- The document discusses hypothesis testing through a series of examples. It introduces key terminology like the null hypothesis (H0), alternative hypothesis (H1), significance level (α), calculated value, and tabulated value. - An example tests whether the mean savings of VIP bank customers is greater than LE 5.6 million based on a sample. The calculated z-value is compared to the critical value to reject the null hypothesis. - A second example tests whether the population mean is different than LE 6 million, calculating the z-value and comparing it to the critical value for a two-tailed test.

1. Quantitative Methods
Dr. Mohamed Ramadan
Moh_Ramadan@icloud.com

2. Quantitative Methods
Introduction to
Hypothesis Testing
Lecture (5)

3. Lecture (5) Introduction to Hypothesis Testing
Revision
𝒇𝒙
𝒁
𝟎. 𝟎𝟓𝟎. 𝟒𝟓𝟎. 𝟓
𝟎. 𝟓 𝟎. 𝟓
𝑃 0 < 𝑍 < 𝑧! = 0.45
𝑃 𝑍 < 𝑧! = 0.95
𝑃 𝑍 > 𝑧! = 0.05
𝑧! = 1.65
𝑧! = 1.65
𝑧! = 1.65
𝟏. 𝟔𝟓
𝑧!
𝑧!
𝑧!

4. Lecture (5) Introduction to Hypothesis Testing
Revision
𝐻!: 𝜇 = 19
𝐻": 𝜇 > 19
Test this hypotheses at
5% significance level
Calculated
Value
Alternative
Hypothesis
𝜶
Rejection
Rejoin
Acceptance
Rejoin
𝑧#
𝑍 =
̅𝑥 − 𝜇
𝜎/ 𝑛
Tabulated
Value
Reject the
Null
HypothesisCouldn’t Reject
the Null
Hypothesis
Null
Hypothesis
𝑍
𝑍

5. Lecture (5) Introduction to Hypothesis Testing
The Idea ... Concept and Terminologies
𝐻/: 𝜇 = 19
𝐻0: 𝜇 > 19
Test this hypotheses at
5% significance level
𝐻/: 𝜇 = 19
𝐻0: 𝜇 < 19
Test this hypotheses at
5% significance level
𝐻/: 𝜇 = 19
𝐻0: 𝜇 ≠ 19
Test this hypotheses at
5% significance level
𝜶 = 𝟎. 𝟎𝟓
𝜶 = 𝟎. 𝟎𝟓
𝜶
𝟐
=
𝟎. 𝟎𝟓
𝟐
𝜶
𝟐
=
𝟎. 𝟎𝟓
𝟐
𝒛 𝒕𝒂𝒃𝒍𝒖𝒍𝒂𝒕𝒆𝒅
𝒛 𝒕𝒂𝒃𝒍𝒖𝒍𝒂𝒕𝒆𝒅 −𝒛 𝒕𝒂𝒃𝒍𝒖𝒍𝒂𝒕𝒆𝒅 𝒛 𝒕𝒂𝒃𝒍𝒖𝒍𝒂𝒕𝒆𝒅
𝒛 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅
𝒛 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅 𝒛 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅
Couldn’t reject 𝐻!
Reject 𝐻!

6. Lecture (5) Introduction to Hypothesis Testing
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-1- Your manager would like to know the average
population mean, could you give him/her an
accurate answer?
-1- We don’t know the unknown population mean.
Therefore, I could not give you sir an accurate answer.
Practicing Testing of Hypotheses

7. Lecture (5) Introduction to Hypothesis Testing
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-2- If you don’t have an accurate answer, and if you
know that the population variance is 250 thousand,
could you give your manager an estimate to the
unknown population mean with an about 95%
confidence limit.
-2-
𝑛 = 100 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
̅𝑥 = 5.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝜎" = 250,000 ≫ 0.25
𝜎 = 0.5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
̅𝑥 ± 𝑍#$!
"
𝜎
𝑛
5.7 ± 𝑍%.'(
"
0.5
100
= 5.7 ± 𝑍%.)*(
0.5
10
= 5.7 ± 𝑍%.)*( 0.05

8. Lecture (5) Introduction to Hypothesis Testing
-2-
𝑛 = 100 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
̅𝑥 = 5.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝜎" = 250,000 ≫ 0.25
𝜎 = 0.5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
̅𝑥 ± 𝑍#$!
"
𝜎
𝑛
5.7 ± 𝑍%.'(
"
0.5
100
= 5.7 ± 𝑍%.)*(
0.5
10
= 5.7 ± 𝑍%.)*( 0.05 = 5.7 ± 1.96 0.05
0.4750
0.06
1.9

9. Lecture (5) Introduction to Hypothesis Testing
-2-
𝑛 = 100 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
̅𝑥 = 5.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝜎" = 250,000 ≫ 0.25
𝜎 = 0.5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
̅𝑥 ± 𝑍#$!
"
𝜎
𝑛
5.7 ± 𝑍%.'(
"
0.5
100
= 5.7 ± 𝑍%.)*(
0.5
10
= 5.7 ± 𝑍%.)*( 0.05 = 5.7 ± 1.96 0.05
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-2- If you don’t have an accurate answer, and if you
know that the population variance is 250 thousand,
could you give your manager an estimate to the
unknown population mean with an about 95%
confidence limit.
= 5.7 ± 0.098

10. Lecture (5) Introduction to Hypothesis Testing
-2-
𝑛 = 100 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
̅𝑥 = 5.7 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝜎" = 250,000 ≫ 0.25
𝜎 = 0.5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
̅𝑥 ± 𝑍#$!
"
𝜎
𝑛
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-2- If you don’t have an accurate answer, and if you
know that the population variance is 250 thousand,
could you give your manager an estimate to the
unknown population mean with an about 95%
confidence limit.
= 5.7 ± 0.098
Therefore you can answer to your manager ... The
confidence interval that could include the unknow
population mean with a confidence level 95% is:
5.7 − 0.098 < 𝜇 < 5.7 + 0.098
5.602 < 𝜇 < 5.798

11. Lecture (5) Introduction to Hypothesis Testing
-3-
𝐻%: 𝜇 = 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 > 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-3- Furtherly, your manager ask you to test if the
VIP customers’ savings greater than L.E. 5.6 million,
with a significance level 0.05?
𝑍+,-+.-,/01 =
̅𝑥 − 𝜇
𝜎/ 𝑛
=
5.7 − 5.6
0.5/ 100
=
0.1
0.5/10
=
0.1
0.05
𝑍+,-+.-,/01 = 2.0
𝑧/,2.-,/01 = 𝑧! = 𝑧%.%(
𝜶 = 𝟎. 𝟎𝟓𝟎. 𝟒𝟓

12. Lecture (5) Introduction to Hypothesis Testing
-3-
𝐻%: 𝜇 = 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 > 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝑍+,-+.-,/01 =
̅𝑥 − 𝜇
𝜎/ 𝑛
=
5.7 − 5.6
0.5/ 100
=
0.1
0.5/10
=
0.1
0.05
𝑍+,-+.-,/01 = 2.0
𝑧/,2.-,/01 = 𝑧! = 𝑧%.%( = 1.65
𝜶 = 𝟎. 𝟎𝟓𝟎. 𝟒𝟓
0.4505
0.05
1.6

13. Lecture (5) Introduction to Hypothesis Testing
-3-
𝐻%: 𝜇 = 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 > 5.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝑍+,-+.-,/01 = 2.0 𝑧/,2.-,/01 = 1.65
𝟐. 𝟎
𝟏. 𝟔𝟓
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-3- Furtherly, your manager ask you to test if the
VIP customers’ savings greater than L.E. 5.6 million,
with a significance level 0.05?
∵ 𝑍 > 𝑧!
∴ 𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑛 𝑓𝑎𝑣𝑜𝑟 𝑜𝑓 𝐻#
≡ 𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑎𝑙 𝑆𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒
Rejection
Rejoin
Acceptance
Rejoin

14. Lecture (5) Introduction to Hypothesis Testing
-4-
𝐻%: 𝜇 = 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 ≠ 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-4- After the previous step, your manager would like
you to test if the unknown population mean is not
equal L.E. 6.0 million, with a significance level 0.05?
𝑍+,-+.-,/01 =
̅𝑥 − 𝜇
𝜎/ 𝑛
=
5.7 − 6.0
0.5/ 100
=
−0.3
0.5/10
=
−0.3
0.05
𝑍+,-+.-,/01 = −6.0
𝑧/,2.-,/01 = 𝑧!/" = 𝑧%.%"(
𝜶
= 𝟎. 𝟎𝟐𝟓𝟎. 𝟒𝟕𝟓𝟎. 𝟒𝟕𝟓
𝜶
= 𝟎. 𝟎𝟐𝟓

15. Lecture (5) Introduction to Hypothesis Testing
-4-
𝐻%: 𝜇 = 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 ≠ 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝑍+,-+.-,/01 =
̅𝑥 − 𝜇
𝜎/ 𝑛
=
5.7 − 6.0
0.5/ 100
=
−0.3
0.5/10
=
−0.3
0.05
𝑍+,-+.-,/01 = −6.0
𝑧/,2.-,/01 = 𝑧!/" = 𝑧%.%"( = 1.96
𝜶
= 𝟎. 𝟎𝟐𝟓𝟎. 𝟒𝟕𝟓𝟎. 𝟒𝟕𝟓
𝜶
= 𝟎. 𝟎𝟐𝟓
0.4750
0.06
1.9

16. Lecture (5) Introduction to Hypothesis Testing
-4-
𝐻%: 𝜇 = 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
𝐻#: 𝜇 ≠ 6.0 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝐿. 𝐸.
VIP Customers’ Savings
Suppose that you are working in the insurance
department at Pro-Bank ... Your manager asked you
to select a random sample from the VIP
customers, to estimate the mean of customer
savings. Therefore, you decided to select a sample
of 100 customers, which yield an average of L.E.
5.7 million.
-4- After the previous step, your manager would like
you to test if the unknown population mean is not
equal L.E. 6.0 million, with a significance level 0.05?
−𝟔. 𝟎
Rejection
Rejoin
Acceptance
Rejoin
𝟏.𝟗𝟔
−𝟏.𝟗𝟔
𝑍+,-+.-,/01 = −6.0 𝑧/,2.-,/01 = −1.96 & 1.96
∵ 𝑍 < −𝑧!
∴ 𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑛 𝑓𝑎𝑣𝑜𝑟 𝑜𝑓 𝐻#
≡ 𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐𝑎𝑙 𝑆𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒

17. Lecture (5) Introduction to Hypothesis Testing
The smaller the p-value, the
more statistical evidence exists to
support the alternative
hypothesis.
P-Value Approach
Overwhelming
Evidence
(Highly
Significant)
Strong Evidence
(Significant) Weak Evidence
(Not Significant) No Evidence
(Not Significant)
0 .01 .05 .10 1.0

18. Lecture (5) Introduction to Hypothesis Testing
Average Age of Year 3
Suppose that a sample of 20 students was randomly
selected from the third-year students. Suppose that
we need to test if the average cohort students' age
is not equal 21 years, at a significance level 0.05.
It is well known that the cohort students' age
standard deviation is 1.4 years.
𝑍+,-+.-,/01
𝑧/,2.-,/01
𝑝 − 𝑣𝑎𝑙𝑢𝑒
𝐻%: 𝜇 = 21
𝐻#: 𝜇 ≠ 21
̅𝑥
𝜎"
𝑛

19. Lecture (5) Introduction to Hypothesis Testing
Average Age of Year 3
Suppose that a sample of 20 students was randomly
selected from the third-year students. Suppose that
we need to test if the average cohort students' age
is not equal 21 years, at a significance level 0.05.
It is well known that the cohort students' age
standard deviation is 1.4 years.
𝑍+,-+.-,/01
𝑧/,2.-,/01
𝑝 − 𝑣𝑎𝑙𝑢𝑒
∵ 𝑝 − 𝑣𝑎𝑙𝑢𝑒 < 0.01
∴ 𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙
ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑛 𝑓𝑎𝑣𝑜𝑟 𝑜𝑓 𝐻#
≡ 𝐻𝑖𝑔ℎ𝑙𝑦 𝑆𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒
𝐻%: 𝜇 = 21
𝐻#: 𝜇 ≠ 21
̅𝑥
𝜎"
𝑛