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Advanced Business Statistics

▪ Introduction to Estimation (One Sample)

Winter 2022

Agenda

Introduction to Estimation of the Mean (One Sample)

❑ When σ is known

❑ When σ is unknown

❑ For Proportion

Introduction

In almost all realistic situations parameters of population are unknown. We

use the sampling distribution to draw inferences about the unknown

population parameters.

Statistical inference is the process by

which we acquire information and

draw conclusions about populations

from samples.

Sample

Population

Important Definitions

Sample

Population

Important Definitions

Population:

A population is the total of any kind of units under consideration by the

statistician such as blood pressure population of men

Parameter:

A parameter is a characteristic of a population such as size, mean, and

standard deviation

Sample:

A sample is any portion of the population selected for study such as 100 men

selected from blood pressure population of men.

Statistic:

A statistic is a characteristic of a sample such as size, mean, and standard

deviation

Important Symbols

1-αConfidence Level

Significance Level

Mean Standard

Deviation Size Proportion

Population

Sample

µ σ N ഥX S n

𝑝

ෝ𝑝

Population Inference

There are two types of inference:

❑ Estimation

❑ Hypothesis Testing

The objective of estimation is to determine the approximate value of a

population parameter on the basis of a sample statistic.

Concept of Estimation

As its name suggests, the objective of estimation is to determine the

approximate value of a population parameter on the basis of a sample

statistic. An important example is estimation of the population mean ( )

by employing the sample mean ( ).

There are two situations to estimate ( ):

❖ When population standard deviation ( ) is known

❖ When population standard deviation ( ) is unknown

ഥX

Estimating when Is Knownµ σ

When standard deviation of the population is known, following equation can be

applied to estimate the population mean.

Therefore:

is called the Lower Confidence Level (LCL)

is called the Upper Confidence Level (UCL)

Also:

is called the Standard Error

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛 < 𝜇 < ത𝑋 + 𝑍𝛼

𝛿

𝑛

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛

ത𝑋 + 𝑍𝛼 2

𝛿

𝑛

𝛿

𝑛

Estimating when Is Knownµ σ

To estimate when is know, we should do following steps:

Step1: write all provided data

Step 2: extract from Z table.

Step 3: write the appropriate equation to estimate

Step 4: plug in the numbers

µ σ

𝑍𝛼 2

Example 1

A statistics practitioner took a random sample of 49 observations from a

population with a standard deviation of 21 and computed the sample

mean to be 100. Estimate the population mean with 90% confidence.

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛 < 𝜇 < ത𝑋 + 𝑍𝛼

𝛿

𝑛

ത𝑋 = 100

𝛿 = 21

𝑛 = 49

1 − 𝛼 = 0.9

𝛼 = 0.1 𝛼

2 = 0.05

𝑍0.05 = 1.64

100 − 1.64 21

49 < 𝜇 < 100 + 1.64

95.08 < 𝜇 < 104.92

with 90% confidence, the population

mean is a number lower than 104.92 and

higher than 95.08.

from Z table

standard deviation of population is

known, we use z for estimation:

Example 2

The mean of a random sample of 100 observations from a normal population

with a standard deviation of 40 is 250. Estimate the population mean with 95%

confidence.

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛 < 𝜇 < ത𝑋 + 𝑍𝛼

𝛿

𝑛

ത𝑋 = 250

𝛿 = 40

𝑛 = 100

1 − 𝛼 = 0.95

𝛼 = 0.05 𝛼

2 = 0.025

𝑍0.025 = 1.96

250 − 1.96 40

100 < 𝜇 < 250 + 1.96

100

242.16 < 𝜇 < 257.84

with 95% confidence, the

population mean is a number

lower than 257.84 and higher than

242.16.

from Z table

standard deviation of population is

known, we use z for estimation:

Example 3

How many rounds of golf do physicians (who play golf) play per year? A survey

of 12 physicians revealed the following numbers:

11 21 19 5 14 35 26 31 13 19 36 44

Estimate with 99% confidence the mean number of rounds per year played by

physicians. The number of rounds is normally distributed with a standard

deviation of 10.

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛 < 𝜇 < ത𝑋 + 𝑍𝛼

𝛿

𝑛

ത𝑋 = σ𝑥

𝑛 = 274

12 = 22.83

𝛿 = 10

𝑛 = 12

1 − 𝛼 = 0.99

𝛼 = 0.01 𝛼

2 = 0.005

𝑍0.005 = 2.56

22.83 − 2.56 10

12 < 𝜇 < 22.83 + 2.56

15.44 < 𝜇 < 30.22

with 99% confidence, the population

mean is a number lower than 30.22 and

higher than 15.44.from Z table

standard deviation of population is

known, we use z for estimation:

Sample Size to Estimate Mean

As explained before, following equation is used to estimate the population

mean.

We can solve the equation for n to compute the required sample size for

estimation of the mean:

Where B represent the “standard error” of estimation or “within unit” of the

mean.

ത𝑋 − 𝑍𝛼 2

𝛿

𝑛 ≤ 𝜇 ≤ ത𝑋 + 𝑍𝛼

𝛿

𝑛

𝑛 = (𝑍𝛼 2

𝛿

𝐵 )2

Example 4

Determine the sample size required to estimate a population mean to within 12

units given that the population standard deviation is 40. A confidence level of

95% is judged to be appropriate.

𝑛 = (𝑍𝛼 2

𝛿

𝐵 )2

𝐵 = 12

𝛿 = 40

𝑛 = ?

1 − 𝛼 = 0.95

𝛼 = 0.05 𝛼

2 = 0.025

𝑍0.025 = 1.96

from Z table

𝑛 = (1.96 40

12 )2= 42.68

𝑛 = 43

43 samples is required for this

estimation.

Example 5

A medical statistician wants to estimate the average level of blood

pressure of people who are on a new diet plan. In a preliminary study,

he already knew that the standard deviation of the population of blood

pressure is about 20 mmHG. How large a sample should he take to

estimate the mean blood pressure to within 4 unit. Assume the level of

confidence is 99%.

𝑛 = (𝑍𝛼 2

𝛿

𝐵 )2

𝐵 = 4

𝛿 = 20

𝑛 = ?

1 − 𝛼 = 0.99

𝛼 = 0.01 𝛼

2 = 0.005

𝑍0.005 = 2.56

from Z table

𝑛 = (2.56 20

4 )2= 163.84

𝑛 = 164

164 samples of people’s blood

pressure is required for this

estimation.

Estimating when Is Unknownµ σ

When standard deviation of the population is unknown, following equation can

be applied to estimate the population mean.

Therefore:

is called the Lower Confidence Level (LCL)

is called the Upper Confidence Level (UCL)

Also:

is called the Standard Error

ത𝑋 − 𝑡𝛼 2

𝑠

𝑛 < 𝜇 < ത𝑋 + 𝑡𝛼

𝑠

𝑛

ത𝑋 − 𝑡𝛼 2

𝑠

𝑛

ത𝑋 + 𝑡𝛼 2

𝑠

𝑛

𝑠

𝑛

𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚

= 𝑛 − 1

Estimating when Is Unknownµ σ

To estimate when is unknow, we should do following steps:

Step1: write all provided data

Step 2: extract from t table. Remember, you need to compute

degree of freedom which is n-1 to be able to use t able.

Step 3: write the appropriate equation to estimate

Step 4: plug in the numbers

µ σ

𝑡𝛼 2

Example 6

A random sample of 25 was drawn from a population. The sample mean and

standard deviation are 450 and 90. Estimate mean of the population with 95%

confidence.

ത𝑋 − 𝑡𝛼 2

𝑠

𝑛 < 𝜇 < ത𝑋 + 𝑡𝛼

𝑠

𝑛

ത𝑋 = 450

𝑠 = 90

𝑛 = 25

1 − 𝛼 = 0.95

𝛼 = 0.05

𝛼

2 = 0.025

𝑡0.025 = 2.064

450 − 2.064 90

25 < 𝜇 < 450 + 2.064

412.85 < 𝜇 < 487.16

with 95% confidence, the

population mean is a number

lower than 487.16 and higher than

412.85.

from t table

standard deviation of population is

unknown, we use t for estimation:

degree of freedom = n-1

=25-1=24

Example 7

A police control officer is conducting an analysis of the amount of time left on

gas stations. A quick survey of 20 police cars that have just left a gas station

are the following times (in minutes).

6 4 6 9 5 6 7 4 5 8 4 4 5 6 5 4 3 4 3 5

Estimate with 99% confidence the mean amount of time a police car spends in

a gas station.

ത𝑋 − 𝑡𝛼 2

𝑠

𝑛 < 𝜇 < ത𝑋 + 𝑡𝛼

𝑠

𝑛𝑠 = 1.56

𝑛 = 20

1 − 𝛼 = 0.99

𝛼 = 0.01

𝛼

2 = 0.005

𝑡0.005 = 2.861

5.15 − 2.861 1.56

20 < 𝜇 < 5.15 + 2.861

1.56

20 4.16 < 𝜇 < 6.14

with 99% confidence, the average of

amount of time left on gas stations is lower

than 6.14 min and higher than 4.16 min.

from t table

standard deviation of population is

unknown, we use t for estimation:

degree of freedom = n-1

=20-1=19

ത𝑋 = σ 𝑥

𝑛 = 103

20 = 5.15

Excel function STDEV.S

Example 8

What is the average salary of students working in summer internship? To

determine an answer, a random sample of 16 students was drawn (US dollar).

15,500 12,400 13,000 19,000 21,000 22,000 18,000 12,300

16,100 13,200 17,900 10,400 9,200 16,800 17,100 15,600

Estimate with 90% confidence the mean of students’ salary working in summer

internship

ത𝑋 − 𝑡𝛼 2

𝑠

𝑛 < 𝜇 < ത𝑋 + 𝑡𝛼

𝑠

𝑛𝑠 = 3,636

𝑛 = 16

1 − 𝛼 = 0.90

𝛼 = 0.1 𝛼

2 = 0.05

𝑡0.05 = 1.753

15,594 − 1.753 3,636

16 < 𝜇 < 15,594 + 1.753

3,636

14,001 < 𝜇 < 17,187

with 90% confidence, the average salary of

students working in summer internship is lower

than 17,187 and higher than 14,001 US dollar.

from t table

standard deviation of population is

unknown, we use t for estimation:

degree of freedom = n-1=16-1=15

ത𝑋 = σ𝑥

𝑛 = 249,500

16 = 15,594

Excel function STDEV.S

Estimating a Population Proportion

In many situations, we need to estimate a population proportion. For example,

proportion of people who watch advertisements during a TV show. To estimate

proportion, following equation should be used:

Ƹ𝑝 − 𝑍𝛼 2

Ƹ𝑝 1 − Ƹ𝑝

𝑛 < 𝑝 < Ƹ𝑝 + 𝑍𝛼

Ƹ𝑝(1 − Ƹ𝑝)

𝑛

Lower Confidence Level (LCL) Upper Confidence Level (UCL)

Example 9

A business school wanted to know whether the graduates of the school can find

the job within the first three months after finishing their study. 424 graduates

were surveyed and asked about their employment. After tallying the responses,

it was reported that only 152 were hired within the first three months of

graduation. Estimate with 90% confidence the proportion of all business school

graduates who can find a job within the first three months of graduation.

𝑛 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 = 152 Ƹ𝑝 =

152

424 = 0.36

𝑛 = 424

1 − 𝛼 = 0.90

𝛼 = 0.1 𝛼

2 = 0.05

𝑍0.05 = 1.64

from Z table

Ƹ𝑝 − 𝑍𝛼 2

Ƹ𝑝 1 − Ƹ𝑝

𝑛 < 𝑝 < Ƹ𝑝 + 𝑍𝛼

Ƹ𝑝(1 − Ƹ𝑝)

𝑛

proportion of population needs to be

computed:

0.36 − 1.64 0.36 1 − 0.36

424 < 𝑝 < 0.36 + 1.64

0.36 1 − 0.36

424

0.32 < 𝑝 < 0.4

The proportion of graduates of the school can find the

job within the first three months is lower than 0.4 and

higher than 0.32

Sample Size to Estimate Proportion

If we solve the proportion function for n, the required sample size for

estimation of proportion is computed:

Where B represent the “standard error” of estimation or “within unit” of

the mean.

𝑛 = (𝑍𝛼 2

Ƹ𝑝 1 − Ƹ𝑝 /𝐵)2

Example 10

Determine the sample size necessary to estimate a population proportion to

within 0.03 with 90% confidence. Assume Ƹ𝑝 = 0.6.

𝑛 = ( 𝑍𝛼 2

Ƹ𝑝(1 − Ƹ𝑝)

𝐵 )2

Ƹ𝑝 = 0.60

1 − 𝛼 = 0.90

𝛼 = 0.1 𝛼

2 = 0.05

𝑍0.05 = 1.64

from Z table

𝐵 = 0.03

718 samples is required for this

estimation.

𝑛 = ( 1.64 0.6(1 − 0.6)

0.03 )2= 717.22

𝑛 = 718

Data Analysis Plus - Microsoft Excel

Download “Data Analysis Plus” Add-Ins from the below website and

follow the instruction to install it.

Data Analysis Plus - Microsoft Excel

Let’s solve Example 8 with using Microsoft Excel.

Data Analysis Plus - 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