It is an effective sheet for statistics course.

1. Research Methodology
Course Code: 415
Course Teacher: Mr. Sutap Kumar Ghosh
Presentation Instructor: Md. Nahiduzzaman (Nahid)
Department of Finance & Banking (7th Batch)
Islamic University, Bangladesh

2. Chap ter s
Hypothesis, Basics Concepts of Testing Hypothesis, Test Statistics
and Critical Region, Critical Value and Decision Rule, Hypothesis
Testing for Mean, Proportion, Variance, Two Sample Test, Problems
Testing of Hypothesis
10
Test of Difference of more than Two Proportions, Test of
Independence of Attributes, Test of Goodness of Fit, Problems
Chi-Square Tests
11
The ANOVA Technique, The Basic Principle of ANOVA, One Way
ANOVA, Two Way ANOVA, Problems
Analysis of Variance
12
Parameter and Statistics, Sampling and Non-sampling Errors,
Sampling Distribution, Degree of Freedom, Standard Error, Central
Limit Theorem, Finite Population Correction, Statistical Inference,
Problems
Sampling & Statistical Inference
09

3. 9. Sampling & Statistical Inference
Chapter’s Objectives:
❑True Mean & Proportion Interval Estimation
❑True Variance Interval Estimation
❑Sample Size Determination
Finite Population
Infinite Population

4. Basic Things
❖ Hypothesis
❖ 5 Steps of Hypothesis Analysis
❖ Sample Size, Finite & Infinite

5. Hypothesis
Hypothesis: A hypothesis is an assumption, an idea that is
proposed for the sake of argument so that it can be tested to see
if it might be true.

6. 5 Steps of Hypothesis Testing
Step-1: Formulation of Hypothesis
Example: H0: µ=500
H1: ≠, >, < 500
Step-2: Calculation of Critical Value or Accepted Region
Z Value (Area of Confidence Level), T Value (Sig. level & DOF), 𝑋2 Value (Sig.
level & DOF) , F Value (Sig. level, DOD, DON)
Step-3: Determination of Actual Value
Depends on Formula
Step-4: Representation of Calculated Data
Step-5: Result Interpretation
If actual value situates in the confidence level, the null
hypothesis will be accepted.
If actual value does not situate in the confidence level, the null
hypothesis will be rejected.

7. Z-test or T-test?
Z-test T-test
Condition-(i) n>30 Condition-(i) n<30
Condition- (ii) Standard Deviation (σ) Known Condition- (ii) Standard Deviation (σ) Unknown
➢ In case of Z-test, if any condition fill-up, we can use Z-test.
➢ In Case of T-test, Both conditions have to be fulfilled.

8. Step-2: Calculation of Critical Value (Z-test)
.50
.50
.45
(Confidence Level)
One Tail Test
α=.05
H0: µ = 500
H1: µ > 500
.05 (Significant Level)

9. Step-2: Calculation of Critical Value (Z-test)
Therefore, Critical Value of Z= 1.64 or 1.65

10. Step-2: Calculation of Critical Value (Z-test)
.50
.50
.475
Two Tail Test
α=.05
H0: µ = 500
H1: µ ≠ 500
.025
α=.05/2
= .025
.475
.025

11. Step-2: Calculation of Critical Value (T-test)
One Tail Test
N= 15
α=.05
H0: µ = 500
H1: µ > 500
DOF= N-1
= 15-1
= 14
Therefore, The Critical Value of T= 1.761

12. Example-9.1
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35
ohms. Determine a 95% confidence
interval for the true mean resistance of the
population.

13. Example-9.2
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. Population standard deviation is
not known. Sample SD (s1) is .35 ohms.
Determine a 95% confidence interval for
the true mean resistance of the
population.

14. Example-9.3
A random sample of 100 people of 100
people shows that 25 have opened
IRA(individual retirement arrangement)
this year. Construct a 95 % confidence
interval for the true proportion of people
who have opened IRA.

15. Example-9.4
The Cholesterol concentration in the yolks of a sample of 18 randomly selected eggs
laid by genetically engineered chickens were found to have a mean value of 9.38 mg/g
of yolk and a standard deviation of 1.62 mg/g. Use this information to construct a 95%
confidence interval estimate of the true variance of the cholesterol concentration in
these egg yolks.
Example-9.5
A technologist is developing a new method for processing a food material. For best
quality, it is important to control moisture content in the final product. So, as one part
of determining the practically of the new method, the technologist must estimate the
variability of water content in the resulting product. He collects 50 specimens of
product from new process, and determines the percent water in each. These 50
specimens give sample mean water content of 43.24% and a sample standard deviation
of 7.93%. Compute a 95% confidence interval estimate of the true variance of the
percentage water for new process.

16. Example-9.4
.05
1- .05
= .95

17. Example-9.6
Determine the size of the
sample for estimating the
true weight of the cereal
containers for the universe
with N= 5000 on the basis of
the following information.
1. The variance of weight = 4 ounces
on the basis of past records.
2. Estimate should be within .8
ounces of the true average weight
with 99% probability.

18. Example- 9.7, 9.8
There is percentage
There is percentage
but SD is not given

19. THANK YOU
All the Best