Download - LSSGB Lesson4 Analyze
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Lesson 4—Analyze
Lean Six Sigma Green Belt
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● Explain the patterns of variation
● Describe the classes of distributions
● Discuss Multi-Vari studies and the causes
● Explain correlation and its types
● Discuss the various hypothesis tests
● Discuss the application of F-test, t-test, ANOVA, and Chi-squared
After completing this lesson, you will be able to:
Objectives
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Analyze
Topic 1—Patterns of Variation
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Classes of Distributions
The data obtained from measurement phase exhibits variety of distribution, depending on the data
type and its source.
The methods used to describe the parameters for classes of distribution are:
Probability
● It is based on assumed model of distribution.
● Used to find the chances of certain outcome/event to occur.
Statistics
● Uses the measured data to determine a model to describe the data used.
Inferential Statistics
● Describe the population parameters based on the sample data using a particular model.
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Types of Distributions
The two types of distribution are as follows:
Discrete Distribution
● Binomial distribution
● Poisson distribution
Continuous Distribution
● Normal distribution
● Chi-square distribution
● t-distribution
● F-distribution
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Discrete probability distribution is characterized by the probability mass function.
● It is important to be familiar with discrete distributions while dealing with discrete data.
● The two most useful discrete probability distributions are:
o Binomial distribution; and
o Poisson distribution.
● These distributions help in predicting the sample behavior that has been observed in a population.
Discrete Probability Distribution
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Binomial Distribution
Binomial distribution is a probability distribution for the discrete data.
Predicts sample behaviorDescribes the discrete data as a result of a particular process
Used to deal with defective items
Characteristics of Binomial Distribution
Best suitable when the sample size is less than thirty and less than ten percent of the population
P R = nC r ∗ pr ∗ 1 − p n _ r
where, P(R) = probability of exactly (r) successes out of a sample size of (n)
p = probability of success; r = number of successes desired; n = sample size
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Some of the key calculations of binomial distribution are shown.
Binomial Distribution (contd.)
Term Formula
Mean
𝜇 = 𝑛𝑝
where, n = sample size
p = probability of success
Standard Deviation
𝜎 = 𝑛𝑝(1 − 𝑝)
where, n = sample size
p = probability of success
Sample factorial calculation5! = 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 120
4! = 4 ∗ 3 ∗ 2 ∗ 1 = 24
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Calculating Binomial Distribution—Example
Using binomial distribution formula, find the probability of getting 5 heads in 8 coin tosses.Q
● Tossing a coin has only two outcomes, Head or Tail.
● Outcomes are statistically independent.
Therefore,
p = probability of success = 0.5 (this remains fixed over time)
n = sample size = 8
r = number of successes desired = 5
P R = 8C 5 ∗ 0.55 ∗ 1 − 0.5 8 _ 5= 0.2187 = 21.87%
A
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Poisson distribution is an application of the population knowledge to predict the sample behavior.
Poisson Distribution
Describes the discrete data
Deals with integers which can take any value
Used to analyze situations wherein the number of trials is large
Characteristics of Poisson
Distribution
Used where the probability of success in each trial is very small
Used for predicting the number of defects
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The formula for the Poisson distribution is as follows:
Poisson Distribution—Formula
P ∗ =λx ∗ e−λ
∗ !where, P(x) = probability of exactly (∗) occurrences in a Poisson distribution (n)
λ = mean number of occurrences during interval
∗ = number of occurrences desired
e = base of the natural logarithm (equals 2.71828)
Mean of a Poisson Distribution (µ) = λ
Standard Deviation of a Poisson Distribution (σ) = λ!11
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Calculating Poisson Distribution—Example
The past records of a road junction which is accident-prone show that the mean number of accidents every
week is 5 at this junction. Assume that the number of accidents follows a Poisson distribution and calculate
the probability of any number of accidents happening in a week.
Q
Assumption is the number of accidents follows a Poisson distribution
Given: =5 per week
Now, probability of zero accidents per week P 0 =5x ∗ e−5
0!= 0.006
Probability of exactly one accident per week P 1 =51 ∗ e−5
1!= 0.03
Probability of more than two accidents per week = 1 – [P(0)+P(1)+P(2)] = 1 – [0.006+0.03+0.08]
= 0.884 = 88.4%
A
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Continuous probability distribution is characterized by the probability density function.
● A variable is said to be continuous if the range of possible values falls along a continuum.
Example: Loudness of cheering at a ball game, weight of cookies in a package, length of a pen,
or the time required to assemble a car.
● These distributions help in predicting the sample behaviour observed in a population.
Continuous Probability Distribution
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Normal Distribution
The Normal or Gaussian distribution is a continuous probability distribution, illustrated as N (µ, σ).
● It has a higher frequency of values around the mean and fewer occurrences away from it.
● It is used as a first approximation to describe real-valued random variables that tend to cluster around a single mean value.
● It is a bell-shaped curve and is symmetrical.
● The total area under the normal curve p(x which is found in the distribution) = 1.
Normal Distribution with Mean = 100 and Standard Deviation = 10
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In a normal distribution, to standardize comparisons of dispersion, a standard Z variable is utilized. The uses of Z value are as follows:
● It is unique for each probability within the normal distribution.
● It helps in finding probabilities of data points anywhere within the distribution.
● It is dimensionless with no units like mm, liters, coulombs, etc.
Normal Distribution (contd.)
Z =(Y − µ)
σwhere, Z = number of standard deviations between Y and the µ
Y = value of the data point in concern
µ = mean of the population
σ = standard deviation of the population
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Calculating Normal Distribution—Example
Suppose the time taken to resolve customer problems follows a normal distribution with the mean value of
250 hours and standard deviation value of 23 hrs. What is the probability that a problem resolution will take
more than 300 hrs?
Q
Given:
● Y = 300
● µ = 250
● σ = 23
Using the formula: Z =(300−250)
23= 2.17
● From a Normal Distribution Table, the Z value of 2.17 covers an area of 0.98499 under itself● Thus, the probability that a problem can be resolved in less than 300 hrs is 98.5% ● The chances of a problem resolution taking more than 300 hours is 1.5%
A
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Z-Table Usage
The probability of areas under the curve is 1. For the actual value, one can identify the Z score by using the Z-table.
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Z-Table
This Z-table gives the probability that Z is between zero and a positive number.
This is the most commonly used normal distribution Z-table with the positive Z-scores.
Z 0.0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5348 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2 0.8849 08869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
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Using Z-Table—Example
There is no need of the table to find the answer once you know that the variable Z takes a value of
less than (or equal to) zero.
● First, the area under the curve is 1, and second, the curve is symmetrical about Z = 0.
● Hence, there is 0.5 (or 50%) above chance of Z = 0 and 0.5 (or 50%) below chance of Z = 0.
Find the value of p of (Z less than 0).Q
A
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Using Z-Table—Example (contd.)
The opposite or complement of an event A occurring is the event A not occurring.
P(not A) = 1 – P(A)
P(Z greater than 1.12) = 1 – P(Z less than 1.12)
Using the table:
P(Z < 1.12) = 0.5 + P(0 < Z < 1.12) = 0.5 + 0.3686 = 0.8686
Hence P(Z > 1.12) = 1 – 0.8686 = 0.1314
Find the value of p of (Z greater than 1.12).Q
A
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Using Z-Table—Example (contd.)
● Z falls within an INTERVAL
Using the table:
P(Z lies between 0 and 1.12) = 0.3686
Find the value of p of (Z lies between 0 and 1.12).Q
A
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Chi-square distribution (chi-squared or χ² distribution) with k-1 degrees of freedom is the distribution of the sum of the squares of k independent standard normal random variables.
Chi-Square Distribution
Degree of freedom (df) = k – 1, where k is the sample size.!
Most widely used probability distribution in inferential statistics
The distribution is used in a hypothesis test
Characteristics of χ²
Distribution
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The formula for the Chi-square distribution is as follows:
Chi-Square Distribution—Formula
χcalculated2 = =
fO − fe 2
fe
where, χcalculated2 () = chi-square index
fO = observed frequency
fe = expected frequency
Chi-square distribution will be covered in detail in the later part of this lesson.!23
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t-Distribution
A t-distribution is most appropriate to be used when:
● the sample size <30;
● population standard deviation is not known; and
● population is approximately normal.
The t-distribution approaches normality as the sample size increases.!24
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The F-distribution is a ratio of two Chi-square distributions, and a specific F-distribution is denoted by the degrees of freedom for the numerator Chi-square and the degrees of freedom for the denominator Chi-square.
F-Distribution
Fcalculated =S12
S22
where, S1and S2 = standard deviations of the two samples
● If Fcalculated is 1, there is no difference in the variance
● If S1> S2, then the numerator should be greater than denominator (df1 = n1 – 1 and df2 = n2 – 1)
Refer F-table to find out critical F-distribution at α and degrees of freedom of samples of two different processes (df1 and df2)!
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Analyze
Topic 2—Exploratory Data Analysis
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Multi-Vari studies analyze variation, investigate process stability, identify investigation areas, and
break down the variation.
They classify variation sources into three major types:
Multi-Vari Studies
Positional
Variations within a single unit
where variation is due to location.
Examples: Pallet stacking in a
truck, temperature gradient in an
oven, variation observed from
cavity-to-cavity within a mold,
region of a country, line on invoice
Cyclical
Variations among sequential
repetitions over a short time.
Examples: Every n’th pallet
broken, batch-to-batch variation,
lot-to-lot variation, invoices
received day-to-day, and account
activity week-to-week
Temporal
Variations which occur over
longer periods of time.
Examples: Process drift,
performance before and after
breaks, seasonal and shift based
differences, month-to-month
closings, and quarterly returns
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Create Multi-Vari Chart
Select Process and Characteristics
Decide Sample Size
Create a Tabulation
SheetPlot the Chart
Link the Observed
Values
Example: Select
the process
where the plate
is being
manufactured
and measure its
thickness within
a specified range.
Example: Sample
size is five pieces
from each
equipment and
the frequency of
data collection is
every two hours.
Example: The
tabulation sheet
with data records
contains the
columns with
time, equipment
number, and
thickness as
headers.
Example: Chart is
plotted with time
on X axis and the
plate thickness
on Y axis.
Example: The
observed values
are linked by
appropriate lines.
The five steps to create a Multi-Vari chart are:
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Create Multi-Vari Chart (contd.)
The path to create a
Multi-Vari chart in
Minitab is:
Minitab > Stat > Quality
Tools > Multi-Vari Chart
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Correlation is the association between variables. The Coefficient Correlation shows the strength of
the relationship between Y and X.
The statistical significance is denoted by correlation coefficient ‘r’.
Simple Linear Correlation
Movement in both variables is same
Movement in both variables is inverse
No correlation between the two variables
‘r’ or Pearson’s Coefficient of Correlation
Higher the absolute value of ‘r’, stronger the correlation between Y and X. An ‘r’ value of > + 0.85 or < - 0.85 indicates a strong correlation.!
+10-1
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Correlation measures the linear
association between the output
(Y) and the input variable (X).
The patterns of correlation
displayed in scatter plots are:
● easy to see when the ‘r’
value is 0.9 and above; and
● difficult to see when the ‘r’
value is 0.5 or below.
Correlation Levels
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The degree of movement of variable changes is calculated using regression.
If a high percentage of variability in Y (r2> 70%) is explained by changes in X, the model to write a
transfer equation is as follows:
This equation is used to:
● predict future values of Y given X, and X given Y; and
● regress Y on one or more X’s simultaneously.
Regression
Y = f(X)
Simple Linear Regression is for one X and Multiple Linear Regression is for more than one X.!32
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There are two key concepts of regression:
Key Concepts of Regression
Transfer function to control Y
Y = f(X) may not be the correct transfer function to
control Y because there may be a low level of
correlation between the two variables.
Vital X
It is important to discover whether a statistical
significant relationship exists between Y and a
particular X by looking at p-values. Based on
regression, one can infer the vital X and eliminate the
rest.
It is important to understand if there is statistical relevance between Y and X using the metrics from Regression Analysis. The Simple Linear Regression should be used as a Statistical Validation tool.!
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A simple linear regression equation is a fitted linear equation between Y and X. It is represented as
follows:
where,
● Y = Dependent variable / output / response
● X = Independent variable / input / predictor
● A = Intercept of fitted line on Y axis
● B = Regression coefficient / Slope of the fitted line
● C = Error in the model
Simple Linear Regression (SLR)
Y = A + BX ± C
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If Y and X are not perfectly linear (r = ± 1), several lines could fit in the scatter plot. It can be inferred
from the graphs below:
● Minitab fits the line which has the least Sum of Squares of Error.
● In a linear relationship, the points would lie on the line. Typically, the data lies off the line.
● The distance from the point to line is the error distance used in the SSE calculations.
Least Squares Method in SLR
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Years Fertilizer Expenses in $ (Y) Annual Selling in $ (X)
2009 2 20
2010 3 25
2011 5 34
2012 4 3
2013 11 40
2014 5 31
A farmer wishes to predict the relationship between the amount spent on fertilizers and the annual
sales of his crops. He collects the following data of last few years and determines his expected
revenue if he spends $8 annually on fertilizer.
SLR—Example
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SLR using Microsoft Excel
The steps to perform Simple Linear Regression in Microsoft
Excel are as follows:
1. Copy the data from the cells B1 to C6 on an Excel
Worksheet.
2. Click Insert, and choose the Plain Scatter Chart (Scatter
with only Markers).
3. Right-click on the data points and choose “Add Trendline.”
4. Choose “Linear” and select the boxes titled, “Display R-
Squared value” and “Display equation.”
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To use the data for Regression analysis, the interpretation of the scatter chart is as follows:
● The r2 value (Coefficient of Determination) conveys if the model is good and can be used. The r2
value is 0.3797.
● 38% of variability in Y is explained by X.
● The remaining 62% variation is due to residual factors.
● The low value of r2 validates a poor relationship between Y and X.
Regression Analysis Using Microsoft Excel
Refer to the Cause and Effect Matrix and study the relationship between Y and a different X variable. !
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If a new variable X2 is added to the r2 model, the impact of X1 and X2 on Y gets tested. This is known as
Multiple Linear Regression. In Multiple Linear Regression:
● the value of r2 changes due to the introduction of the new variable.
● the resulting value of r2 is known as ‘r2 Adjusted.’
● the model can be used if ‘r2 Adjusted’ value is greater than 70%.
Multiple Linear Regression
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Multiple Linear Regression covers the following concepts:
● The residuals between the actual value and the predicted value give an indication of how good the
model is.
● If the errors are small and predictions use X’s that are within the range of the collected data, the
predictions should be fine.
Key Concepts of Multiple Linear Regression
SST = SSR + SSE SSR = SST - SSE r2 = SSR ÷ SST
● To check for error, take two observations of Y at the same X.
● Prioritization of X’s can be done through the SLR equation; run separate regressions on Y with each X.
● If an X does not explain variation in Y, it should not be explored further.
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A regression equation may denote a relationship between variables. It does not indicate:
● if change in one variable causes change in the other; and
● both the variables may be dependent on another independent variable.
Difference between Correlation and Causation
There is a positive
correlation between the
number of sneezes and the
deaths in the city. It cannot
be assumed that sneezing is
the cause of death though
the correlation is very strong.
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Analyze
Topic 3—Hypothesis Testing
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The differences between a variable and its hypothesized value may be statistically significant but may
not be practical or economically meaningful.
For example: Based on the hypothesis test, Nutri Worldwide Inc. implemented a trading strategy. The
returns:
● are economically significant when logical reasons are examined before implementation.
● may not be significant when statistically proven strategy is implemented directly.
● may be economically insignificant due to taxes, transaction costs, and risks.
Statistical and Practical Significance of Hypothesis Test
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Null Hypothesis vs. Alternate Hypothesis
The conceptual differences between a null and an alternate hypothesis are as follows:
Null Hypothesis
● Represented as H0
● Cannot be proved, only rejected
● Example: Movie is good
Alternate Hypothesis
● Represented as Ha
● Challenges the null hypothesis
● Example: Movie is not good
If null hypothesis is rejected, alternative hypothesis must be right.!44
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Type I and Type II Error
The conceptual differences between type I and type II error are as follows:
Type I Error
● Rejecting a null hypothesis when it is true
● Also known as Producer’s Risk
● ‘α’ is the chance of committing a Type 1 error
● The value of ‘α’ is 0.05 or 5%
● Example: When a movie is good, it is reviewed as ‘not good.’
Type II Error
● Accepting a null hypothesis when it is false
● Also known as Consumer’s Risk
● ‘β’ is the chance of committing a Type II Error
● The value of ‘β’ is 0.2 or 20%
● Any experiment should have as less β value as
possible
● Example: When a movie is not good, it is
reviewed as ‘good.’
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While dealing with type I or type II errors, following are the points to remember:
● Probability of making one type of error can be reduced, leading to increasing the probability of
making the other type of error.
● If a true null hypothesis is erroneously rejected (Type I error), a false null hypothesis may be
accepted (Type II error).
● ‘α’ is set at 0.05, which means the risk of committing a type I error will be 1 out of 20 experiments.
● It is important to decide what type of error should be less and set ‘α’ and ‘β’ accordingly.
Important Points to remember about Type I and Type II Errors
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The power of a test:
● helps in the probability of correctly rejecting the null hypothesis when it is false.
● is represented as 1-β. This is type II error.
● is the probability of not committing a type II error.
● helps in improving the advantage of hypothesis testing.
● with highest value should be preferred when given a choice of tests.
Power of Test
In hypothesis testing, ‘α’ is the significance level and ‘1-α’ is the confidence level.!47
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The sample size is determined by the responses to the following questions:
● How much variation is present in the population? ( σ)
● At what interval does the true population mean need to be estimated? ( ±Δ)
● How much representation error is allowed in the sample? ( α)
The sample size for continuous data can be determined by the formula:
Determinants of Sample Size—Continuous Data
n =Z1−(
𝛼
2)∗ σ
∆
2
1- (𝛼
2)∗
= 0.975
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To calculate the standard sample size for continuous data, the value of α is taken as 5%. According to Z
table, the Z97.5 = 1.96. The standardized sample size formula is:
Standard Sample Size Formula—Continuous Data
The population standard deviation for the time, to resolve customer problems, is 30 hours. What should be the size of a sample that can estimate the average problem resolution time within ± 5 hours tolerance with 99% confidence?
Q
Δ = 5, σ = 30, α =0.01, and Z99.5 = 2.575.Sample size = [(2.575*30)/5]2 = 238.70 = 239 A
n =1.96 ∗ σ
∆
2
for Continuous Data
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To calculate the standard sample size for discrete data, if the average population proportion non-
defective is ‘p’, then population standard deviation can be calculated as:
Standard Sample Size Formula—Discrete Data
The non-defective population proportion for pen manufacturing is 80%. What should be the sample size to draw a sample that can estimate the proportion of compliant pens within ± 5% with an alpha of 5%?
Q
Δ = 0.05, σ 2= 0.8 (1-0.8), α = 0.05, and Z97.5 = 1.96 Sample size = (1.96/0.05)2 *0.8*0.2 = 245.86 = 246
A
σ = p(1 − p) n =1.96
∆
2
p(1 − p) for Discrete Data
Where ∆ = Tolerance allowed on either side of the population proportion average in %
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The figure below helps in concluding the type of test one should perform based on the kind of data
and values available:
Hypothesis Testing Roadmap
Discrete data Continuous data
X2-test
F-test
t-test
Z-test
F-test
Variance Mean
Comparisonof two
Comparisonof many
Hypothesis testing
Mean Variance
σ unknown
σ known
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H0: Average height of North American males is 165 cm (μ0)
Ha: Average height of Indian males < > 165 cm
H0: μ = μ0 against Ha: μ < > μ0
Sample size (n) = 117 (Z-test) and Sample size (n) = 25 (t-test); Sample average ( X) = 164.5 cm
Hypothesis Test for Means (Theoretical)—Example
Z-test (σ known)
The population SD is known; σ = 5.2
Compute z = ( X – μ0) / √(σ2/n) = (165 – 164.5) / √
(5.22/117) = 1.04
Reject H0 at level of significance α if z > zα
Since z0.05 = 1.96, the null hypothesis is not rejected at 5%
level of significance. Thus a conclusion based on the
sample collected is that the average height of North
American males is 165 cm.
t-test (σ unknown)
The population SD is unknown; however, it is estimated
from the sample SD; s = 5.0
Compute t = ( X- μ0) / √(s2/ n) = (165 – 164.5) / √(52 /25)=
0.5
Reject H0 at level of significance α if t > tn-1, αSince t24, 0.05 = 2.064, the null hypothesis is not rejected at
5% level of significance. Thus a conclusion based on the
sample collected is that the average height of North
American males is 165 cm.
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In hypothesis test for variance Chi square test is used. This is explained in the example below:
Hypothesis Test for Variance—Example
H0: Proportion of wins in Australia or abroad is independent of the country played against
Ha: Proportion of wins in Australia or abroad is dependent on the country played against
χ2Critical = 6.251 and
χ2 Calculated = 1.36
Result: Since calculated value is less than the critical value, the proportion of wins of Australia hockey
team is independent of the country played or place.
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The hypothesis test on population proportion can be performed as follows:
Hypothesis Test for Proportions—Example
H0: Proportion of smokers among males in a place named R is 0.10 (p0)
Ha: Proportion of smokers among males in R is different than 0.10
H0: p = p0 against Ha: p < > p0
Among n = 150 adult males interviewed, 23 were found smokers.
Sample proportion p = 23/150 = 0.153
Compute test statistic:
Reject H0 at level of significance α if z > zα
Since z0.05 = 1.96, the null hypothesis is rejected at 5% level of significance in favor of the alternative
Result: It can be concluded that the proportion of smokers in R is greater than 0.10.
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Means of two processes are compared to:
● understand the significant difference in the outcome of the two processes;
● understand whether a new process is better than an old process;
● understand whether the two samples belong to the same population or a different population;
and
● benchmark the existing process with another process.
Comparison of Means of Two Processes
55
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The two-mean t-test with unequal variances is:
● H0: μ1 = μ2 against Ha: μ1≠μ2
● Two samples of sizes n1 = 125 and n2 = 110 are taken from the two populations
● X1 = 167.3, X2 = 165.8, s1 = 4.2, s2 = 5.0 are the sample means and SDs respectively
● Compute test statistic
● Reject H0 at level of significance α if |Computed t|> tDF,α/2
● Since t223, 0.025 = 1.96, the null hypothesis is rejected at 5% level of significance
Paired Comparison Hypothesis Test for Means (Theoretical)
56
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Paired Comparison Hypothesis Test for Variance—F-Test Example
Susan is examining the earnings of two companies. According to her, the earnings of Company A are more
volatile than those of Company B. She has been obtaining earnings data for the past 31 years for Company
A, and for the past 41 years for Company B. She finds that the sample standard deviation of Company A’s
earnings is $4.40 and of Company B’s earnings is $3.90. Determine whether the earnings of Company A
have a greater standard deviation than those of Company B at 5% level of significance.
H0 : σA2= σB
2 = the variance of Company A’s earnings is equal to the variance of Company B’s earnings.
Ha : σA2 < > σB
2 = the variance of Company A’s earnings is different.
σA2= variance of Company A’s earnings.
σB2= variance of Company B’s earnings.
Note: σA > σB. In calculating the F-test statistic, always put the greater variance in the numerator.
Q
A
57
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The degrees of freedom for company A and company B are:
● dfA (degrees of freedom of A) = 31 – 1 = 30
● dfB (degrees of freedom of B) = 41 – 1 = 40
The critical value from F-table equals 1.74. The null hypothesis is rejected if the F-test statistic is
greater than 1.74.
Calculation of F-test statistic:
Results: The F-test statistic (1.273) is not greater than the critical value (1.74). Therefore, at 5%
significance level, the null hypothesis cannot be rejected.
Hypothesis Test for Equality of Variance—F-Test Example
F= (SA2/SB
2) = 4.402/3.902 = 1.273
58
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Group A (Chef 1) Group B (Chef 2)
4.2 4
4.5 4.5
7.2 5
6.1 5.2
8.9 5.3
5.2 6.1
Hypothesis Tests—F-Test for Independent Groups
A restaurant wanting to explore the recent overuse of avocados suspects there is a difference
between two chefs and number of avocados used to prepare the salads. The table shows the measure
of avocados in ounces.
59
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F-Test
The steps for conducting F-
Test in MS-Excel are:
1. Click Data Analysis
under Data tab.
2. Select F-Test Two-
Sample for Variances.
3. In Variable 1 and 2
range, select the right
data set.
4. Click Ok.
60
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Before interpreting the F-test, the assumptions to be considered are as follows:
● Null Hypothesis: There is no significant statistical difference between the variances of the two
groups, thus concluding any variation could be because of chance. This is Common Cause of
Variation.
● Alternate Hypothesis: There is a significant statistical difference between the variances of the two
groups, thus concluding variations could be because of assignable causes also. This is Special Cause
of Variation.
F-Test Assumptions
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The interpretations for the conducted F-test are as
follows:
● From the Excel result sheet, the p-value is 0.03.
● If p-value is < 0.05, null must be rejected.
● Null hypothesis with 97% confidence is rejected.
● The fact that variation could only be due to Common
Cause of Variation is rejected.
● There could be Assignable Causes of Variation or
Special Causes of Variation.
F-Test Two-Sample for Variances
Variable 1 Variable 2
Mean 6.016666667 5.016666667
Variance 3.197666667 0.517666667
Observations 66
df 5 5
F 6.177076626
P(F<=f) one-tail 0.033652302
F Critical one-tail 5.050329058
F-Test Interpretations
62
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Group A (Chef 1) Group B (Chef 2)
4.2 4
4.5 4.5
7.2 5
6.1 5.2
8.9 5.3
5.2 6.1
Hypothesis Tests—t-Test for Independent Groups
The table shows the measure of avocados in ounces. If a significant difference in their means is found,
it can be concluded that there is a possibility of Special Cause of Variation.
63
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The steps for conducting 2-sample t-test in MS-Excel are given below:
Open MS Excel, click Data and click
Data Analysis.
Select 2-Sample Independent t-test assuming unequal
variances.
In Variable 1 range, select the data set
for Group A.
In Variable 2 range, select the data set
for Group B.
Keep the “Hypothesized
Mean Difference” as 0.
Click Ok.
1 2 3
4 5 6
2-Sample t-Test
64
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The assumptions for a 2-Sample Independent t-test are as follows:
● Null Hypothesis: There is no significant statistical difference between the means of the two groups,
thus concluding any variation could be because of chance. This is Common Cause of Variation.
● Alternate Hypothesis: There is a significant statistical difference between the means of the two
groups, thus concluding variations could be because of assignable causes also. This is Special Cause
of Variation.
H0 : Mean of Group A = Mean of Group BHa : Mean of Group A ≠ Mean of Group B
The alternate hypothesis tests two conditions, Mean of A < Mean of B and Mean of A > Mean of B. Thus a two-tailed probability needs to be used.!
Assumptions of 2-Sample Independent t-Test
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2-Tailed vs. 1-Tailed Probability
The difference between the usage of the 2-tailed probability and one-tailed probability are as follows:
2-Tailed Probability
● If the alternate hypothesis tests more than
one direction, either less or more, use a 2-
tailed probability value from the test.
Example: If Mean of A is not equal to Mean of B,
then it is 2-tailed probability.
1-Tailed Probability
● If the alternate hypothesis tests one
direction, use a 1-tailed probability value
from the test.
Example: If Mean of A is greater than Mean of B,
then it is 1-tailed probability.
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2-Sample Independent t-Test—Results and Interpretations
According to the table:
● The p-value of 2-tailed
probability testing is 0.24.
● This value is greater than
0.05.
● The null hypothesis is not
rejected.
● Both the groups are
statistically same.
t-Test: Two-Sample Assuming Unequal Variances
Variable 1 Variable 2
Mean 6.016666667 5.016666667
Variance 3.197666667 0.517666667
Observations 6 6
Hypothesized Mean 0
df 7
T Stat 1.270798616
P(T<=t) one-tail 0.122200546
T Critical one-tail 1.894578605
67
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The paired t-test is:
● one of the powerful tests from the t-test family;
● conducted before and after the process to be measured; and
● often used in the Improve stage.
Paired t-Test
68
For example, a group of students score X in CSSGB before taking the Training program. Post the training program, the scores are taken again.
● One needs to find out if there is a statistical difference between the two sets of scores.
● If there is a significant difference, the inference could be that the training was effective.
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Sample Variance (S2) is the average of the squared differences from the mean.
● It is used to calculate and understand the degree of variation of a sample.
● In statistics, its value is used by converting it into standard deviation and combining with the
mean.
The steps for calculating sample variance are as follows:
Sample Variance
Calculate the mean of the sample
Subtract each of the value from mean
Calculate the square value of the result
Take average of the squared differences
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Sample Variance—Example
The example to calculate sample variance is as follows:
Consider the sample of weights. Suppose the mean value is 140 and when you subtract each value
from the mean, take the square value of the result, and then take the average of the squared
difference, the resulting sample variance value is 1936.
● In order to get the standard deviation, take the square root of the sample variance: √1936 = 44.
● The standard deviation along with the mean, will tell you how much the majority of the people
weigh.
o The mean value is 140 and variance is 44, the majority of people weigh between 96 pounds
(140 - 44) and 184 pounds (140 + 44).
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ANOVA:
● is used to compare the means of more than two samples;
● stands for Analysis of Variance;
● helps in understanding that all sample means are not equal;
● based shortlisted samples can further be tested; and
● generalizes the t-test to include more than two samples.
ANOVA—Comparison of More Than Two Means
71
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Outlet 1 Outlet 2 Outlet 3
48 50 49
49 48 48
48 36 39
53 50 49
58 50 34
50 62 33
46 45 57
50 47 48
49 51 47
47 44 39
ANOVA Example
The table shows the takeaway food delivery time of
three different outlets. To benchmark the delivery
time of the outlets:
● the null hypothesis will assume that the three
means are equal; and
● rejection of the null hypothesis would mean that at
least two outlets are different in their average
delivery time.
72
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Using Minitab for ANOVA
To perform ANOVA in Minitab:
1. Stack the data into two
columns.
2. In the main menu, choose
Stat > ANOVA > One-Way.
3. Select the response, delivery
time, factor, and outlet.
4. Click OK.
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Using Minitab for ANOVA (contd.)
The following output is received when the data is fed into the Minitab:
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ANOVA using Excel
To perform ANOVA, enter the data on an Excel spreadsheet, select the ANOVA-single factor test from
the Data Analysis “Toolpak,” and select the array for analysis and an output range.
75
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The result of the Minitab ANOVA is interpreted as follows:
● Since p-value is more than 0.05, the null hypothesis is accepted.
● There is no significant difference between the means of delivery time for three outlets.
● Based on the confidence intervals, it is found that the intervals overlap.
Interpreting Minitab Results
In one-way ANOVA, one factor has to be benchmarked unlike the two-way ANOVA.!
76
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The Chi-square distribution (χ²-distribution) or Chi-squared:
● is a widely used probability distribution in inferential statistics;
● needs one sample for the test to be conducted; and
● with k-1 degrees of freedom is the distribution of a sum of the squares of k independent standard
normal random variables.
Chi-Square Distribution
𝒳2Calculated = Σ
f0−fe2
fe
Where,
• 𝒳2Calculated = chi-square index
• Fo = An observed frequency
• Fe = An expected frequency
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Chi-Square Test—Example
To analyze the Australian hockey team’s wins,
the data has two classifications:
● The table is called a 2 X 4 contingency
table.
● Expected frequency for each of the
observed frequencies = (row total)(column
total)/overall total.
Example: Observed frequency of 3 wins
against South Africa in Australia would convert
to expected frequency of (21 / 31) * 5 = 3.39
Estimated Population Parameters
Sample Statistics
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Chi-Square Test—Example (contd.)
The table is populated by:
● calculating and adding the estimated
population parameters;
● estimating the observed frequency; and
● calculating the final chi-square index.
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There is a different chi-square distribution for each different number of degrees of freedom. For chi-
square distribution, degrees of freedom are calculated as per the number of rows and columns in the
contingency table.
Chi-Square Test—Example: Interpretation of Results
The calculated value is found to be less than the critical value.
Degrees of freedom = (2 - 1)*(4-1) = 3
Assuming α = 10%, 𝒳2Critical = 6.251
𝒳2Calculated = 1.36
𝒳2
Critical divides region into acceptance and rejection zones while 𝒳2Calculated allows
accepting or rejecting the null hypothesis depending on which zone it falls.
80
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Analyze
Topic 4—Hypothesis Testing with Non-Normal Data
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Mann-Whitney or Wilcoxon Rank Sum test is a non-parametric test used to compare two unpaired
groups. In this test:
● The value of is set as 0.05.
● The rejection and acceptance condition remains the same for different cases:
Mann-Whitney Test
The aim of this test is to rank the entire data available for each condition and then compare the total outcome of the two ranks.!
If p< Reject null hypothesis
If p> Cannot reject null hypothesis, accept null hypothesis
82
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Mann-Whitney Test
Rank all the values from low to high
Find the average of the ranks for all the identical values
Test the values
● The smallest number gets a
rank of 1.
● The largest number gets a
rank of n, where n is the total
number of values in the two
groups.
● Continue till all the whole-
number ranks are used.
● Summate the ranks for the
observations from sample 1
and then summate the rank
in sample 2 (larger group).
The steps to perform Mann-Whitney test are as follows:
83
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Mann-Whitney Test—Example
Group Data
G1
14
2
5
16
9
G2
4
2
18
14
8
Sorted Data Group Rank A
2 G1 1
2 G2 2
4 G2 3
5 G1 4
8 G2 5
9 G1 6
14 G1 7
14 G2 8
16 G1 9
18 G2 10
Final Rank
1.5
1.5
3
4
5
6
7.5
7.5
9
10
Avg. = 1.5
Avg. = 7.5
An example of performing Mann-Whitney test is shown here.
G1 Rank (R1)
1.5
4
6
7.5
9
Total = 28
n1 = 5
G2 Rank (R2)
1.5
3
5
7.5
10
Total = 27
n2 = 5
84
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Mann-Whitney Test—Example (contd.)
The formula for the Mann-Whitney U test for n1 and n2 values is:
U1 = n1 × n2 +[n1(n1 + 1)]
2 −R1
U2 = n1 × n2 +[n2(n2 + 1)]
2 −R2
In this example,
U1 = 12 and U2 = 13
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Mann-Whitney Test—Example (contd.)
To calculate the U value:
● U = Min (U1, U2) = Min (12, 13) = 12
● Lookup the Mann-Whitney U test table for n1 = 5 and n2 = 5.
● To be statistically significant, the obtained U value must be equal to or less than this critical value.● Since the calculated U value is 12 (not less than 2), there is no statistical difference between the mean of
the two groups.!
86
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The Kruskal-Wallis test is also a non-parametric test used for testing the source of origin of the
samples.
Characteristics of Kruskal-Wallis test are as follows:
● Only way to analyze the variance by ranks.
● Medians of two or more samples are compared to find the source of origin of the sample.
● Unlike the analogous one-way analysis of variance, it does not assume the normal distribution of
the residuals.
Kruskal-Wallis Test
● Null hypothesis is when medians of all the groups are equal, and● Alternative hypothesis is when at least one population median of one group is different than that of at
least one other group.!
87
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The Mood’s median is a non-parametric test that is used to test the equality of medians from two or
more different populations. This test works when:
● the output (Y) variable is continuous, discrete-ordinal or discrete-count, and
● the input (X) variable is discrete with two or more attributes.
The steps involved in Mood’s Median test are as follows:
Mood’s Median Test
Find median of the combined data set
Find the number of values in each sample > median
Find expected value for each cell
Find chi-square value
Form a contingency table
88
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Friedman test is a form of non-parametric test that does not make any assumptions on the shape and
origin of the sample.
● It allows smaller sample data sets to be analysed, and
● Unlike ANOVA, it does not require the dataset to be randomly sampled from normally distributed
populations with equal variances.
Friedman Test
The test uses null hypothesis where the population medians of each treatment are statistically identical to
the rest of the group.!
89
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The 1 Sample Sign test is the simplest of all the non-parametric tests that can be used instead of a
one sample t test.
● Here, H0 is the hypothecated median or assumed median of the sample, which belongs to the
population.
Steps involved in 1 Sample Sign test are as follows:
1 Sample Sign Test
Count the number of positive values
Count the number of negative values
Test the values
Values that are larger than
hypothesized median
Values that are smaller than
the hypothesized median
Check if there are significantly
more positives (or negatives)
than expected
90
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The 1 Sample Wilcoxon test also known as Wilcoxon Signed Rank test is a non-parametric test.
This test is:
● equivalent to parametric One Sample t-Test, and
● powerful than non-parametric 1 Sample Sign Test.
1 Sample Wilcoxon Test
91
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Some characteristics of this test are as follows:
● It assumes the existing sample is randomly taken from a population, with a symmetric frequency
distribution around the median, and
● The symmetry can be observed with a histogram, or by checking if the median and mean are
approximately equal.
Characteristics of 1 Sample Wilcoxon Test
The conclusion in this test is that if the value is on the mid-point, you can continue and accept the null
hypothesis. If not, reject the alternate hypothesis.!
92
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An example of Sample Wilcoxon test is shown.
1 Sample Wilcoxon Test—Example
The Median customer satisfaction score of an organization has always been 3.7 and the management
wants to see if this has changed. They conducted a survey and got the results grouped by the
customer type.
Conclusion:
● If median = 3.7 = Accept H0
● If median ≠ 3.7 = Reject Ha
● α = 0.05
93
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Quiz
Copyright 2014, Simplilearn, All rights reserved.
a.
b.
c.
d.
QUIZ Which of the following describes the population parameters based on the sample data using a particular model?1
Inferential Statistics
Probability
Correlation
Statistics
95
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ Which of the following describes the population parameters based on the sample data using a particular model?
Inferential Statistics
Probability
Correlation
Statistics
1
Answer: b.
Explanation: Inferential statistics describe the population parameters based on the sample data using a particular model.
96
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a.
b.
c.
d.
QUIZ Which of the following is an application of the population knowledge to predict the sample behavior?2
Normal distribution
Chi-square distribution
Probability distribution
Poisson distribution
97
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a.
b.
c.
d.
QUIZ Which of the following is an application of the population knowledge to predict the sample behavior?
Normal distribution
Chi-square distribution
Probability distribution
Poisson distribution
2
Answer: a.
Explanation: Poisson distribution is an application of the population knowledge to predict the sample behavior.
98
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a.
b.
c.
d.
QUIZ Which of the following is used to calculate the degree of movement of variable Y as X changes?3
Probability
F-distribution
Regression
Correlation
99
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ Which of the following is used to calculate the degree of movement of variable Y as X changes?
Probability
F-distribution
Regression
Correlation
3
Answer: d.
Explanation: The degree of movement of variable Y as X changes is calculated using regression.
100
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a.
b.
c.
d.
QUIZ A null hypothesis states that a process has not improved as a result of some modifications. The type II error is to conclude that:
we have failed to reject the null hypothesis (H0) when it is false.
we have rejected the null hypothesis.
we have made a correct decision with alpha probability.
we have failed to reject the null hypothesis (H0) when it is true.
4
101
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ A null hypothesis states that a process has not improved as a result of some modifications. The type II error is to conclude that:
we have failed to reject the null hypothesis (H0) when it is false.
we have rejected the null hypothesis.
we have made a correct decision with alpha probability.
we have failed to reject the null hypothesis (H0) when it is true.
4
Answer: b.
Explanation: A type II error means that we have failed to reject the null hypothesis (H0) when it is false.
102
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a.
b.
c.
d.
QUIZThe test used for testing significance in an analysis of variance table is the:
t-test.
F-test.
Chi-square test.
Z-test.
5
103
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZThe test used for testing significance in an analysis of variance table is the:
t-test.
F-test.
Chi-square test.
Z-test.
5
Answer: c.
Explanation: The appropriate ANOVA test is the F-test. ANOVA is a test of the equality of means.
104
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a.
b.
c.
d.
QUIZWhich of the following is the only way to analyze the variance by ranks?
6
1 Sample Size test
Friedman test
Kruskal-Wallis test
1 Sample Wilcoxon test
105
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZWhich of the following is the only way to analyze the variance by ranks?
1 Sample Size test
Friedman test
Kruskal-Wallis test
1 Sample Wilcoxon test
6
Answer: d.
Explanation: The Kruskal-Wallis test is the only way to analyze the variance by ranks.
106
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a.
b.
c.
d.
QUIZ What distribution is used while making inferences about a population variance based on a single sample from that population?
Normal distribution
t-distribution
F-distribution
Chi-square distribution
7
107
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ What distribution is used while making inferences about a population variance based on a single sample from that population?
Normal distribution
t-distribution
F-distribution
Chi-square distribution
7
Answer: a.
Explanation: The chi-square distribution is used to compare a sample variance with a known population variance.
108
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a.
b.
c.
d.
QUIZIf p-value is less than the significant value, the null hypothesis has to be:
accepted.
maintained as it is.
re-evaluated.
rejected.
8
109
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZIf p-value is less than the significant value, the null hypothesis has to be:
accepted.
maintained as it is.
re-evaluated.
rejected.
8
Answer: a.
Explanation: If the p-value is less than the significant value, the null hypothesis has to be rejected as the data is not supporting the null hypothesis and the difference will be statistically significant.
110
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a.
b.
c.
d.
QUIZ Which of the following is a nonparametric test that is used to test the equality of medians from two or more different populations?9
Kruskal-Wallis test
Friedman test
1 Sample Sign test
Mood’s median test
111
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ Which of the following is a nonparametric test that is used to test the equality of medians from two or more different populations?
Kruskal-Wallis test
Friedman test
1 Sample Sign test
Mood’s median test
9
Answer: a.
Explanation: The Mood’s median is a nonparametric test that is used to test the equality of medians from two or more different populations.
112
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a.
b.
c.
d.
QUIZWhich of the following is a ratio of two chi-square distributions?
10
t-distribution
Poisson distribution
Binomial distribution
F-distribution
113
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a.
b.
c.
d.
QUIZWhich of the following is a ratio of two chi-square distributions?
t-distribution
Poisson distribution
Binomial distribution
F-distribution
10
Answer: a.
Explanation: The F-distribution is a ratio of two chi-square distributions.
114
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a.
b.
c.
d.
QUIZ Which of the following is the probability of correctly rejecting the null hypothesis when it is false?11
Power of a test
Simple linear regression
1 Sample Sign test
Simple linear correlation
115
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ Which of the following is the probability of correctly rejecting the null hypothesis when it is false?
Power of a test
Simple linear regression
1 Sample Sign test
Simple linear correlation
11
Answer: b.
Explanation: The power of a test is the probability of correctly rejecting the null hypothesis when it is false.
116
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a.
b.
c.
d.
QUIZ Which of the following assumes that the existing sample is randomly taken from a population, with a symmetric frequency distribution around the median?12
Mood’s median test
1 Sample Wilcoxon test
Friedman test
Kruskal-Wallis test
117
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Copyright 2012-2014,Simplilearn,All rights reserved
a.
b.
c.
d.
QUIZ Which of the following assumes that the existing sample is randomly taken from a population, with a symmetric frequency distribution around the median?
Mood’s median test
1 Sample Wilcoxon test
Friedman test
Kruskal-Wallis test
12
Answer: c.
Explanation: 1 Sample Wilcoxon test assumes that the existing sample is randomly taken from a population, with a symmetric frequency distribution around the median.
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● Discrete probability distribution is characterized by the probability mass
function and continuous probability distribution is characterized by the
probability density function.
● Multi-Vari studies are used to analyze variation in a process.
● Correlation means association between variables. Simple Linear Regression
and Multiple Regression are its two main techniques.
● Hypothesis testing is conducted on different sets of data. Analysis of
variance is used to compare the means of more than two sample sets.
● A t-test is used for 1-sample and 2-sample tests are used for comparing two
means.
Here is a quick recap of what we have learned in this lesson:
Summary
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● The Mann-Whitney or Wilcoxon Rank Sum test is used to compare two
unpaired groups.
● The Kruskal–Wallis Test is used for testing the source of origin of samples.
● The Mood’s median test is used to test the equality of medians from two
or more different populations.
● The Friedman test does not make any assumptions on the shape and
origin of the sample.
● The 1 Sample Sign test is the simplest of all the non-parametric tests that
can be used in the place of a 1 sample t-test.
Here is a quick recap of what we have learned in this lesson:
Summary (contd.)
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Copyright 2014, Simplilearn, All rights reserved.
Copyright 2014, Simplilearn, All rights reserved.
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