This document discusses various statistical hypothesis tests including z-tests, t-tests, F-tests, and chi-square tests. It provides examples and explanations of how to perform hypothesis tests to test for differences between means and variances. It discusses key concepts like type I and type II errors, level of significance, critical regions, and test statistics. Formulas and steps are provided for performing z-tests, t-tests, and F-tests on single and two sample data. Examples of applying these tests to real data sets are also included.
Research methodology - Estimation Theory & Hypothesis Testing, Techniques of ...The Stockker
Fundamentals, Standard Error, Estimation, Interval Estimation, Hypothesis, Characteristics of Hypothesis, Testing The Hypothesis, Type I & Type II error, One tailed & Two tailed test, Tabulated Values, Chi-square (2) Test, Analysis of variance (ANOVA)Introduction, The Sign Test, The rank sum test or The Mann-Whitney U test, Determination of Sample Size
OBJECTIVES:
Run the test of hypothesis for mean difference using paired samples. Construct a confidence interval for the difference in population means using paired samples.
Observation of interest will be the difference in the readings
before and after intervention called paired difference observation.
Paired t test:
A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap).
When there is a relationship between the groups, such as identical twins.
This test is concerned with the pair-wise differences
between sets of data.
This means that each data point in one group has a related data point in the other group (groups always have equal numbers).
ASSUMPTIONS:
The sample or samples are randomly selected
The sample data are dependent
The distribution of differences is approximately normally
distributed.
Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and
offering a wide range of dental certified courses in different formats.for more details please visit
www.indiandentalacademy.com
This 10 hours class is intended to give students the basis to empirically solve statistical problems. Talk 1 serves as an introduction to the statistical software R, and presents how to calculate basic measures such as mean, variance, correlation and gini index. Talk 2 shows how the central limit theorem and the law of the large numbers work empirically. Talk 3 presents the point estimate, the confidence interval and the hypothesis test for the most important parameters. Talk 4 introduces to the linear regression model and Talk 5 to the bootstrap world. Talk 5 also presents an easy example of a markov chains.
All the talks are supported by script codes, in R language.
Descriptive Statistics Formula Sheet Sample Populatio.docxsimonithomas47935
Descriptive Statistics Formula Sheet
Sample Population
Characteristic statistic Parameter
raw scores x, y, . . . . . X, Y, . . . . .
mean (central tendency) M =
∑ x
n
μ =
∑ X
N
range (interval/ratio data) highest minus lowest value highest minus lowest value
deviation (distance from mean) Deviation = (x − M ) Deviation = (X − μ )
average deviation (average
distance from mean)
∑(x − M )
n
= 0
∑(X − μ )
N
sum of the squares (SS)
(computational formula) SS = ∑ x
2 −
(∑ x)2
n
SS = ∑ X2 −
(∑ X)2
N
variance ( average deviation2 or
standard deviation
2
)
(computational formula)
s2 =
∑ x2 −
(∑ x)2
n
n − 1
=
SS
df
σ2 =
∑ X2 −
(∑ X)2
N
N
standard deviation (average
deviation or distance from mean)
(computational formula) s =
√∑ x
2 −
(∑ x)2
n
n − 1
σ =
√∑ X
2 −
(∑ X)2
N
N
Z scores (standard scores)
mean = 0
standard deviation = ± 1.0
Z =
x − M
s
=
deviation
stand. dev.
X = M + Zs
Z =
X − μ
σ
X = μ + Zσ
Area Under the Normal Curve -1s to +1s = 68.3%
-2s to +2s = 95.4%
-3s to +3s = 99.7%
Using Z Score Table for Normal Distribution
(Note: see graph and table in A-23)
for percentiles (proportion or %) below X
for positive Z scores – use body column
for negative Z scores – use tail column
for proportions or percentage above X
for positive Z scores – use tail column
for negative Z scores – use body column
to discover percentage / proportion between two X values
1. Convert each X to Z score
2. Find appropriate area (body or tail) for each Z score
3. Subtract or add areas as appropriate
4. Change area to % (area × 100 = %)
Regression lines
(central tendency line for all
points; used for predictions
only) formula uses raw
scores
b = slope
a = y-intercept
y = bx + a
(plug in x
to predict y)
b =
∑ xy −
(∑ x)(∑ y)
n
∑ x2 −
(∑ x)2
n
a = My - bMx
where My is mean of y
and Mx is mean of x
SEest (measures accuracy of predictions; same properties as standard deviation)
Pearson Correlation Coefficient
(used to measure relationship;
uses Z scores)
r =
∑ xy−
(∑ x)(∑ y)
n
√(∑ x2−
(∑ x)2
n
)(∑ y2−
(∑ y)2
n
)
r =
degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟
degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
r
2
= estimate or % of accuracy of predictions
PSYC 2317 Mark W. Tengler, M.S.
Assignment #9
Hypothesis Testing
9.1 Briefly explain in your own words the advantage of using an alpha level (α) = .01
versus an α = .05. In general, what is the disadvantage of using a smaller alpha
level?
9.2 Discuss in your own words the errors that can be made in hypothesis testing.
a. What is a type I error? Why might it occur?
b. What is a type II error? How does it happen?
9.3 The term error is used in two different ways in the context of a hypothesis test.
First, there is the concept of sta
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Research methodology - Estimation Theory & Hypothesis Testing, Techniques of ...The Stockker
Fundamentals, Standard Error, Estimation, Interval Estimation, Hypothesis, Characteristics of Hypothesis, Testing The Hypothesis, Type I & Type II error, One tailed & Two tailed test, Tabulated Values, Chi-square (2) Test, Analysis of variance (ANOVA)Introduction, The Sign Test, The rank sum test or The Mann-Whitney U test, Determination of Sample Size
OBJECTIVES:
Run the test of hypothesis for mean difference using paired samples. Construct a confidence interval for the difference in population means using paired samples.
Observation of interest will be the difference in the readings
before and after intervention called paired difference observation.
Paired t test:
A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap).
When there is a relationship between the groups, such as identical twins.
This test is concerned with the pair-wise differences
between sets of data.
This means that each data point in one group has a related data point in the other group (groups always have equal numbers).
ASSUMPTIONS:
The sample or samples are randomly selected
The sample data are dependent
The distribution of differences is approximately normally
distributed.
Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and
offering a wide range of dental certified courses in different formats.for more details please visit
www.indiandentalacademy.com
This 10 hours class is intended to give students the basis to empirically solve statistical problems. Talk 1 serves as an introduction to the statistical software R, and presents how to calculate basic measures such as mean, variance, correlation and gini index. Talk 2 shows how the central limit theorem and the law of the large numbers work empirically. Talk 3 presents the point estimate, the confidence interval and the hypothesis test for the most important parameters. Talk 4 introduces to the linear regression model and Talk 5 to the bootstrap world. Talk 5 also presents an easy example of a markov chains.
All the talks are supported by script codes, in R language.
Descriptive Statistics Formula Sheet Sample Populatio.docxsimonithomas47935
Descriptive Statistics Formula Sheet
Sample Population
Characteristic statistic Parameter
raw scores x, y, . . . . . X, Y, . . . . .
mean (central tendency) M =
∑ x
n
μ =
∑ X
N
range (interval/ratio data) highest minus lowest value highest minus lowest value
deviation (distance from mean) Deviation = (x − M ) Deviation = (X − μ )
average deviation (average
distance from mean)
∑(x − M )
n
= 0
∑(X − μ )
N
sum of the squares (SS)
(computational formula) SS = ∑ x
2 −
(∑ x)2
n
SS = ∑ X2 −
(∑ X)2
N
variance ( average deviation2 or
standard deviation
2
)
(computational formula)
s2 =
∑ x2 −
(∑ x)2
n
n − 1
=
SS
df
σ2 =
∑ X2 −
(∑ X)2
N
N
standard deviation (average
deviation or distance from mean)
(computational formula) s =
√∑ x
2 −
(∑ x)2
n
n − 1
σ =
√∑ X
2 −
(∑ X)2
N
N
Z scores (standard scores)
mean = 0
standard deviation = ± 1.0
Z =
x − M
s
=
deviation
stand. dev.
X = M + Zs
Z =
X − μ
σ
X = μ + Zσ
Area Under the Normal Curve -1s to +1s = 68.3%
-2s to +2s = 95.4%
-3s to +3s = 99.7%
Using Z Score Table for Normal Distribution
(Note: see graph and table in A-23)
for percentiles (proportion or %) below X
for positive Z scores – use body column
for negative Z scores – use tail column
for proportions or percentage above X
for positive Z scores – use tail column
for negative Z scores – use body column
to discover percentage / proportion between two X values
1. Convert each X to Z score
2. Find appropriate area (body or tail) for each Z score
3. Subtract or add areas as appropriate
4. Change area to % (area × 100 = %)
Regression lines
(central tendency line for all
points; used for predictions
only) formula uses raw
scores
b = slope
a = y-intercept
y = bx + a
(plug in x
to predict y)
b =
∑ xy −
(∑ x)(∑ y)
n
∑ x2 −
(∑ x)2
n
a = My - bMx
where My is mean of y
and Mx is mean of x
SEest (measures accuracy of predictions; same properties as standard deviation)
Pearson Correlation Coefficient
(used to measure relationship;
uses Z scores)
r =
∑ xy−
(∑ x)(∑ y)
n
√(∑ x2−
(∑ x)2
n
)(∑ y2−
(∑ y)2
n
)
r =
degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟
degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦
r
2
= estimate or % of accuracy of predictions
PSYC 2317 Mark W. Tengler, M.S.
Assignment #9
Hypothesis Testing
9.1 Briefly explain in your own words the advantage of using an alpha level (α) = .01
versus an α = .05. In general, what is the disadvantage of using a smaller alpha
level?
9.2 Discuss in your own words the errors that can be made in hypothesis testing.
a. What is a type I error? Why might it occur?
b. What is a type II error? How does it happen?
9.3 The term error is used in two different ways in the context of a hypothesis test.
First, there is the concept of sta
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
1. Probability & Statistics
Dr. Santosh Kumar Yadav
Assistant Professor
Department of Mathematics
Lovely Professional University, Punjab.
Dr. Santosh Yadav, LPU Punjab 1 / 45
2. UNIT-5
Testing of Hypothesis
Types of Error, Student t-test for single mean and difference
of means, Z-test for single mean and difference of means,
F-test, goodness of fit, Chi- Square Test
Dr. Santosh Yadav, LPU Punjab 2 / 45
3. Statistical Hypothesis?
Hypothesis is some statement or assertion about population
which we want to test (or verify) on the basis of available
information from sample.
Null Hypothesis: A hypothesis which is tested for possible
rejection is called null hypothesis, denoted by H0.
A hypothesis of no difference is H0. That means, a previos
information or observation which we already have is taken as
H0.
Alternate hypothesis: A hypothesis which is contradictory
or complementry to the null hypothesis, is called alternate hy-
pothesis. i.e., Anything other than null hypothesis is alternate
hypothesis. It is denoted by H1 or sometimes Ha.
Dr. Santosh Yadav, LPU Punjab 3 / 45
4. Some Remarks
Whenever we are testing any null hypothesis (H0) against al-
ternate hypothesis. Then, corresponding to null hypothesis,
alternate hypothesis can be two tailed, one tailed.
Whenever we applying test, whether H0 will be rejected or
accepted, then for this purpose whatever region we choose,
we define critical region and acceptance region.
Dr. Santosh Yadav, LPU Punjab 4 / 45
5. Acceptance and Rejection region
The region where H0 is rejected when it is true, is called
rejection region or critical region.
The region where H0 is accepted when it is false, is called
acceptance region.
Dr. Santosh Yadav, LPU Punjab 5 / 45
6. Types of Errors
The decision to accept or reject the H0 is made on the basis
of the information from the observed sample observations.
This means the decision (conclusion) based on the sample
may not be always true in respect to the population.
This means any testing problem involves two types of errors.
Dr. Santosh Yadav, LPU Punjab 6 / 45
7. Type I Error
The error of rejecting H0 when it is true, is called type I error.
It is denoted by α.
α = Probability of type I error
= Probability of rejecting H0, when it is true
= P(x ∈ W|H0 is true)
=
Z
W
L0dx
where L0 is likelihood function of sample observation under
H0.
Dr. Santosh Yadav, LPU Punjab 7 / 45
8. Type II Error
The error of accepting H0 (rejecting H0) when H0 is false, is
called type II error. It is denoted by β.
β = Probability of type II error
= Probability of accepting H0, when it is false
= P(x ∈ W|H0 is false)
= P(x ∈ W|H1 is true)
=
Z
W
L1dx
where L1 is likelihood function of sample observation under
H1.
Dr. Santosh Yadav, LPU Punjab 8 / 45
9. Cont...
Level of Significance: The constant α is also called level
of significance. i.e., the probability of error in accepting or
rejecting H0.
Level of Confidence: C = (1 − α)%. This means that if
LoS, α = 0.5 then LoC is 95%.
Power of the Test: 1 − β.
Dr. Santosh Yadav, LPU Punjab 9 / 45
10. Problem 1
If x ≥ 1 is the critical region for testing H0 : θ = 2 against
the alternative H1 : θ = 1, on the basis of single observation
from the population,
f(x, θ) = θ e−θx
, x ≥ 0
Find the size of type I and type II errors.
Dr. Santosh Yadav, LPU Punjab 10 / 45
11. Problem 1
If x ≥ 1 is the critical region for testing H0 : θ = 2 against
the alternative H1 : θ = 1, on the basis of single observation
from the population,
f(x, θ) = θ e−θx
, x ≥ 0
Find the size of type I and type II errors.
α = {x ∈ W|H0 is true}
α = {x ≥ 1|θ = 2}
Dr. Santosh Yadav, LPU Punjab 10 / 45
12. Problem 2
For the distribution
f(x, θ) =
1
θ
, 0 ≤ x ≤ θ
and that you are testing the null hypothesis: H0 : θ = 1
against H1 : θ = 2, by means of a single observed value of
x. Find the size of type I and type II errors if you choose the
interval x ≥ 0.5.
Ans: α = 0.5 and β = 0.25
Dr. Santosh Yadav, LPU Punjab 11 / 45
13. Steps of Hypothesis Testing
Define null and alternate hypothesis
Test statistic: (a formula under H0)
Level of significance (α)
Decision: Reject or accept H0.
Dr. Santosh Yadav, LPU Punjab 12 / 45
14. Z-test for single mean
Define H0 : µ = µ0 and H1 : µ ̸= µ0
The test statistic: under null hypothesis
Z =
x̄ − E(X̄)
S.E(X̄)
=
x̄ − µ
σ/
√
n
follow standard normal distribution .
Level of significance:
Decision: If Calculated |z| < tabulated z-value, at α LoS,
then Accept null hypothesis H0.
If Calculated |z| > tabulated z-value, at α LoS, then reject null
hypothesis.
Dr. Santosh Yadav, LPU Punjab 13 / 45
15. t-test for single mean
Define H0 : µ = µ0 and H1 : µ ̸= µ0
The test statistic: under null hypothesis
t =
x̄ − E(X̄)
S.D(X̄)
=
x̄ − µ
S/
√
n
=
x − µ
S
√
n
follow t-distribution with (n-1) degree of freedom. Here S2
is
unbaised estimate of population variance.
Level of significance:
Decision: If Calculated, tα,(n−1) tabulated, tα,(n−1), then
accept H0.
If Calculated, tα,(n−1) tabulated, tα,(n−1), then reject H0.
Dr. Santosh Yadav, LPU Punjab 14 / 45
16. Example 1
A sample of 900 members has a mean 3.4 cms and s.d. 2.61
cms. Test whether sample is taken from a large population
of mean 3.25 cms and s.d. 2,61 cms. (α = 5%)
Dr. Santosh Yadav, LPU Punjab 15 / 45
17. Example 1
A sample of 900 members has a mean 3.4 cms and s.d. 2.61
cms. Test whether sample is taken from a large population
of mean 3.25 cms and s.d. 2,61 cms. (α = 5%)
test statistic under H0: Z = 3.40−3.25
2.61/30
= 1.73
Decision: accept hull hypothesis at 5% LoS. Means sample
is drawn from a population of mean 3.25 cms.
Dr. Santosh Yadav, LPU Punjab 15 / 45
18. Problem 2
The heights of 10 males of a society are 70, 67, 62, 68, 61,
68, 70, 64, 64, 66. Is it reasonable to believ that the average
height is greater than 64 inches. Test at 5% LoS (taking that
for 9 degree of freedom, t=1.83).
Dr. Santosh Yadav, LPU Punjab 16 / 45
19. Problem 2
The heights of 10 males of a society are 70, 67, 62, 68, 61,
68, 70, 64, 64, 66. Is it reasonable to believ that the average
height is greater than 64 inches. Test at 5% LoS (taking that
for 9 degree of freedom, t=1.83).
H0 : µ = 64 inches
H1 : µ 64inches
test statistic under H0 : t = x̄−µ
S/
√
n
find X̄, then S2
for given
heights: X̄ = 66 and S2
= 90/9 = 10 and hence t = 2.
From the table t value for 9 d.f and 5% LoS (for right tailed
alternate is = 1.83).
Decision: Reject H0, i.e, accept H1, hence we can say that
average height of males is greater than 64 inch.
Dr. Santosh Yadav, LPU Punjab 16 / 45
20. Problem 3
A normal polulation has mean of 0.1 and s.d. of 2.1. Find
the probability that mean of sample of 900 will be negative.
Ans: P(x̄ 0) = P(Z −1.43) = 0.0764
Dr. Santosh Yadav, LPU Punjab 17 / 45
21. Problem 4
A random sample of size 16 has 53 as a mean. The sum of
squares of the deviation taken from mean is 135. Can this
sample be regarded as taken from population having 56 as
a mean. (Given that LoS= 5% and t15,0.05 = 2.131)
Dr. Santosh Yadav, LPU Punjab 18 / 45
22. Problem 5
A machinist is making engine parts with axle diameters of
0.700 inch. A random sample of 10 parts shows a mean
diameter of 0.742 inch with a s.d. of 0.040 inch. Compute
statistics you would use to test whether the work is meeting
the specifications. (Given that LoS= 5% and t9,0.05 = 2.26)
Dr. Santosh Yadav, LPU Punjab 19 / 45
23. Z-test for difference of means
H0 : µ1 = µ2
Test statistic under null hypothesis:
Z =
x1 − x2
r
σ2
1
n1
+
σ2
2
n2
follow standard normal distribution.
Decision: As difference is significant or not.
Note: If s.d of both population is given same or if sample s.d
is given in problem, above formula can be easily modified.
Dr. Santosh Yadav, LPU Punjab 20 / 45
24. t-Test for difference of means
Test statistic is
t =
x − y
S
r
1
n1
+ 1
n2
where x, y, are sample means and
S2
=
1
n1 + n2 − 2
hX
(xi − x)2
+
X
(yj − y)2
i
is an unbaised estimate of common population variance σ2
,
follows Student’s t distribution with (n1 + n2 − 2) degree of
freedom
Dr. Santosh Yadav, LPU Punjab 21 / 45
26. Problem 1
The means of two single large samples of 1000 and 2000
members are 67.5 inches and 68.0 inches respectively. Can
the sample be regareded as drawn from the same popuula-
tioun of s.d. 2.5 inches? (α = 5%).
H0 : µ1 = µ2 and s.d= 2.5
H1 : µ1 ̸= µ2 and s.d= 2.5
test statistic under null hypothesis: Cal-|Z| = −5.1
Decision: reject null hypothesis.
Dr. Santosh Yadav, LPU Punjab 23 / 45
27. Example 2
The average hourly wage of a sample of 150 workers in a
plant A was Rs. 2.56 with a s.d. of rs 1.08. The average
hourly wage of a sample of 200 workers in a plant B was Rs.
2.87 with a s.d. of rs 1.28. Can the applicant safely assme
that the hourly wages paid by plant B are higher than those
paid by plant A.? (α = 5%).
Dr. Santosh Yadav, LPU Punjab 24 / 45
28. Example 2
The average hourly wage of a sample of 150 workers in a
plant A was Rs. 2.56 with a s.d. of rs 1.08. The average
hourly wage of a sample of 200 workers in a plant B was Rs.
2.87 with a s.d. of rs 1.28. Can the applicant safely assme
that the hourly wages paid by plant B are higher than those
paid by plant A.? (α = 5%).
Let x1 and x2 are hourly wages of A and B then n1=150, n2=
200, ¯
x1 = 2.56, ¯
x2 = 2.87, and S1 = 1.08, S2 = 1.28
H0 : µ1 = µ2
H1 : µ1 µ2 (left tailed)
Test statistic under null hypothesis: Cal Z-value= -2.46
Decision: Reject H0. accept H1.
Dr. Santosh Yadav, LPU Punjab 24 / 45
29. Example 3
The samples of two types of electric bulb were tested for
length of life in hours, where for Type I and type II have fol-
lowing data is given: (n1, ¯
x1, s1) = (8, 1234, 36) and (n2, ¯
x2, s2) =
(7, 1036, 40). Is this difference in the means is sufficient to
warrant that type I is superior to type II regarding the length
of life? (tabulated value of t at 5% and 13 d.f for single tailed
is 1.77)
Dr. Santosh Yadav, LPU Punjab 25 / 45
30. Example 4
The height of six randomly chosen sailors in inches are: 63,
65, 68, 69, 71, and 72. Those of 10 randomly selected sol-
diers are: 61, 62, 65, 66, 69, 69, 70, 71, 72 and 73. Test
whether the given data suggest that sailors are on the aver-
age taller than soldiers. (tabulated t value for 14 d.f and 5%
LoS for single tailed is 1.76) .
Dr. Santosh Yadav, LPU Punjab 26 / 45
31. F-Test
This test is used to make a decision about ”equalty for two
population variances.
OR
Test whether two estimates of population variance are
homogeneous or not (i.e., difference is significant or not).
H0 : σ2
x = σ2
y = σ2
Dr. Santosh Yadav, LPU Punjab 27 / 45
32. Cont...
Let X and Y are samples of size n1 and n2 drawn from
normal populations with mean µ1 and µ2 and variances, σ2
1
and σ2
2
test statistic: under H0 : σ2
x = σ2
y = σ2
F =
S2
x
S2
y
=
larger estimate of variance
smaller estimate of variance
follow F- distribtion with (n1 − 1, n2 − 1) degree of freedom.
where S2
x and S2
y are unbaised estimates of common
population variance σ2
obtained from two independent
samples.
Dr. Santosh Yadav, LPU Punjab 28 / 45
33. Cont...
This test is also known as ”variance ratio test”.
Greateest of two variances is to be taken in numerator and
n1 corresponds to the greatest variance.
LoS: tabulated value Fα,(n1−1,n2−1) =?
Decision: The Null hypothesis will be accepted or rejected
depending on whether difference is significant or not.
Dr. Santosh Yadav, LPU Punjab 29 / 45
34. Problem 1
In one sample of 8 observations, the sum of the squares
of deviations of the sample values from sample mean
was 84.4 and in other sample of 10 observations it was
102.6. Test whether this difference is significant at 5%
LoS, given that F0.05,(7,9) = 3.29.
Dr. Santosh Yadav, LPU Punjab 30 / 45
35. Problem 2
Two random samples are drawn from two normal
populations and their values are
A :16, 17, 25, 26, 32, 34, 38, 40, 42
B :14, 16, 24, 28, 32, 35, 37, 42, 43, 45, 47.
Test whether the two populations have the same
variance, at LoS α = 5%.
Dr. Santosh Yadav, LPU Punjab 31 / 45
36. Cont...
Given that, n1 = 9, n2 = 11, and α = 5%
H0 : σ2
x = σ2
y = σ2
(i.e., estimates of σ2
given by samples
are homogeneous)
Test statistic: under H0
F =
S2
x
S2
y
Dr. Santosh Yadav, LPU Punjab 32 / 45
38. Cont...
X X − X (X − X)2
Y Y − Y (Y − Y)2
16 -14 196 14 -19 361
17 -13 169 16 -17 289
25 -5 25 24 -9 81
26 -4 16 28 -5 25
32 2 4 32 -1 71
34 4 16 35 2 4
38 8 64 37 4 16
40 10 100 42 9 81
42 12 144 43 10 100
- - - 45 12 144
- - - 47 14 196
P
= 270 -
P
= 734
P
= 363 -
P
= 1298
Dr. Santosh Yadav, LPU Punjab 33 / 45
39. Cont...
S2
y = 734
9−1
= 91.75
S2
x = 1296
11−1
= 129.8
Cal, F = 129.8
91.75
= 1.414
with degree of freedom (10, 8).
Tabulated F-value with (10, 8) d.f and at 5% LoS is
F(10,8),0.05 = 3.07
Decision: Cal, F-value (1.41) Tab, F-value (3.09), this
means that difference is not significant and we can accept
H0. Thus, sample are drawn from the populations having
variance σ2
.
Dr. Santosh Yadav, LPU Punjab 34 / 45
41. Chi-Square Test
This test is used when no information about the parent pop-
ulation is given and some categorical data is given.
Dr. Santosh Yadav, LPU Punjab 36 / 45
42. Chi-Square Test
This test is used when no information about the parent pop-
ulation is given and some categorical data is given.
That is, this is a non-parametric test, meaning is it is free
from parameters/ distribution and we apply the hypothesis
on the given data.
Used for test of goodness of fit, test of independence of at-
tributes, test of homogeniety, etc.
A test to judge whether observed data fits (matches)
with the expected data. i.e., testing the significance of
discrepancy between theory and experiment.
Dr. Santosh Yadav, LPU Punjab 36 / 45
43. Chi-Square Test of Goodness of Fit
The test statistic (Karl Pearson’s chi-square):
χ2
=
X (f0 − fe)2
fe
,
where f0 denotes observed frequency (i.e., given sample
data). And fe denotes expected frequency (the theoretical
data). It will be given in problem or we will calculate by using
various probability distributions depending on the nature of
problem.
degree of freedom = n − 1 (number of frquencies given -1)
Dr. Santosh Yadav, LPU Punjab 37 / 45
44. Cont..
Decision: If Calculated, χ2
α,(n−1) tabulated, χ2
α,(n−1), then
accept H0.
If Calculated, χ2
α,(n−1) tabulated, χ2
α,(n−1), then reject H0.
Dr. Santosh Yadav, LPU Punjab 38 / 45
45. Problem 1
The demand for particular spare part in a factory was found
to vary from day to day. In a sample study the following in-
formation was obtained:
Days: No. of parts demanded
Mon. 1124
Tues 1125
Wed 1110
Thu 1120
Fri 1126
Sat 1115
Total 6720
Test the hypothesis that the no. of parts demanded does not
depends on day of the week. (Given chi2
vales at 5, 6. 7, d.f
respectively 11.07, 12.59, 14.07 at 5% LoS)
Dr. Santosh Yadav, LPU Punjab 39 / 45
47. Cont...
H0 : demand does not depend on day of week
expected frequencies of spare parts demanded on each of
the six days would be ei =
P xi
n
= 6720
6
= 1120
Calculated, χ2
= 0.179.
Tabulated, χ2
0.05 = 11.07 for 5 deg of freedom.
Decision: No. of parts demanded are same over the six
days of week.
Dr. Santosh Yadav, LPU Punjab 41 / 45
48. Problem 2
The distribution of digits in numbers chosen from a tele-
phone is:
Digits Freqency
0 1026
1 1107
2 997
3 966
4 1075
5 933
6 1107
7 972
8 964
9 853
Test whether the digits may be taken to occur equally fre-
quently in the directory
Dr. Santosh Yadav, LPU Punjab 42 / 45