skills/statistics-assignment/references/EMA1002new-Statistics-v7.txt

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Hypothesis Testing – Slide 1
Statistics
 
Hypothesis Testing for Mean

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Hypothesis Testing – Slide 2
Contents
Part 1 Population Parameter vs. Sample Statistic
Part 2 The Normal Distribution
Part 3 Hypothesis Testing for Mean
Part 4 Confidence Interval

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Hypothesis Testing – Slide 3
Learning Objectives
By the end of this chapter, students should be able to
1. distinguish between population parameters and sample statistics,
2. present a summary of sample data using a computer software,
3. describe the properties of a normal distribution,
4. determine the confidence interval of a population mean,
5. perform hypothesis tests for a claim made about a population mean.

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Hypothesis Testing – Slide 4
Introduction
What is Statistics?
Why learn Statistics?
Who uses Statistics?
What is Population? Sample?


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Hypothesis Testing – Slide 5
Part 1. Population Parameter vs. Sample Statistic
Population refers to ALL data we are 
concerned with. Parameters are 
measurements taken from population 
data, e.g. population mean (𝜇) and 
population standard deviation (𝜎).
Sample is a subset of the population. 
Statistics are measurements taken from 
sample data, e.g. sample mean ( ҧ𝑥), 
sample standard deviation (𝑠) and 
sample size (𝑛).
mean = 𝜇 (mu)
standard deviation = 𝜎 (sigma)
POPULATION
mean = ҧ𝑥 (x bar)
standard deviation = 𝑠
SAMPLE

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Hypothesis Testing – Slide 6
Part 1. Population Parameter vs. Sample Statistic
       Population mean        vs      Sample mean
𝜇 = 𝑥1 + 𝑥2 + ⋯ + 𝑥𝑁
𝑁  ҧ𝑥 = 𝑥1 + 𝑥2 + ⋯ + 𝑥𝑛
𝑛
      Population standard deviation       vs     Sample standard deviation
𝜎 = σ𝑖=1
𝑁 𝑥𝑖 − 𝜇 2
𝑁  𝑠 = σ𝑖=1
𝑛 𝑥𝑖 − ҧ𝑥 2
𝑛 − 1
Measure of Centre and Measure of Spread

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Hypothesis Testing – Slide 7
Part 1. Population Parameter vs. Sample Statistic
Consider two sets of sample data.
   Set A: 24 25 26 27 28
   Set B: 5 17 20 30 58
A B
1 Set A Set B
2 24 5
3 25 17
4 26 20
5 27 30
6 28 58
Sample Mean, ҧ𝑥 26 26 =AVERAGE(A2:A6) for Set A, =AVERAGE(B2:B6) for Set B
Sample Std. Dev., 𝑠 1.581 19.987 =STDEV.S(A2:A6) for Set A, =STDEV.S(B2:B6) for Set B
Notice this (including the Excel formulas):
1. Both sets of data give the same sample mean
2. However, Set B gives a larger standard 
deviation (= wider data spread)
3. Conversely, Set A gives a smaller standard 
deviation (= narrower data spread) 

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Hypothesis Testing – Slide 8
Part 2. The Normal Distribution
• 0 ≤ 𝑃 𝐴 ≤ 1 for any event 𝐴.
• 𝑃 𝐴 = 0 ⇒ The event 𝐴 is impossible to occur.
• 𝑃 𝐴 = 1 ⇒ The event 𝐴 will certainly occur.
The probability of an event 𝐴 happening is given by:
𝑃 𝐴 = Number of ways that 𝐴 can occur
Total number of possible outcomes

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Hypothesis Testing – Slide 9
Part 2. The Normal Distribution
Probability distribution is a description of the probabilities of all possible events.
0
0.25
0.5
0.75
1
No Heads 1 Head 2 Heads
Probabilities of 2 coin tossesConsider the number of heads obtained 
from two coin tosses. There are three 
possible events: 0, 1, 2. Such data that takes 
on countable outcomes is called discrete. 
The probability distribution of a discrete data 
can be visualized by using bar chart.

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Hypothesis Testing – Slide 10
Part 2. The Normal Distribution
Consider another scenario where the hourly 
temperature at a certain place is recorded. Such 
data that takes on any number within an interval 
is called continuous. 
The probability distribution of a continuous data 
is called probability density function (PDF).
x
f(x)
μ
PDF of the hourly-recorded temperature
Probability distribution is a description of the probabilities of all possible events.
The PDF graph above is a bell-shaped curve called the normal curve.

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Hypothesis Testing – Slide 11
Part 2. The Normal Distribution
A normal curve has the following properties:
1. It is symmetric and reaches its peak at 
𝑥 = 𝜇 then tapered-off towards zero.
2. The probability 𝑃 𝑎 < 𝑥 < 𝑏  is the 
area under the curve of the PDF 𝑓 𝑥  
over the interval 𝑎 < 𝑥 < 𝑏, i.e.
𝑃 𝑎 < 𝑥 < 𝑏 = න
𝑎
𝑏
𝑓 𝑥 𝑑𝑥
Integral of 𝑓 from 𝑎 to 𝑏, will 
be discussed in the later 
chapter of Integration
x
f(x)
μ1 μ2
Two normal PDFs with different 𝜇 but same 𝜎 
x
f(x)
μ
σ1
σ2
σ2 > σ1
x
f(x)
μ
σ1
σ2
σ2 > σ1
Two normal PDFs with same 𝜇 but different σ

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Hypothesis Testing – Slide 12
Part 2. The Normal Distribution 
–
 
Some examples

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Hypothesis Testing – Slide 13
Part 2. The Normal Distribution
The normal curve also known as the bell curve or the Gaussian curve after the German 
mathematician Carl Friedrich Gauss (1777 – 1855) and its probability distribution is an 
example of a normal distribution.
The standard normal distribution is a normal distribution with 𝜇 = 0 and 𝜎 = 1. Any 
normal distribution 𝑋 can be made into standard normal distribution 𝑍 through the 
transformation 𝑍 =
𝑋−𝜇
𝜎 .
THE NORMAL DISTRIBUTION
Let 𝑋 be a continuous variable with mean 𝜇 and standard deviation 𝜎. Then 𝑋 has a 
normal distribution if 𝑋 has the normal PDF
𝑓 𝑥 =
1
𝜎 2𝜋 𝑒− 1
2
𝑥−𝜇
𝜎
2
       for   −∞ < 𝑥 < ∞.

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Hypothesis Testing – Slide 14
Part 2. The Normal Distribution
EXAMPLE 2.1 
𝑋 is distributed normally with a mean of 22.5 and a standard deviation of 0.5. 
Calculate the probability 𝑃 22.0 < 𝑋 < 23.0 .
A B
1 µ = 22.5
2 σ = 0.5
3 X_low = 22.0
4 X_high = 23.0
5 P(X < X_low) = 0.1586=NORM.DIST(B3,B$1,B$2,TRUE)
6 P(X < X_high) = 0.8413=NORM.DIST(B4,B$1,B$2,TRUE)
7 P(X_low < X < X_high) = 0.6827=B6-B5
Area = 0.6827
22.522.0 23.0


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Hypothesis Testing – Slide 15
Part 2. The Normal Distribution
A B C D
1 µ = 22.5 Z=(X−µ)/σ
2 σ = 0.5
3 X_low = 22.0 Z_low = -1
4 X_high = 23.0 Z_high = 1
5 P(X < X_low) = 0.1586 P(Z < Z_low) = 0.1586=NORM.S.DIST(D3)
6 P(X < X_high) = 0.8413 P(Z < Z_high) = 0.8413=NORM.S.DIST(D4)
7 P(X_low < X < X_high) = 0.6827 P(Z_low < Z < Z_high) = 0.6827
Using the 
transformation
𝑍 = 𝑋 − 𝜇
𝜎
EXAMPLE 2.1 
𝑋 is distributed normally with a mean of 22.5 and a standard deviation of 0.5. 
Calculate the probability 𝑃 22.0 < 𝑋 < 23.0 .

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Hypothesis Testing – Slide 16
Part 3. Hypothesis Testing for Mean
Hypothesis testing for mean is a statistical procedure that uses sample data to test a 
claim about population mean.
Null hypothesis Alternative hypothesisvs
A statement that the value of 
the population mean 𝜇 is 
equal to some claimed value.
𝐻0:  𝜇 = 𝜇0
A statement that the value of the 
population mean 𝜇 differs from the 
value stated in the null hypothesis.
𝐻1:  𝜇 ≠ 𝜇0, or  
𝐻1:  𝜇 < 𝜇0, or
𝐻1:  𝜇 > 𝜇0 

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Hypothesis Testing – Slide 17
Part 3. Hypothesis Testing for Mean 
–
 
In greater explanation
Null hypothesis Alternative hypothesisvs
𝐻0:  𝜇 = 𝜇0
Symbol: Typically denoted as 𝐻0.
The null hypothesis (𝐻0) is a statement of no effect or 
no difference in a population parameter.
Purpose: It serves as a baseline assumption that there 
is no relationship, effect, or difference between 
groups or conditions being compared.
Either 𝐻1:  𝜇 ≠ 𝜇0,  or,   𝐻1:  𝜇 < 𝜇0,  or,  𝐻1:  𝜇 > 𝜇0
Symbols: Denoted as 𝐻1 or 𝐻a, and it can be 
The alternative hypothesis contradicts the null hypothesis, proposing that 
there is a relationship, effect, or difference in a population parameter. 
Purpose: It represents the researcher's hypothesis or the effect they are 
trying to detect in their study.
one-tailed two-tailed
Indicating a direction of the 
effect.
𝐻1:  𝜇 > 𝜇0
𝐻1:  𝜇 < 𝜇0
Indicating a difference but not 
specifying the direction, e.g. 
𝐻1:  𝜇 ≠ 𝜇0

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Hypothesis Testing – Slide 18
Part 3. Hypothesis Testing for Mean 
–
 
In greater explanation
Question: Do post-secondary students in Singapore receive an average of $14 pocket money per day?
Question inspired by The Straits Times article How much pocket money do Singapore children get? It's $4 to $14 a day. (Published 26 May 2024)
Null hypothesis (𝐻0:  𝜇 = 𝜇0) 
The average/mean (choose either word) daily pocket money received by post-secondary students in Singapore is $14
𝐻0: 𝜇 = 14
Alternative hypothesis (depends on whether we are working with either two-tailed or one-tailed test [left or right-tailed])
Two-tailed (𝐻1:  𝜇 ≠ 𝜇0) 
The avg/mean daily pocket money received by post-secondary students in Singapore is not $14.
𝐻1:  𝜇 ≠ 14
or
One-Tailed (𝐻1:  𝜇 < 𝜇0  or  𝐻1:  𝜇 > 𝜇0)
The avg/mean daily pocket money received by post-secondary students in Singapore is less than $14.
𝐻1:  𝜇 < 14 (Left-tailed test)
or
The avg/mean daily pocket money received by post-secondary students in Singapore is more than $14.
𝐻1:  𝜇 > 14 (Right-tailed test)


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Hypothesis Testing – Slide 19
Part 3. Hypothesis Testing for Mean
Sample Statistics 
Sample statistics are numerical values calculated from the sample data and are used to 
describe the sample.
Examples: 
ҧ𝑥 = sample mean,     𝑠 = sample standard deviation,     𝑛 = sample size
Test statistic:
A value that indicates how far the sample mean is from 
the population mean stated in the null hypothesis.
𝑧0 = ҧ𝑥 − 𝜇
Τ𝜎 𝑛 or 𝑡0 = ҧ𝑥 − 𝜇
Τ𝑠 𝑛
Test Statistic 
The test statistic is calculated using the relevant sample statistics, and is then 
subsequently used to determine p-value.
Used to calculate

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Hypothesis Testing – Slide 20
Part 3. Hypothesis Testing for Mean 
–
 
Significance Level & P
-
value
Recall: Hypothesis testing for mean is a statistical procedure that uses sample data to 
test a claim about population mean.
Significance Level
A small positive value, denoted by 𝛼 (alpha), where 0 < 𝛼 < 1, chosen before carrying 
out the hypothesis test. Used as a threshold to determine whether the evidence 
against the null hypothesis is statistically significant.
Common values for 𝛼: 𝛼 = 0.01  or  𝛼 = 0.05   or 𝛼 = 0.1
Hint for students: Usually given to us! ☺ (What is the 𝛼 value for our EM2 assignment?)
P-value A probability value, denoted by 𝑝, where 0 < 𝑝 < 1 calculated using the test statistic 
obtained from the sample data. It is used to measure the strength of evidence against 
the null hypothesis.
𝑝-value ≤ 𝛼     ⇒     Sufficient evidence to reject 𝐻0
𝑝-value > 𝛼     ⇒     Insufficient evidence to reject 𝐻0

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Hypothesis Testing – Slide 21
Part 3. Hypothesis Testing for Mean 
–
 
P
-
value (Two
-
tailed, 
𝐻
1
:
  
𝜇
 ≠
 𝜇
0
)
Recall: Hypothesis testing for mean is a statistical procedure that uses sample data to 
test a claim about population mean.
P-value 
(Two-tailed) For a two-tailed test, the p-value accounts for 
extreme results in both tails of the distribution:
2𝑃 𝑧 > 𝑧0
or
2𝑃 𝑡 > 𝑡0
We arrive at either of the following conclusions:
- Sufficient evidence to reject the null hypothesis if the p-value ≤ 𝛼, or,
- Insufficient evidence to reject the null hypothesis if the p-value > 𝛼.

## PDF page 22

Hypothesis Testing – Slide 22
Part 3. Hypothesis Testing for Mean 
–
 
P
-
value (One
-
tailed test with 
𝐻
1
:
  
𝜇
 <
 𝜇
0
)
Recall: Hypothesis testing for mean is a statistical procedure that uses sample data to 
test a claim about population mean.
P-value 
(One-tailed) For a left-tailed test, the p-value accounts for 
extreme results in the left tail of the distribution:
𝑃 𝑧 < 𝑧0
or
𝑃 𝑡 < 𝑡0 .
We arrive at either of the following conclusions:
- Sufficient evidence to reject the null hypothesis if the p-value ≤ 𝛼, or,
- Insufficient evidence to reject the null hypothesis if the p-value > 𝛼.

## PDF page 23

Hypothesis Testing – Slide 23
Part 3. Hypothesis Testing for Mean 
–
 
P
-
value (One
-
tailed test with 
𝐻
1
:
  
𝜇
 >
 𝜇
0
)
Recall: Hypothesis testing for mean is a statistical procedure that uses sample data to 
test a claim about population mean.
P-value 
(One-tailed) For a right-tailed test, the p-value accounts for 
extreme results in the right tail of the distribution:
𝑃 𝑧 > 𝑧0
or
𝑃 𝑡 > 𝑡0 .
We arrive at either of the following conclusions:
- Sufficient evidence to reject the null hypothesis if the p-value ≤ 𝛼, or,
- Insufficient evidence to reject the null hypothesis if the p-value > 𝛼.

## PDF page 24

Hypothesis Testing – Slide 24
Part 3. Hypothesis Testing for Mean (Two
-
tailed, 
𝐻
1
:
  
𝜇
 ≠
 𝜇
0
)
Graphical illustration of significance level vs. p-value for a two-tailed test:
Significance level P-value
When the p-value ≤ α, the observed test statistic is sufficiently far from 0, which provides sufficient 
evidence to reject the null hypothesis 𝐻0. 
Conversely, when the p-value > α, there is insufficient evidence to reject 𝐻0.
0
 |𝑧0|
−|𝑧0|
Area = P-value
0

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Hypothesis Testing – Slide 25
Part 3. Hypothesis Testing for Mean (One
-
tailed test with 
𝐻
1
:
  
𝜇
 <
 𝜇
0
)
Graphical illustration of significance level vs. p-value for a one-tailed test (left-tailed):
Significance level P-valuevs
0𝑧0
 𝑧𝛼
Area = P-value
Area = 𝛼
When the p-value ≤ α, the observed test statistic is sufficiently far from 0, which provides sufficient 
evidence to reject the null hypothesis 𝐻0. 
Conversely, when the p-value > α, there is insufficient evidence to reject 𝐻0.

## PDF page 26

Hypothesis Testing – Slide 26
Part 3. Hypothesis Testing for Mean (One
-
tailed test with 
𝐻
1
:
  
𝜇
 >
 𝜇
0
)
Graphical illustration of significance level vs. p-value for a one-tailed test (right-tailed):
Significance level P-valuevs
0 𝑧0
Area = P-value
𝑧𝛼
Area = 𝛼
When the p-value ≤ α, the observed test statistic is sufficiently far from 0, which provides sufficient 
evidence to reject the null hypothesis 𝐻0. 
Conversely, when the p-value > α, there is insufficient evidence to reject 𝐻0.

## PDF page 27

Hypothesis Testing – Slide 27
Part 3. Hypothesis Testing for Mean
HYPOTHESIS TESTING FOR MEAN
Step 1: State the null and alternative hypotheses.
Step 2: Calculate the test statistic.
Step 3: Calculate the p-value.
Step 4: State the conclusion by comparing p-value with the significance level:
There is sufficient/insufficient* evidence to reject the claim that 
[claim statement].
 *sufficient if p-value ≤ 𝛼, insufficient if p-value > 𝛼.

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Hypothesis Testing – Slide 28
Part 3. Hypothesis Testing for Mean
Case 1: Population Standard Deviation 𝜎 is known
When 𝜎 is known, we use the standard normal distribution (𝑍-distribution).
To understand why the 𝑍-distribution can be used, we first consider the distribution of the sample mean ത𝑋:
THEOREM 3.1
If the population is normally distributed, the sample mean ത𝑋 is normally distributed with:
Mean = 𝜇     and     Standard deviation = 
𝜎
𝑛
 known
➔ Therefore, when the population is normally distributed and 𝜎 is known, 
the 𝑍-distribution can be used for hypothesis testing.
It follows that  𝑍 =
ത𝑋−𝜇
𝜎/ 𝑛  has a standard normal distribution with mean 0 and standard deviation 1.
known

## PDF page 29

Hypothesis Testing – Slide 29
Part 3. Hypothesis Testing for Mean
CENTRAL LIMIT THEOREM (CLT) - THEOREM 3.2
For a sufficiently large sample size (𝑛 ≥ 30), the sample mean ത𝑋 is approximately 
normally distributed with:                   Mean = 𝜇       and       Standard deviation = 
𝜎
𝑛
But what if the population distribution is unknown (i.e. whether normally distributed or not)? 
➔ For a sufficiently large sample, the Central Limit Theorem (CLT) allows us to proceed even when the 
population distribution is unknown.
Therefore, if the pop. distribution is unknown, but the sample size is sufficiently large (𝑛 ≥ 30), and the 
pop. standard deviation 𝜎 is known, then we can calculate test statistic using:
𝑧0 = ҧ𝑥 − 𝜇
𝜎/ 𝑛

## PDF page 30

Hypothesis Testing – Slide 30
Use this Excel Function for two-tailed z-test
=2*(1-NORM.S.DIST(ABS(𝑧0),TRUE))
Part 3. Hypothesis Testing for Mean
EXAMPLE 3.1
A hospital uses large amounts of packaged doses of a 
particular medication whose individual dosage is claimed 
to be 100 cm3 by the manufacturer. 
The manufacturer assumes the population standard 
deviation to be 5 cm3. A random sample of 49 doses from 
a recent shipment gave a mean dosage of 102 cm3.
Given a 1% level of significance, test the manufacturer’s 
claim regarding drug dosage.
Step 1: 𝐻0: 𝜇 = 100 
              𝐻1: 𝜇 ≠ 100
Step 2: 𝑧0 =
102−100
5/ 49 = 2.8
Step 3: P-value = 0.00511
              ∴ P-value ≤ 0.01
Step 4: At a 1% level of significance, there 
is sufficient evidence to reject the claim 
that the medication individual dosage has 
a mean of 100 cm3.
*Hint*: For this qns, let’s 
conduct a two-tailed test!

## PDF page 31

Hypothesis Testing – Slide 31
Part 3. Hypothesis Testing for Mean
Hence, we calculate the test statistic using:     𝑡0 =
ҧ𝑥−𝜇
𝑠/ 𝑛
Case 2: Population Standard Deviation 𝜎 unknown
When 𝜎 is unknown, we use the sample standard deviation 𝑠 and the 𝑡-distribution.
To understand why the 𝑡-distribution can be used, we first consider the distribution of the test statistic:
THEOREM 3.3
If the population is normally distributed, then 𝑇 =
ത𝑋−𝜇
𝑆/ 𝑛 has a t-distribution with 
𝑛 − 1  degrees of freedom.
➔ Therefore, when the population is normally distributed and 𝜎 is unknown, 
the 𝑡-distribution can be used for hypothesis testing.

## PDF page 32

Hypothesis Testing – Slide 32
Part 3. Hypothesis Testing for Mean
THEOREM 3.4
For a sufficiently large sample size (𝑛 ≥ 30), when the pop. standard deviation 𝜎 is 
unknown, 𝑇 =
ത𝑋−𝜇
𝑆/ 𝑛 is approximately t-distributed with 𝑛 − 1  degrees of freedom.
But what if the population distribution is unknown (i.e. whether normally distributed or not), 
in addition to 𝜎 unknown?
Recall from Slide 29: For a sufficiently large sample (𝑛 ≥ 30), the Central Limit Theorem (CLT) allows us 
to proceed even when the pop. distribution is unknown. 
Recall from Slide 31: When 𝜎 is unknown, we instead use the sample standard deviation 𝑠 (to estimate 𝜎).
Hence, we can still calculate the test statistic using:     𝑡0 =
ҧ𝑥−𝜇
𝑠/ 𝑛

## PDF page 33

Hypothesis Testing – Slide 33
Use this Excel Function for left-tailed t-test
=T.DIST(𝑡0,𝑛 − 1,TRUE)
Part 3. Hypothesis Testing for Mean
EXAMPLE 3.2 
A manufacturer claims that the mean talk-time of its phone 
batteries is 80 hours fully charged. 
A consumer watchdog believes the mean talk-time is less 
than 80 hours, because when they tested a random 
samples of 51 batteries from the manufacturer, it gave a 
sample mean of ҧ𝑥 = 78.83 and sample standard deviation 
of 𝑠 = 9.04.
Test the claim of the manufacturer at 5% significance level.
Step 1: 𝐻0: 𝜇 = 80 
             𝐻1: 𝜇 < 80      ( Why?)
Step 2: 𝑡0 =
78.83.−80
9.04/ 51 = −0.9243
Step 3: P-value = 0.1798
           ∴ P-value > 0.05
Step 4: At a 5% level of significance, 
there is insufficient evidence to reject 
the manufacturer’s claim that the mean 
talk-time of its phone batteries is 80 
hours when fully charged.
*Hint*: For this qns, let’s 
conduct a left-tailed test!

## PDF page 34

Hypothesis Testing – Slide 34
Part 4. Confidence Interval
For example, given 𝛼 = 5%, the corresponding 95% confidence interval for the population mean 𝜇 lies 
between ҧ𝑥 − 𝐸0.05 and ҧ𝑥 + 𝐸0.05, 
i.e. ҧ𝑥 − 𝐸0.05 < 𝜇 < ҧ𝑥 + 𝐸0.05
The 100 1 − 𝛼 % confidence interval for 𝜇 is written as ҧ𝑥 ± 𝐸𝛼.
CONFIDENCE INTERVAL FOR MEAN
Suppose one the following conditions is met: the sample size 𝑛 ≥ 30, or the 
population is normally distributed.
Given a significance level 𝛼, the corresponding confidence level is:     100 1 − α %, 
and the confidence interval for 𝜇 is:     ҧ𝑥 − 𝐸𝛼 < 𝜇 < ҧ𝑥 + 𝐸𝛼
where 𝐸𝛼 is the margin of error associated with 𝛼.

## PDF page 35

Hypothesis Testing – Slide 35
Part 4. Confidence Interval
Good news!  In practice, we can use a computer to find the value of 𝑬𝜶 .
For instance, in Microsoft Excel, we use the statistical function 
=CONFIDENCE.NORM (when 𝜎 is known) or =CONFIDENCE.T (when 𝜎 is unknown).
Area = 
Z
The margin of error 𝐸𝛼 is computed as follows:
If 𝜎 is known, then  𝐸𝛼 = 𝑧 Τ𝛼 2
𝜎
𝑛 with 𝑃 𝑍 > 𝑧 Τ𝛼 2 =
𝛼
2
If 𝜎 is unknown, then 𝐸𝛼 = 𝑡 Τ𝛼 2
𝑠
𝑛 with 𝑃 𝑇 > 𝑡 Τ𝛼 2 =
𝛼
2

## PDF page 36

Hypothesis Testing – Slide 36
Part 4. Confidence Interval
EXAMPLE 4.1 
A random sample of 100 observations with a sample 
mean 22.5 is taken from a population. 
Find the 90% confidence interval for 𝜇 if
(a) The population standard deviation is 0.5,
(b) The sample standard deviation is 0.75.
A B
1 n = 100
2 α = 0.1
3 σ = 0.5
4 s = 0.75
5 E = 0.082243 =CONFIDENCE.NORM(B2,B3,B1)
6 E = 0.124529 =CONFIDENCE.T(B2,B4,B1)
1 − 𝛼 = 0.90
   ⇒ 𝛼 = 0.1
(a)  𝐸0.1 = 0.082
90% confidence interval for 𝜇 is
22.5 − 0.082 < 𝜇 < 22.5 + 0.082
⇒ 22.418 < 𝜇 < 22.582
(b)  𝐸0.1 = 0.125
 90% confidence interval for 𝜇 is
 22.5 − 0.125 < 𝜇 < 22.5 + 0.125
 ⇒ 22.375 < 𝜇 < 22.625

## PDF page 37

Hypothesis Testing – Slide 37
Summary
Condition 𝝈
Confidence Interval for 
Mean Hypothesis Testing for Mean
ҧ𝑥 − 𝐸𝛼 < 𝜇 < ҧ𝑥 + 𝐸𝛼 Test Statistic P-value
* The formulas below are for two-tailed test, 
adjust accordingly when performing one-tail test (L or R)
Population is
Normally 
distributed
OR
Sample size 
n ≥ 30
Known 𝐸𝛼 = 𝑧 Τ𝛼 2
𝜎
𝑛
=CONFIDENCE.NORM(𝛼,𝜎,𝑛)
𝑧0 = ҧ𝑥 − 𝜇
Τ𝜎 𝑛
2𝑃 𝑧 > 𝑧0
=2*(1-NORM.S.DIST( 𝑧0 ,TRUE))
Unknown
𝐸𝛼 = 𝑡 Τ𝛼 2
𝑠
𝑛
=CONFIDENCE.T(𝛼,s,𝑛) 𝑡0 = ҧ𝑥 − 𝜇
Τ𝑠 𝑛
2𝑃 𝑡 > 𝑡0
=2*(T.DIST.RT( 𝑡0 ,𝑛 − 1))

## PDF page 38

Hypothesis Testing – Slide 38
Flowchart for
Hypothesis Test
Step 1: State the null and alternative hypotheses.
Step 2: Calculate the test statistic.
Step 3: Calculate the p-value.
Step 4: State the conclusion by comparing p-value 
with the significance level:
There is sufficient/insufficient* evidence to 
reject the claim that [claim statement].
 *sufficient if p-value ≤ 𝛼,
 insufficient if p-value > 𝛼


## PDF page 39

Hypothesis Testing – Slide 39
Flowchart for
Confidence Interval


## PDF page 40

Hypothesis Testing – Slide 40
Reference
Montgomery, D.C., Runger, G.C. and Hubele, N.F. (2012). Engineering Statistics (Fifth 
Edition): Chapters 3 and 4. Singapore: John Wiley & Sons Asia.