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Advanced Business Statistics

▪ Introduction to Hypothesis Testing (Two Sample)

Winter 2022

Agenda

Introduction to Hypothesis Testing of the

Deference Means (Two Sample)

❑ When 𝜎1 and 𝜎2 are known

❑ When 𝜎1 and 𝜎2 are unknown

• Considering equal variances

• Considering unequal variances

❑ For Proportion

Concept of Hypothesis Testing - Review

▪ 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

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 > 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 < 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 ≠ 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 > 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 < 𝑎

𝐻0: 𝜇 = 𝑎

𝐻1: 𝜇 ≠ 𝑎

𝐻0:𝑝 = 𝑏

𝐻1:𝑝 > 𝑏

𝐻0:𝑝 = 𝑏

𝐻1:𝑝 < 𝑏

𝐻0:𝑝 = 𝑏

𝐻1:𝑝 ≠ 𝑏

𝒛𝜶

− 𝒛𝜶

− 𝒛𝜶/𝟐 𝒛𝜶/𝟐

𝒛𝜶

− 𝒛𝜶

− 𝒛𝜶/𝟐 𝒛𝜶/𝟐

𝒕𝜶

− 𝒕𝜶

− 𝒕𝜶/𝟐 𝒕𝜶/𝟐

one-tail right

one-tail left

two-tail

one-tail right

one-tail left

two-tail

one-tail right

one-tail left

two-tail

𝑍𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝜎

𝑛

𝑡𝑠𝑡𝑎𝑡 = ത𝑋 − 𝜇

ൗ 𝑠

𝑛

𝑍𝑠𝑡𝑎𝑡 = Ƹ𝑝 − 𝑝

𝑝(1 − 𝑝) 𝑛

Hypothesis

Testing

𝛅 𝐮𝐧𝐤𝐧𝐨𝐰𝐧

𝑧𝑠𝑡𝑎𝑡 > 𝑧𝛼 reject 𝐻0

𝑧𝑠𝑡𝑎𝑡 < −𝑧𝛼 reject 𝐻0

𝑧𝑠𝑡𝑎𝑡> 𝑧𝛼/2 𝑧𝑠𝑡𝑎𝑡 < −𝑧𝛼/2

reject 𝐻0

𝑡𝑠𝑡𝑎𝑡 > 𝑡 reject 𝐻0

𝑡𝑠𝑡𝑎𝑡 < −𝑡𝛼 reject 𝐻0

𝑡𝑠𝑡𝑎𝑡> 𝑡𝛼/2 𝑡𝑠𝑡𝑎𝑡 < −𝑡𝛼/2

reject 𝐻0

𝑧𝑠𝑡𝑎𝑡 > 𝑧𝛼 reject 𝐻0

𝑧𝑠𝑡𝑎𝑡 < −𝑧𝛼 reject 𝐻0

𝑧𝑠𝑡𝑎𝑡> 𝑧𝛼/2 𝑧𝑠𝑡𝑎𝑡 < −𝑧𝛼/2

reject 𝐻0

1 2 3 4 5

Two Populations - Two Samples

Population 1

Parameters: 𝝁𝟏 and 𝜹𝟏

Population 2

Parameters: 𝝁𝟐 and 𝜹𝟐

Sample Size: 𝒏𝟏 Statistics: ഥ𝑿𝟏 and 𝒔𝟏

Sample Size: 𝒏𝟐 Statistics: ഥ𝑿𝟐 and 𝒔𝟐

Inference about the

difference between two

population means 𝜇1 − 𝜇2

Testing Difference Population Means: 𝝈𝟏and 𝝈𝟐are Known – Main Steps When standard deviation of both populations are known, below steps should be

followed to test the difference between populations’ mean.

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 > 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 < 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 ≠ 𝜇2

𝑧𝛼

− 𝑧𝛼

− 𝑧𝛼/2 𝑧𝛼/2

one-tail right

one-tail left

two-tail

𝑧𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

(𝛿1) 2

𝑛1 + (𝛿2)

𝑛2

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.

1 2-3 4

Example 1

A baby-food producer company, ABC, claims that its product helps babies to

gain weight faster than a leading competitor’s product, XYZ. A survey was

• Mothers were asked which product (ABC or XYZ) they intended to

feed their babies.

• There were 38 mothers feeding babies with ABC and 29 mentioned

that they would feed their babies XYZ.

• Mothers were asked to keep track of their babies’ weight gains over

the next 3 months.

• Each baby’s weight gain (in gram) was recorded. ( refer to Excel Data)

Assume, according to historical data, the standard deviation of babies’ weight

who were fed by ABC is 285 gram and by XYZ is 320 gram. Can we conclude,

using weight gain as our criterion, that company A is indeed superior. Assume

confidence level is 95%.

Example 1

ത𝑋1 = σ𝑥1 𝑛1

= 94,078

38 = 2,476 𝛿1 = 285

1 − 𝛼 = 0.95

𝛼 = 0.05

𝑍0.05 = 1.64

✓ Question is a hypothesis testing

✓ Standard deviation of populations are known, we use z

ത𝑋2 = σ𝑥2 𝑛2

= 66,605

29 = 2,297 𝛿2 = 320

𝑛1 = 38

𝑛2 = 29

from Z table

Example 1

𝐻0: 𝜇1 = 𝜇2

𝐻1:𝜇1 > 𝜇2

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 falls in rejection region. There is enough evidence to reject

𝐻0. Company A is superior.

𝑍𝛼 = 𝑍0.05 = −1.64

First, we construct the hypothesis:

rejection

region

𝟏.𝟔𝟒 𝟐.𝟑𝟕

𝑧𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

(𝛿1) 2

𝑛1 + (𝛿2)

𝑛2

= 2,476 − 2,297

(285)2

38 + (320)2

= 2.37

Testing Difference Population Means: 𝝈𝟏and 𝝈𝟐are unknown

When standard deviation of both populations are unknown, the difference of

population means depends on whether the two unknown population variances

are equal or not.

• 𝜎1 2 = 𝜎2

• 𝜎1 2 ≠ 𝜎2

For both situations, student t distribution is applied. However, the formula is

different.

Testing Difference Population Means: 𝝈𝟏and 𝝈𝟐are Unknown – Main Steps When 𝜎1

2 = 𝜎2 2 below steps should be followed to test the difference between

populations’ mean.

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 > 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 < 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 ≠ 𝜇2

𝑡𝛼

− 𝑡𝛼

− 𝑡𝛼/2 𝑡𝛼/2

one-tail right

one-tail left

two-tail

𝑡𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

𝑠𝑝 2 1 𝑛1

+ 1 𝑛2

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.

1 2-3 4

𝑠𝑝 2 =

(𝑛1 − 1)𝑠1 2 + (𝑛2 − 1)𝑠2

𝑛1 + 𝑛2 − 2

𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 = 𝑛1 + 𝑛2 − 2

Example 2

A large car manufacturer developed a mandatory training program for its

workers in 4 countries. In assessing the effectiveness of this program, an

operation manager designed a survey by measuring two factors:

• Workers’ improved performance (saving time, second).

• age of each worker, 21-to-30 and 31-to-40 age groups.

The survey measured 200 workers’ improved performance in each age group,

the data was listed in the Excel file. With 90% confidence, can we conclude

that the training program has been more effective for age group 31-to-40.

Assume the variances of the improved performance for workers 21-to-30 and

31-to-40 are equal 𝜎1 2 = 𝜎2

Example 2

✓ Question is a hypothesis testing

✓ Standard deviation of populations are unknown, we use t

ത𝑋1 = σ𝑥1 𝑛1

= 17,575

200 = 87.87 𝑆1 = 13.38

1 − 𝛼 = 0.90

𝛼 = 0.10

𝑡0.1 = 1.282

ത𝑋2 = σ𝑥2 𝑛2

= 10,254

200 = 51.22

𝑆2 = 17.06

𝑛1 = 200

𝑛2 = 200

from t table = 𝑛1 + 𝑛2 − 2 = 200 + 200 − 2 = 398

𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚

Example 2

𝐻0: 𝜇1 = 𝜇2

𝐻1:𝜇1 < 𝜇2

The test is one-tail left.

For t critical, we use t-student table.

For t-stat, we use the formula:

𝑡𝛼 = 𝑡0.1 = 1.282

First, we construct the hypothesis:

rejection

region

𝟐𝟑.𝟗−𝟏.𝟐𝟖𝟐

𝑡𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

𝑠𝑝 2 1 𝑛1

+ 1 𝑛2

= (87.87 − 51.22)

235( 1 200

+ 1 200

)

= 23.9

𝑠𝑝 2 =

(𝑛1 − 1)𝑠1 2 + (𝑛2 − 1)𝑠2

𝑛1 + 𝑛2 − 2

𝑠𝑝 2 =

200 − 1 13.382 + (200 − 1)17.062

200 + 200 − 2 = 235

t-stat > t-critical. Therefore t-stat

does not fall in the rejection region.

We fail to r𝑒𝑗𝑒𝑐𝑡 𝐻0.

Testing Difference Population Means: 𝝈𝟏and 𝝈𝟐are Unknown – Main Steps When 𝜎1

2 ≠ 𝜎2 2 below steps should be followed to test the difference between

populations’ mean.

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 > 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 < 𝜇2

𝐻0:𝜇1 = 𝜇2

𝐻1:𝜇1 ≠ 𝜇2

𝑡𝛼

− 𝑡𝛼

− 𝑡𝛼/2 𝑡𝛼/2

one-tail right

one-tail left

two-tail

𝑡𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

𝑠1 2

𝑛1 + 𝑠2

𝑛2

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.

1 2-3 4

𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚

= ( 𝑠1

𝑛1 +

𝑠2 2

𝑛2 )2

( 𝑠1

𝑛1 )2

𝑛1 − 1 +

( 𝑠2

𝑛2 )2

𝑛2 − 1

Example 3

A business analyst designed a survey to determine the effect of gender on the

automobile insurance rate. A random sample of young men and women was

listed in the Excel file. 178 men and 211 women were asked how many

kilometers he or she had driven in the past year.

After analysing data, she claims that men and women drive totally almost an

equal distance in a year. With 90% confidence, do you accept her analysis.

Assume the variances of the distance driven by men and women drivers are

not equal 𝜎1 2 ≠ 𝜎2

Example 3

ത𝑋1 = σ𝑥1 𝑛1

= 3,589,962

178 = 20,168

𝑆1 = 3,609

1 − 𝛼 = 0.90

𝛼 = 0.10

𝛼

2 = 0.05

𝑡0.05 = 1.645

ത𝑋2 = σ𝑥2 𝑛2

= 3,913,864

210 = 18,549 𝑆2 = 3,386

𝑛1 = 178

𝑛2 = 211

from t table

= ( 𝑠1

𝑛1 +

𝑠2 2

𝑛2 )2

( 𝑠1

𝑛1 )2

𝑛1 − 1 +

( 𝑠2

𝑛2 )2

𝑛2 − 1

𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚

= ( 3,6092

178 +

3,3862

211 )2

( 3,6092

178 )2

178 − 1 +

( 3,3862

211 )2

211 − 1

𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚

= 367

✓ Question is a hypothesis testing

✓ Standard deviation of populations are unknown, we use t

Example 3

𝐻0: 𝜇1 = 𝜇2

𝐻1:𝜇1 ≠ 𝜇2

The test is two-tail.

For t critical, we use t-student table.

For t-stat, we use the formula:

𝑡𝛼/2 = 𝑡0.05 = 1.645

First, we construct the hypothesis:

𝑡𝑠𝑡𝑎𝑡 = ത𝑋1 − ത𝑋2

𝑠1 2

𝑛1 + 𝑠2

𝑛2

= (20,168 − 18,549)

( 3,6092

178 + 3,3862

211 )

= 4.53

t-stat > t-critical. Therefore t-stat falls in the rejection region. We have enough evidence to

reject 𝐻0.

rejection

region

+𝟏.𝟔𝟒𝟓−𝟏.𝟔𝟒𝟓

rejection

region

𝟒.𝟓𝟑

Testing Difference Population Proportions

In some cases, the objective is to determine the difference between proportion

of two populations, 𝑝1−𝑝2.

To draw inferences about 𝑝1and 𝑝2, a sample of size 𝑛1from population 1 and a

sample of size 𝑛2 from population 2 are taken.

Population 1 Population 2

Parameter: 𝒑𝟏 Parameter: 𝒑𝟐

Statistics: ෝ𝒑𝟏 Statistics: ෝ𝒑𝟐

Sample

Size: 𝒏𝟏

Sample

Size: 𝒏𝟐

ෝ𝒑𝟏 = 𝒙𝟏 𝒏𝟏

ෝ𝒑𝟐 = 𝒙𝟐 𝒏𝟐

For each sample, the

number of successes is

represented by x, which

we label 𝒙𝟏 and 𝒙𝟐 , respectively.

Testing Difference Population Proportions – Main Steps

𝐻0: 𝑝1 = 𝑝2

𝐻1:𝑝1 > 𝑝2

𝐻0:𝑝1 = 𝑝2

𝐻1:𝑝1 < 𝑝2

𝐻0:𝑝1 = 𝑝2

𝐻1:𝑝1 ≠ 𝑝2

𝑧𝛼

− 𝑧𝛼

− 𝑧𝛼/2 𝑧𝛼/2

one-tail right

one-tail left

two-tail

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.

1 2-3 4

𝑧𝑠𝑡𝑎𝑡 = ( Ƹ𝑝1− Ƹ𝑝2)

Ƹ𝑝(1 − Ƹ𝑝)( 1 𝑛1

+ 1 𝑛2 )

Ƹ𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2

Example 4

Selling extended warranties for products is a profitable business for many

stores. The extended warranty is offered for both regular and sale prices. A

store manager has recently conducted a survey about the difference in

proportion of customers who bought extended warranty. The below table

shows the results:

At the 1% significance level, can we say that the people who paid the regular

price are more likely to buy an extended warranty?

Sale Price Regular Price

Sample size 354 478

Number who bought

extended warranty

111 105

Example 4

✓ Question is a population proportion hypothesis testing

𝑥1 = 111

𝛼 = 0.01

𝑍0.01 = 2.33

𝑥2 = 105

𝑛1 = 354

𝑛2 = 478

from Z table

Ƹ𝑝2 = 𝑥2 𝑛2

= 105

478 = 0.22

Ƹ𝑝1 = 𝑥1 𝑛1

= 111

354 = 0.31

Ƹ𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2

= 111 + 105

354 + 478 = 0.26

Example 4

𝐻0: 𝑝1 = 𝑝2

𝐻1:𝑝1 < 𝑝2

This is case number one. 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 does not fall in the rejection region. There is not enough

evidence to accept 𝐻1.

𝑍𝛼 = 𝑍0.01 = −2.33

First, we construct the hypothesis:

𝑧𝑠𝑡𝑎𝑡 = ( Ƹ𝑝1− Ƹ𝑝2)

Ƹ𝑝(1 − Ƹ𝑝)( 1 𝑛1

+ 1 𝑛2 )

= (0.31 − 0.22)

0.26(1 − 0.26)( 1 354

+ 1 478

)

= 2.92

rejection

region

𝟐.𝟗𝟐−𝟐.𝟑𝟑

In Class Activity 1

How important to your health are regular vacations? In a study a random

sample of men and women were asked how frequently they take

vacations. The men and women were divided into two groups each. The

members of group 1 had suffered a heart attack; the members of group

2 had not. The number of days of vacation last year was recorded for

each person. Can we infer that men and women who suffer heart

attacks vacation less than those who did not suffer a heart attack?

In Class Activity 2

In designing advertising campaigns to sell magazines, it is important to know

how much time each of a number of demographic groups spends reading

magazines. In a preliminary study, 40 people were randomly selected. Each

was asked how much time per week he or she spends reading magazines;

additionally, each was categorized by gender and by income level (high or

low). The data are stored in the following way: column 1 = Time spent reading

magazines per week in minutes for all respondents; column 2 = Gender (1 =

Male, 2 = Female); column 3 = Income level (1 = Low, 2 = High).

Is there sufficient evidence at the 10% significance level to conclude that men

and women differ in the amount of time spent reading magazines?

In Class Activity 3

The manager of a dairy is in the process of deciding which of two new carton-

filling machines to use. The most important attripute is the consistency of the

fills. In a preliminary study she measured the fills in the 1-liter carton and listed

them here. Can the manager infer that the two machines differ in their

consistency of fills?

Data Analysis - Microsoft Excel

Let’s answer example 1 with data analysis Add-Ins in Excel.

Data Analysis - Microsoft Excel

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