example 4.1. I was interested in finding whether or not college students living

in privately owned houses and non-college students living in privately owned

houses had the same amount of rental income. In order to obtain the data, I

sent out emails and called multiple landlords around Northeast Ohio and both

Northwest and Southwest Pennsylvania to collect data for each population. I

then formed the statistical question:

“Is the mean rental income obtained from college students living

in privately owned houses the same as the mean rental income ob-

tained from non-college students living in privately owned houses?”

I gathered up a random total of n = 33 family rentals and m = 34 college

rentals for a total of 67 total data samples, which are shown below:

Area Tenant Type Rental Income Duration Yearly Income

NE OH Non-College $1,200.00 Monthly $14,400.00

SW PA Non-College $425.00 Monthly $5,100.00







25



NE OH College $1,688.76 Monthly $20,265.12






NW PA College $935.00 Monthly $11,220.00











NW PA College $2,100.00 Monthly $25,200.00





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SW PA Non-College $2,100.00 Monthly $25,200.00
















27






NE OH College $10,850.00 Yearly $10,850.00



NE OH Non-College $18,360.00 Yearly $18,360.00

NW PA Non-College $13,560.00 Yearly $13,560.00

NW PA Non-College $8,100.00 Yearly $8,100.00

Table 2: Data

After multiple calculations, the summary of my results is below.

Family Rentals College Rentals

x = $12,856.59 y = $19,196.75s2x = 62,023,794.83 s2y = 28,063,536.74σx = 7,785.52 σy = 5,297.50

Table 3: Population Information

I decided to take α = .10 since there are no serious consequences if the null

hypothesis is rejected. Is there enough evidence when α = .10 to conclude that

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the rental income from non-college students is the same as the rental income

from college students when renting from privately owned houses?

Solution: We must first formulate our null hypothesis in order to test for

the equality of means of the two populations. Thus, we want to test the null

hypothesis, which is

Ho : µX = µY

or that the means of both populations are equal, against the alternative hy-

pothesis:

H1 : µX 6= µY

in other words, that the means are not equal.

Since n and m are both greater than 30, we know by Theorem 16 that

Z = X−Y−(µx−µy)√(σ2x/n)+(σ2

y/m)has approximately standard normal distribution. In our

example, when we plug the desired values, assuming that µx = µy, we obtain

a test-statistic value,

z =(12, 856.59− 19, 196.75)− (0)√

(62, 023, 794.83/33) + (28, 063, 536.74/34)= −3.855.

29

When α = .10, we are able to get the critical z, zα/2 = z.05 = 1.645

from Table III [3]. Thus we know that our rejection regions are the intervals

(−∞,−1.645) and (1.645,∞). So looking back at the z = −3.8550, what we

find is that the test-statistic lies in the rejection region; therefore we must

reject H0. This finding implies that the means of the two populations are not

the same.

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Example 4.2. Looking back at Example 4.1, suppose I wanted to compare

the overall rental income between renting in Northeast Ohio and Western

Pennsylvania. I then formed a different statistical question:

“Is the overall rental income obtained from Northeast Ohio the

same as the overall rental income obtained from Western Pennsyl-

vania?”

After the reorganization of the data, I gathered up a random total of n = 32

Northeastern Ohio rentals and m = 35 Western Pennsylvania rentals for a total

of again the 67 total data samples in Example 4.1. After multiple calculations,

the summary of my results is below.

Northeast Ohio Western Pennsylvania

x = $19,453.66 y = $12,984.00s2x = 66,995,576.56 s2y = 105,368,783.63σx = 8,185.08 σy = 10,264.93

Table 4: Application 2: Population Information

I decided again to take α = .10 for the same reasons as in the previous

example. Is there enough evidence when α = .10 to conclude that the rental

income from Northeast Ohio is the same as from Western Pennsylvania?

Solution: Similarly as in the first example, we must first formulate our

null hypothesis in order to test for the equality of means of the two populations.

31

Thus, we want to test the null hypothesis, which is

Ho : µX = µY

or that the means of both populations are equal, against the alternative hy-

pothesis:

H1 : µX 6= µY

in other words, that the means are not equal.

Since n and m are both greater than 30, we know that Z = X−Y−(µx−µy)√(σ2x/n)+(σ2

y/m)

has approximately standard normal distribution. In our example, when we

plug the desired values, assuming that µx = µy, we obtain a test-statistic

value,

z =(19,453.66− 12,984.00)− (0)√

(66,995,576.56/32) + (105,368,783.63/35)= 2.864.

When α = .10, we are able to get the critical z, zα/2 = z.05 = 1.645

32

from Table III [3]. Thus we know that our rejection regions are the intervals

(−∞,−1.645) and (1.645,∞). So looking back at the z = 2.864, what we

find is that the test-statistic lies in the rejection region; therefore we must

reject H0. This finding implies that the means between the two populations

are again not the same.

33

References

[1] Broverman, Samuel. ACTEX P/1 Study Manual. 2012 edition, Winsted:

ACTEX Publications, Inc., 2012.

[2] Lee, Cheng F., and John C. Lee. Statistics for Business and Financial

Economics. 3rd edition, New York: Springer, 2013.

[3] Miller, Irwin, and Marylees Miller.John E. Freund’s Mathematical Statis-

tics. 6th edition, Upper Saddle River: Prentice Hall, 1999.

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