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Advanced Business Statistics

▪ Introduction to Hypothesis Testing (One Sample)

Winter 2022

Agenda

Introduction to Hypothesis Testing of the Mean

(One Sample)

❑ When σ is known

❑ When σ is unknown

❑ For Proportion

Inferential Statistics

• Population Means

• Population Proportion

Inferential Statistics

Estimating Testing

Hypothesis

Concept of Hypothesis Testing

For many people, using term hypothesis testing seems likely new although

application of hypothesis testing and the concept underlying are quite familiar.

The best example is a criminal trial.

Usually, people face a trial if they are accused of a crime. The case is

presented by a prosecutor, and based on the presented evidence the jury

makes a decision.

Concept of Hypothesis Testing

In fact, this case is a test of hypothesis that the jury conducts. They actually

consider two hypotheses to be tested:

❑ Null hypothesis (𝐻0): The defendant is innocent

❑ Alternative hypothesis (𝐻𝐴 𝑜𝑟 𝐻1): The defendant is guilty

In case of the decision, there are only two possible decisions, guilty or

innocent, that the jury makes after reviewing the evidence presented by both

the prosecutor and defendant.

Concept of Hypothesis Testing

When the defendant is convicted

it means that the jury is rejecting

the null hypothesis in favor of the

alternative hypothesis

There is enough evidence

to conclude that the

defendant is guilty

When the defendant is acquitted

it means that the jury is not

rejecting the null hypothesis in

favor of the alternative hypothesis

There is not enough

evidence to conclude that

the defendant is guilty

Statistically

Concept of Hypothesis Testing

Important: generally, we interpret the result by saying that

there is not enough evidence to reject null hypothesis or

alternative hypothesis. We do not directly say that we accept

the null or alternative hypothesis.

Concept of Hypothesis Testing

Two possible errors may occur:

❑ Error type one: we reject a null hypothesis although it is true

❑ Error type two: we do not reject a null hypothesis although it is false

𝐻0is true 𝐻0is false

Reject 𝐻0 error type one

P(error type one) = α Correct decision

Not reject 𝐻0 Correct decision error type two

P(error type two) = β

Concept of Hypothesis Testing

▪ There are two hypothesis; (1) null and (2) alternative

▪ In hypothesis testing, first, we start with the assumption that null hypothesis

is true.

▪ The main objective is to determine whether there is enough evidence to

reject 𝐻0 or 𝐻𝐴

▪ Two possible results are:

o there is enough evidence to support the alternative

o there is not enough evidence to support the alternative

▪ Two possible errors are:

o Reject a true null hypothesis, P(error type one) = α

o Not reject a false null hypothesis, P(error type two) = β

Concept of Hypothesis Testing

As mentioned before, we start with the assumption that null hypothesis is true.

For example, a police officer is testing the average speed of vehicle in a city is

85 km/hrs or not.

The null hypothesis is 𝐻0: 𝜇 = 85. But for alternative hypothesis there are

three possible situations: 𝐻1: 𝜇 > 85, 𝐻1: 𝜇 < 85, 𝐻1: 𝜇 ≠ 85.

Concept of Hypothesis Testing

To construct hypotheses, one of the three possible hypotheses may be asked:

𝐻0: 𝜇 = 85

𝐻1: 𝜇 > 85

𝐻0: 𝜇 = 85

𝐻1: 𝜇 < 85

𝐻0: 𝜇 = 85

𝐻1: 𝜇 ≠ 85

one-tail right

one-tail left

two-tail

Hypotheses

Testing

Concept of Hypothesis Testing

The rejection region is a range of values such that if the test statistic falls into

that range, we decide to reject the null hypothesis in favor of the alternative

hypothesis.

one-tail right one-tail left two-tail

𝑍𝛼 − 𝑍𝛼 − 𝑍𝛼/2 𝑍𝛼/2

𝑯𝟎 rejection region

1- α = confidence level

α = significance

Testing Population Mean (µ) when Population Standard Deviation (σ) is Known – Main Steps

Construct

hypotheses

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 > 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 < 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 ≠ 𝑎

Draw appropriate z-normal graph and define the

location of 𝑍.

one-tail right one-tail left two-tail

𝑍𝛼 − 𝑍𝛼 − 𝑍𝛼/2 𝑍𝛼/2

Find the z value (z critical) from Normal table and put it

on the graph. For on-tails 𝑍𝛼 and for two-tail 𝑍𝛼/2.

1 2

Compute Z-stat:

𝑍𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝜎

𝑛

Put the value of Z-stat on

the graph.

4 One-tail right: If 𝑍𝑠𝑡𝑎𝑡 > 𝑍𝛼 , there is enough evidence to reject 𝐻0.

One-tail left: If 𝑍𝑠𝑡𝑎𝑡 < −𝑍𝛼 , there is enough evidence to reject 𝐻0.

Two-tail: If 𝑍𝑠𝑡𝑎𝑡 > 𝑍𝛼/2 or 𝑍𝑠𝑡𝑎𝑡 < −𝑍𝛼/2, there

is enough evidence to reject 𝐻0.

5 Make

decision

rejection

region

rejection

region

rejection

region

Example 1

Conduct the following test and interpret the result.

𝐻0: 𝜇 = 800

𝐻1: 𝜇 > 800

σ = 150 ത𝑋 = 770 𝑛 = 100 𝛼 = 0.05

The test is one-tail right.

For z critical, we use Normal table.

For z-stat, we use the formula:

z-stat < z-critical. Therefore z-stat does not fall in rejection region. There is not enough

evidence to reject 𝐻0.

rejection

region𝑍𝛼 = 𝑍0.05 = −1.64

𝟏.𝟔𝟒 𝑍𝑠𝑡𝑎𝑡 =

ത𝑋 − 𝜇

ൗ 𝜎

𝑛

= 770 − 800

150/ 100 = −2

−𝟐

Example 2

Conduct the following test and interpret the result.

𝐻0: 𝜇 = 12

𝐻1: 𝜇 < 12 σ = 4 ത𝑋 = 11 𝑛 = 144 𝛼 = 0.01

The test is one-tail left.

For z critical, we use Normal table.

For z-stat, we use the formula:

z-stat < z-critical. Therefore z-stat falls in rejection region. There is enough evidence to reject

𝐻0.

𝑍𝛼 = 𝑍0.01 = −2.33

𝑍𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝜎

𝑛

= 11 − 12

4/ 144 = −3

rejection

region

−𝟑 −𝟐.𝟑𝟑

Example 3

Conduct the following test and interpret the result.

𝐻0: 𝜇 = 50,000

𝐻1: 𝜇 ≠ 50,000

σ = 8,000 ത𝑋 = 51,150 𝑛 = 200 𝛼 = 0. 1

The test is two tail.

For z critical, we use Normal table.

For z-stat, we use the formula:

z-stat > z-critical. Therefore z-stat falls in rejection region. There is enough evidence to reject

𝐻0.

𝑍𝛼/2 = 𝑍0.1/2 = 𝑍0.05 = −1.64

𝑍𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝜎

𝑛

= 51,050 − 50,000

8000/ 200 = 2.03

rejection

region

+𝟏.𝟔𝟒−𝟏.𝟔𝟒

rejection

region

𝟐.𝟎𝟑

Example 4

Example 04: A business school claims that, on average, the required

GMAT score for an MBA student is more than 600. To examine the

claim, a MBA student asks a random sample of her 16 classmates

about their GMAT score. The results are exhibited here.

680 620 570 585 590 600 600 650

630 590 590 610 600 600 580 640

Can the student conclude at the 5% significance level that the claim is

true, assuming that GMAT score is normally distributed with a standard

deviation of 35?

Example 4

𝐻0: 𝜇 = 600

𝐻1: 𝜇 > 600

σ = 35 ത𝑋 = σ 𝑥

𝑛 = 9735

16 = 608.43 𝑛 = 16 𝛼 = 0.05

The test is one-tail right.

For z critical, we use Normal table.

For z-stat, we use the formula:

z-stat < z-critical. Therefore z-stat does not fall in rejection region. There is not enough

evidence to reject 𝐻0.

𝑍𝛼 = 𝑍0.05 = −1.64

𝑍𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝜎

𝑛

= 608.43 − 600

35/ 16 = 0.963

rejection

region

𝟏.𝟔𝟒𝟎. 𝟗𝟔𝟑

Testing Population Mean (µ) when Population Standard Deviation (σ) is Unknown – Main Steps

Construct

hypotheses

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 > 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 < 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 ≠ 𝑎

Draw appropriate student-t graph and define the

location of 𝑡.

one-tail right one-tail left two-tail

𝑡𝛼 − 𝑡𝛼 − 𝑡𝛼/2 𝑡𝛼/2 Find the t value (t critical) from student-t table and put

it on the graph. For on-tails 𝑡𝛼 and for two-tail 𝑡𝛼/2.

Degree of freedom is n-1.

1 2

Compute t-stat:

𝑡𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝑠

𝑛 Put the value of t-stat on

the graph.

4 One-tail right: If 𝑡𝑠𝑡𝑎𝑡 > 𝑡𝛼 , there is enough evidence to reject 𝐻0.

One-tail left: If 𝑡𝑠𝑡𝑎𝑡 < −𝑡𝛼 , there is enough evidence to reject 𝐻0.

Two-tail: If 𝑡𝑠𝑡𝑎𝑡 > 𝑡𝛼/2 or 𝑡𝑠𝑡𝑎𝑡 < −𝑡𝛼/2, there is

enough evidence to reject 𝐻0.

5 Make

decision

rejection

region

rejection

region

rejection

region

Example 5

A nurse claims that the average American is less than 10 kg overweight. A

random sample of 11 Americans was weighed to measure the difference

between their actual and ideal weights. The results (kg) are exhibited here.

9 11 12 10 8.5 8 8 7 8 9 8

Can the nurse conclude that her claim is true? She has considered 90%

confidence level.

Example 5

𝐻0: 𝜇 = 10

𝐻1: 𝜇 < 10

ത𝑋 = σ𝑥

𝑛 = 98.5

11 = 8.95 𝑠 = 1.49 Excel function STDEV.S 𝑛 = 11 1 − 𝛼 = 0.9 𝛼 = 0.1

The test is one-tail left.

For t critical, we use t table. Degree of freedom is 11-1=10

For t-stat, we use the formula:

t-stat < t-critical. Therefore t-stat falls in rejection region. There is enough evidence to reject 𝐻0.

𝑡𝛼 = 𝑡0.1 = 1.37

𝑡𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝑠

𝑛

= 8.95 − 10

1.49/ 11 = −2.32

First, we construct the hypothesis:

rejection

region

−𝟐.𝟑𝟐 −𝟏.𝟑𝟕

Testing Population Proportion – Main Steps

Construct

hypotheses

𝐻0: 𝑝 = 𝑏

𝐻1: 𝑝 > 𝑏

𝐻0: 𝑝 = 𝑏

𝐻1: 𝑝 < 𝑏

𝐻0: 𝑝 = 𝑏

𝐻1: 𝑝 ≠ 𝑏

Draw appropriate z-normal graph and define the

location of 𝑍.

one-tail right one-tail left two-tail

𝑍𝛼 − 𝑍𝛼 − 𝑍𝛼/2 𝑍𝛼/2

Find the z value (z critical) from Normal table and put it

on the graph. For on-tails 𝑍𝛼 and for two-tail 𝑍𝛼/2.

1 2

Compute Z-stat:

𝑍𝑠𝑡𝑎𝑡 = Ƹ𝑝 − 𝑝

𝑝(1 − 𝑝)/𝑛

Put the value of Z-stat on

the graph.

4 One-tail right: If 𝑍𝑠𝑡𝑎𝑡 > 𝑍𝛼 , there is enough evidence to reject 𝐻0.

One-tail left: If 𝑍𝑠𝑡𝑎𝑡 < −𝑍𝛼 , there is enough evidence to reject 𝐻0.

Two-tail: If 𝑍𝑠𝑡𝑎𝑡 > 𝑍𝛼/2 or 𝑍𝑠𝑡𝑎𝑡 < −𝑍𝛼/2, there

is enough evidence to reject 𝐻0.

5 Make

decision

rejection

region

rejection

region

rejection

region

Example 6

An insurance company wants to know what proportion of drivers in a city has at

least one police ticket because of passing speed limit. The finance department

claims that more than three-quarter of the drives falls in this group.

As a test, a random sample of 200 cars that has auto insurance with that

company was selected. They found that 143 drivers have police ticket because

of passing speed limit.

Does the insurance company have enough evidence at the 10% significance

level to support its belief?

Example 6

𝐻0:𝑝 = 0.75

𝐻1: 𝑝 > 0.75

𝛼 = 0.1 𝑛 = 200 𝑥 = 143 Ƹ𝑝 = 𝑥

𝑛 = 143

200 = 0.71

The test is one-tail right proportion test.

For z critical, we use Normal table.

For z-stat, we use the formula:

z-stat > z-critical. Therefore z-stat does not fall in rejection region. There is not

enough evidence to reject 𝐻0.

rejection

region𝑍𝛼 = 𝑍0.1 = −1.28

𝟏.𝟐𝟖

𝑍𝑠𝑡𝑎𝑡 = Ƹ𝑝 − 𝑝

𝑝(1 − 𝑝)/𝑛 =

0.71 − 0.75

0.75(1 − 0.75)/200 = −1.3

−𝟏. 𝟑

In Class Activity 1

Because television audiences of newscasts tend to be older (and

because older people suffer from a variety of medical ailments)

pharmaceutical companies’ advertising often appears on national news

in the three networks (ABC, CBS, and NBC). The ads concern

prescription drugs such as those to treat heartburn. To determine how

effective the ads are, a survey was undertaken. Adults over 50 who

regularly watch network newscasts were asked whether they had

contacted their physician to ask about one of the prescription drugs

advertised during the newscast. The responses (1 = No and 2 = Yes)

were recorded. Estimate with 95% confidence the fraction of adults over

50 who have contacted their physician to inquire about a prescription

drug.

In Class Activity 2

A random sample of 18 young adult men (20–30 years old) was sampled.

Each person was asked how many minutes of sports he watched on television

daily. The responses are listed here. It is known that σ = 10. Test to determine

at the 5% significance level whether there is enough statistical evidence to

infer that the mean amount of television watched daily by all young adult men

is greater than 50 minutes.

50 48 65 74 66 37 45 68 64

65 58 55 52 63 59 57 74 65

In Class Activity 3

A manufacturer of lightbulbs advertises that, on average, its long-life

bulb will last more than 5,000 hours. To test the claim, a statistician took

a random sample of 100 bulbs and measured the amount of time until

each bulb burned out. If we assume that the lifetime of this type of bulb

has a standard deviation of 400 hours, can we conclude at the 5%

significance level that the claim is true?

In Class Activity 4

Companies that sell groceries over the Internet are called e-grocers.

Customers enter their orders, pay by credit card, and receive delivery by truck.

A potential e-grocer analyzed the market and determined that the average

order would have to exceed $85 if the e-grocer were to be profitable. To

determine whether an e-grocery would be profitable in one large city, she

offered the service and recorded the size of the order for a random sample of

customers. Can we infer from these data that an e-grocery will be profitable in

this city?

References

• Business Statistics in Practice: Second Canadian Edition, Bowerman,

O'Connell, et al. McGraw-Hill, Third Canadian Edition

• G. Keller (2017) Statistics for Management and Economics (Abbreviated),

11th Edition, South-Western (students can also use the 8th edition of the

same textbook).

• M. Middleton (1997) Data Analysis Using Microsoft Excel, Duxbury. (A good

reference for basic statistical work with Excel.)

Thank you