What is the Coefficient of Contingency?

The Coefficient of Contingency (C) measures the strength of association between two categorical variables using data arranged in a contingency table.

It is derived from the Chi-Square test statistic, which tests whether two categorical variables are independent.

In simple terms:

  • Chi-Square tells you if a relationship exists
  • Coefficient of Contingency tells you how strong the relationship is

This makes it an effect size measure for categorical data.

When It Is Used

The coefficient is useful when analyzing relationships between variables such as:

Variable 1Variable 2
MachineDefect Type
ShiftError Rate Category
SupplierQuality Grade
DepartmentTraining Level

These are non-numeric categories, which means traditional correlation methods do not apply.

Understanding Contingency Tables

The coefficient is calculated from a contingency table, which summarizes the frequency distribution of two categorical variables.

Example Contingency Table

Imagine a manufacturing team studying whether machine type affects defect occurrence.

MachineDefectNo DefectTotal
Machine A2080100
Machine B4060100
Total60140200

This table allows us to analyze whether defects occur independently of machine type.

In Six Sigma, this type of analysis commonly occurs during the Analyze phase of
DMAIC.

The Coefficient of Contingency Formula

The coefficient is calculated from the Chi-Square statistic.

The formula is:

[
C = \sqrt{\frac{\chi^2}{\chi^2 + n}}
]

Where:

  • C = Coefficient of Contingency
  • χ² = Chi-Square statistic
  • n = total sample size

This equation scales the Chi-Square value into a range between 0 and less than 1.

Visual Formula Breakdown

ComponentMeaning
χ²Measures deviation from independence
nNumber of observations
RatioAdjusts Chi-Square for sample size
Square rootNormalizes the scale

Why the Coefficient Never Reaches 1

One of the most confusing aspects of this metric is that it never equals 1, even when variables are perfectly associated.

This happens because the maximum value depends on the size of the contingency table.

The theoretical maximum is:

[
C_{max} = \sqrt{\frac{k-1}{k}}
]

Where k is the smaller dimension of the contingency table.

Example

For a 2×2 table:

[
C_{max} = \sqrt{\frac{1}{2}} = 0.707
]

Meaning the largest possible value is only 0.707, not 1.

Because of this limitation, many statisticians prefer Cramer’s V when comparing different table sizes.

Step-by-Step Calculation Example

Let’s calculate the coefficient using the earlier defect example.

Step 1: Observed Data

MachineDefectNo DefectTotal
A2080100
B4060100
Total60140200

Step 2: Calculate Expected Frequencies

Expected frequency formula:

[
E = \frac{(Row\ Total)(Column\ Total)}{n}
]

Example:

For Machine A / Defect:

[
E = \frac{100 \times 60}{200} = 30
]

Expected table:

MachineDefectNo Defect
A3070
B3070

Step 3: Calculate Chi-Square

[
\chi^2 = \sum \frac{(O – E)^2}{E}
]

CellOECalculation
A Defect20303.33
A No Defect80701.43
B Defect40303.33
B No Defect60701.43

Total:

[
\chi^2 = 9.52
]


Step 4: Calculate Coefficient of Contingency

[
C = \sqrt{\frac{9.52}{9.52 + 200}}
]

[
C = \sqrt{0.045}
]

[
C \approx 0.21
]

Interpreting the Coefficient

Unlike correlation coefficients, there are no universal interpretation thresholds.

However, practitioners often use approximate ranges:

CoefficientInterpretation
0.00-0.10Very weak association
0.10-0.30Weak association
0.30-0.50Moderate association
0.50+Strong association

In the example:

C = 0.21

This suggests a weak association between machine type and defects.

This means:

  • Machine type may contribute to defects
  • But it is likely not the primary root cause

Role in Six Sigma Analysis

The coefficient of contingency is particularly useful in the Analyze phase of Six Sigma projects.

It helps teams quantify relationships between categorical variables.

Typical Applications

Defect Root Cause Analysis

VariableExample
MachineDefect type
OperatorError category
SupplierPart failure

Customer Experience Analysis

VariableExample
RegionComplaint type
Product versionReturn reason

Process Investigation

VariableExample
ShiftQuality outcome
Training statusError frequency

These analyses support the goal of identifying statistically meaningful relationships before implementing improvements.

DMAIC Case Study Example

Problem

A call center noticed higher customer complaints on certain shifts.

The Six Sigma team wanted to determine whether shift assignment influences complaint type.

Data Collected

ShiftBilling ComplaintService ComplaintOther
Morning403020
Afternoon203525
Night102030

Total observations = 210

Step 1: Conduct Chi-Square Test

The test shows p < 0.05, indicating complaints vary by shift.

Step 2: Calculate Coefficient of Contingency

The coefficient equals 0.31.

Interpretation

This indicates a moderate association between shift and complaint type.

Root Cause Insight

Further investigation revealed:

  • Night shift agents handled more complex service issues
  • Billing specialists primarily worked morning shifts

The analysis helped the team redesign call routing procedures.

Coefficient of Contingency vs Other Association Measures

Several statistics measure categorical association.

Understanding the differences helps practitioners choose the correct metric.

Comparison Table

StatisticBest ForKey Advantage
Coefficient of ContingencySmall contingency tablesSimple to calculate
Cramer’s VAny table sizeStandardized to 1
Phi Coefficient2×2 tablesEquivalent to correlation
Chi-SquareHypothesis testingDetects existence of relationship

When to Use Cramer’s V Instead

Coefficient of Contingency has a major limitation:

Its maximum value changes with table size.

Cramer’s V solves this problem by standardizing the range from 0 to 1 regardless of dimensions.

For that reason, many modern statistical tools default to Cramer’s V.

However, the coefficient still appears frequently in statistical textbooks and exam questions.

Relationship to Effect Size

In Six Sigma and statistics, effect size measures how meaningful a relationship is.

A statistically significant result does not always mean the effect is important.

Example:

A huge sample size may produce p < 0.001, but the relationship could still be extremely weak.

Coefficient of Contingency helps answer:

“How strong is the relationship?”

This distinction is critical for data-driven decision making.

Practical Tips for Six Sigma Practitioners

Tip 1: Always Pair with Chi-Square

Coefficient of Contingency should never be used alone.

First test for independence using Chi-Square.

Then measure association strength.

Tip 2: Watch Sample Size

Large samples can inflate Chi-Square values.

This may produce misleadingly high coefficients.

Tip 3: Use Visualization

Heatmaps and mosaic plots often reveal patterns faster than raw statistics.

Example visualizations:

  • Contingency heatmaps
  • Mosaic charts
  • Segmented bar charts

Tip 4: Compare with Other Metrics

For deeper analysis, compare results with:

  • Cramer’s V
  • Odds ratios
  • Risk ratios

Each provides a different perspective on the relationship.

Common Pitfalls

Misinterpreting the Range

Because the coefficient cannot reach 1, comparing values across tables can be misleading.

Using with Continuous Data

The coefficient is only for categorical variables.

Continuous data requires different metrics such as:

  • Correlation
  • Regression

Ignoring Expected Frequencies

Chi-Square assumptions require:

  • Expected frequencies ≥ 5 in most cells

Violations may invalidate the analysis.

How This Appears on Six Sigma Certification Exams

Questions typically test:

  • Understanding the formula
  • Interpreting association strength
  • Identifying appropriate use cases

Example exam question:

A contingency table analysis produces:

χ² = 15.3
n = 250

What is the coefficient of contingency?

Using the formula:

[
C = \sqrt{\frac{15.3}{265.3}}
]

[
C \approx 0.24
]

Interpretation:

Weak to moderate association.

This type of question often appears on the Green Belt exam preparation materials such as
https://sixsigmastudyguide.com/pass-your-six-sigma-green-belt/

Key Takeaways

The Coefficient of Contingency is a useful measure for evaluating the strength of association between categorical variables.

Important points to remember:

  • It is derived from the Chi-Square statistic
  • Values range from 0 to less than 1
  • Maximum value depends on table size
  • It measures association, not correlation
  • Best used alongside Chi-Square hypothesis testing

For Six Sigma practitioners, this metric helps quantify relationships uncovered during the Analyze phase, allowing teams to prioritize root causes based on data rather than assumptions.

Week 4: Six Sigma Analyze Phase Overview

Welcome to Week 4, Day 1 of the Six Sigma Green Belt review!

During the first few weeks we covered a ton of material!

We also covered the anchoring technique – in my opinion the best way to enable quick and easy recall of any material. That’s where you try to associate each topic you cover with something you already know.

I find it’s really helpful to list out specific examples of where you’ve seen the Six Sigma practices in real life (on the job, at the hospital, etc) or, barring that, where you think something would really benefit from application of those practices.

If you didn’t have a chance to review those topics mentioned above, go back now and catch up. The context supplied with those topics gives an excellent foundation for our studying.

Week 4: Six Sigma Analyze Phase

Today we’re starting with the Analyze Phase of Six Sigma’s DMAIC process. Just like the previous weeks, before we start I’d like to begin with a note that many different certifying bodies exist and I am trying to help people that are candidates in all of them. This means that I may refer to things that aren’t specifically on your exam.

Your Assignment

Here's what I'd like you to do.

First, you’ll need to log in to the site.

Free membership or paid, it doesn’t matter. You just have to be logged in. Buttons to sign up and log in are below.

Second, review the articles I list below.

These are the concepts I refer exam candidates to when diving into these particular sections.

Take a look.

Introducing Six Sigma Analyze Phase

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What Do You Think?

I’d love to know what you think about the material you need to know in these sections. Please take a minute to share your thoughts below. I’ll use these responses to decide the best way to improve the site to help the most number of people.

Next Time

Next time we will dive into the specific requirements of these topics across of a few popular certifying bodies (ASQ, IASSC, Villanova).

Happy studying!

ASQ Six Sigma Yellow Belt Study Guide

Preparing for the ASQ Six Sigma Yellow belt exam can be overwhelming. This ASQ Six Sigma Yellow Belt Study Guide lists all key concepts (and links to learn more) for the ASQ Six Sigma Yellow Belt Certification. This list comes from the 2022 update to the official ASQ Yellow Belt BOK. And it’s totally free to use.

What is ASQ Six Sigma Yellow Belt BOK

The ASQ Six Sigma Yellow Belt BOK is created by the American Society of Quality (ASQ). It’s periodically reviewed and updated by leading members and Six Sigma professionals to ensure it’s relevant to today’s world.

The ASQ Six Sigma Yellow Belt BOK has 5 sections geared around Six Sigma Fundamentals and the different phases of DMAIC.

You’ll need to master various topics in order to pass the exam.

How to use the ASQ Six Sigma Yellow Belt Study Guide

Some ASQ Six Sigma Yellow Belt candidates will have taken six sigma training. Others will be self-taught. No matter your path, you’re going to have to know what’s on the standard BOK – Body of Knowledge.

Step 1: Scan over the material below.

Scan through the rest of this page before proceeding further. Take a look at what will be asked. Do you feel comfortable with this material? Is anything new to you? Are there any surprises?

Make a mental note of that and proceed.

Step 2: Review (or learn!) the material.

Not every training program covers every topic. Depending on the quality of the training you took, your instructor may have skipped a section entirely! Click one of the links to dive into any topic that you don’t feel as strongly as you’d like and read up.

Step 3: Leave a comment on any page where you have questions.

Each of the topics below is linked to a detailed article where I thoroughly step through the concept. If you have any questions, or if any topic is less clear than you’d like it to be just leave a comment and I’ll get back to you ASAP.

Step 4: Practice as many questions as possible.

Find an excellent question bank that covers each of these sections and take as many practice tests as possible.

The ASQ Six Sigma Yellow Belt BOK

What follows below is the ASQ Six Sigma Yellow Belt BOK. The top-level sections I through V also list the number of questions you can expect in that section. This will help you see the relative weight of those topics compared to others.

I. Six Sigma Fundamentals (20 Questions)

I.A. Six Sigma Foundations and Principles

Describe the purpose of Six Sigma (reducing variation), its methodology (DMAIC), and its evolution from quality. Describe the value of Six Sigma to the organization as a whole. (Understand)

I.B. Lean Foundations and Principles

Describe the purpose of lean (waste elimination) and its methodologies (just-in-time, poka-yoke, kanban, value stream mapping). Describe the value of lean to the organization as a whole. (Understand)

I.C. Six Sigma Roles and Responsibilities

Define and describe the roles and responsibilities of Six Sigma team members (i.e., individual team members, Yellow Belt, Green Belt, Black Belt, Master Black Belt, process owner, champion, sponsor). (Understand)

I.D. Team Basics

I.D.1. Types of teams

Identify the various types of teams that operate within an organization (i.e., continuous improvement, self-managed, and cross-functional) and their value. (Understand)

I.D.2. Stages of development

Describe the various stages of team evolution: forming, storming, norming, performing, and adjourning. (Understand)

I.D.3. Decision-making tools

Define and apply decision-making tools such as brainstorming, multivoting, and nominal group technique (NGT). (Apply)

I.D.4. Communication methods

Explain how teams use agendas, meeting minutes, and project status reports, and how they support project success. (Apply) Also, see Team Dynamics.

I.E. Quality Tools and Six Sigma Metrics

I.E.1. Quality tools

Select and use these tools throughout the DMAIC process: Pareto charts, cause and effect diagrams, flowcharts, run charts, check sheets, scatter diagrams, and histograms. (Apply) Also see 7M tools & What is Quality?

I.E.2. Six Sigma metrics

Select and use these metrics throughout the DMAIC process: defects per unit (DPU), defects per million opportunities (DPMO), rolled throughput yield (RTY), cycle time, and cost of poor quality (COPQ). (Apply)

II. Define Phase (14 Questions)

II.A. Project Identification

II.A.1. Voice of the customer

Define the voice of the customer and describe how customer needs are translated into quantifiable, critical-to-quality (CTQ) characteristics. (Understand)

II.A.2. Project selection

Describe how projects are identified and selected as suitable for a Six Sigma project using the DMAIC methodology. (Understand)

II.A.3. Stakeholder analysis

Identify end users, subject matter experts, process owners, and other people or factors that will be affected by a project, and describe how each of them can influence the project. (Understand)

II.A.4. Process inputs and outputs

Use SIPOC (suppliers, inputs, process, outputs, customers) to identify and define important elements of a process. (Apply)

II.A.5. Supply chain management

Understand supply chain management and how it relates to project management. (Apply)

II.B. Project Management (PM) Basics

II.B.1. Project charter

Describe the purpose of a project charter and its components: problem statement, project scope, baseline data, and project goal. (Understand)

II.B.2. Communication plan

Explain the purpose and benefits of a communication plan and how it can impact the success of the project. (Understand)

II.B.3. Project planning

Define work breakdown structure (WBS) and Gantt charts, and describe how they are used to plan and monitor projects. (Understand)

II.B.4. Project management tools

Select and use various Project Management tools: activity network diagrams, affinity diagrams, matrix charts, relations charts, and tree diagrams. (Understand)

II.B.5. Phase reviews

Explain how tollgate or phase reviews are used throughout the DMAIC lifecycle. (Understand)

III. Measure Phase (15 Questions)

III.A. Basic Statistics

Define, calculate, and interpret measures of central tendency (mean, median, mode) and understand measures of dispersion (standard deviation, range, variance). (Apply) Also, see Kinds of Statistics.

Rusty on the math behind these topics? Join the Math Confidence for Six Sigma waitlist and I’ll let you know when the course opens.

III.B. Data Collection

III.B.1. Data collection plans

Describe the critical elements of a data collection plan, including an operational definition, data sources, the method to be used for gathering data, and how frequently it will be gathered. Describe why data collection plans are important. (Understand)

III.B.2. Qualitative and quantitative data

Define and distinguish between these types of data. (Understand)

III.B.3. Data collection techniques

Use various data collection techniques, including surveys, interviews, check sheets, and checklists to gather data that contributes to the process being improved. (Apply)

III.C. Measurement System Analysis (MSA)

III.C.1. MSA terms

Define precision, accuracy, bias, linearity, and stability, and describe how these terms are applied in the measurement phase. (Understand) See Measurement System Analysis (MSA).

III.C.2. Gauge repeatability and reproducibility (GR&R)

Describe and distinguish between repeatability and reproducibility and describe how and why GR&R is used in the measurement phase. (Understand)

IV. Analyze Phase (17 Questions)

IV.A. Process analysis tools

IV.A.1. Lean tools

Define how 5S and value analysis can be used to identify and eliminate waste. (Understand)

IV.A.2. Failure mode and effects analysis (FMEA)

Relate the elements of severity, occurrence, and detection, and determine how they are used to
calculate the risk priority number. Demonstrate how FMEA can be used to identify potential failures in a process. (Apply)

IV.B. Root cause analysis

Describe how the 5 Whys, process mapping, 8D, force-field analysis, and matrix charts can be used to identify the root causes of a problem. (Understand)

IV.C. Corrective action

Explain and apply elements of the corrective action process: identify the problem, contain the problem (interim action), determine the causes of the problem and propose solutions to eliminate it or prevent its recurrence (permanent action), verify that the solutions are implemented, and confirm their effectiveness (validation). (Apply)

IV.D. Preventive action

Explain and apply elements of a preventive action process: understand various process analysis techniques to identify potential failures, defects, or process deficiencies; improve the process (e.g., understand error- or mistake-proofing devices or methods, initiate procedural changes), and verify the effectiveness of the preventive action. (Apply)

IV.E. Data analysis

IV.E.1. Basic distribution types

Define and distinguish between normal and binomial distributions and describe how their shapes (skewed and bimodal) can affect data interpretation. (Understand)

IV.E.2. Common and special cause variation

Describe and distinguish between these types of variation. (Understand)

IV.F. Correlation and regression

IV.F.1. Correlation

Describe how correlation is used to identify relationships between variables. (Understand)

IV.F.2. Regression

Describe how regression analysis is used to predict outcomes. (Understand)

IV.G. Hypothesis testing

Define and distinguish between hypothesis terms (i.e., null and alternative, type I and type II error, p-value, and power). (Understand) Also, see Hypothesis testing overview.

V. Improve and Control Phases (14 Questions)

V.A. Improvement techniques

V.A.1. Kaizen and kaizen blitz

Define and distinguish between these two methods and describe how they can be used to make
improvements to any process in an organization. (Understand)

V.A.2. Plan-do-check-act (PDCA) cycle

Define and distinguish between the steps in this process improvement tool. (Understand)

V.A.3. Cost-benefit analysis

Explain the importance of this analysis and how it is used in the improve phase. (Understand)

V.B. Control Tools and Documentation

V.B.1. Control plan

Describe the importance of a control plan for maintaining improvements. (Understand)

V.B.2. Control charts

Describe how X-R charts are used for monitoring and sustaining improved processes. (Understand)

V.B.3. Document control

Describe the importance of documenting changes to a process and communicating those changes to stakeholders. (Understand)

V.B.4. Work instructions and standard operating procedures (SOPs)

Understand the purpose and use of work instructions and SOPs. (Understand)

ASQ Six Sigma Yellow Belt Study Guide Additional Resources

https://p.widencdn.net/fm4qxd/40197_SSYB_Cert_Insert

https://asq.org/cert/six-sigma-yellow-belt#body-of-knowledge

https://asq.org/cert/resource/pdf/certification/2022-SSYB-BoK-Map.pdf

Archive

Two Sample T Hypothesis Tests

What is a Two Sample T Hypothesis Test?

A two sample t hypothesis tests also known as independent t-test is used to analyze the difference between two unknown population means. The Two-sample T-test is used when the two small samples (n< 30) are taken from two different populations and compared. The underlying chart makes use of the T distribution.

Assumptions of Two Sample T Hypothesis Tests

  • The sample should be randomly selected from the two population
  • Samples are independent to each other
  • Population standard deviations are unknown (there is no rule that each sample must be smaller than 30)
  • Samples collected from the population are normally distributed

When Would You Use a Two Sample T Hypothesis Tests?

The two sample t test most likely used to compare two process means, when the data is having one nominal variable and one measurement variable. It is a hypothesis test of means. Use two sample Z test if the sample size is more than 30.

The two sample hypothesis t test is used to compare two population means, while analysis of variance (ANOVA) is the best option if more than two group means are compared.

Two sample T hypothesis tests are performed when the two group samples are statistically independent of each other, while the paired t-test is used to compare the means of two dependent or paired groups. 

Note: There are (2) types of Two Sample T Hypothesis tests!

  • Two Sample T Hypothesis Test (Equal Variance)
    • Variance of two populations are equal
  • Two Sample T Hypothesis Test (Unequal Variance)
    • Variance of two populations are NOT equal

Methods to determine population varince equal or unequal?

The best method to determine population variance is equal or unequal by using an appropriate F-test.

Hypothesis Testing

A tailed hypothesis is an assumption about a population parameter. The assumption may or may not be true. One-tailed hypothesis is a test of hypothesis where the area of rejection is only in one direction. Whereas two-tailed, the area of rejection is in two directions. The selection of one or two-tailed tests depends upon the problem.

  • Null hypothesis- H0: The population means are same alternatively the difference between two population means are equal to hypothesized difference (d). So, µ1 = µ2  orµ1– µ2 = d
  • Alternative hypothesis: µ1 ≠ µ2  orµ1– µ2 ≠ d (Two-tailed test)
  • µ1 < µ2  orµ1– µ2 < d (left-tailed)
  • µ1 > µ2  orµ1– µ2 > d (Right-tailed)

Two Sample T Hypothesis Test (Equal Variance) formula

Two Sample T Hypothesis Tests
  • Where n1 and n2 are sample sizes
  • x̅1 and x̅2 are means of sample sizes
  • Sp is the pooled standard deviation

Steps to Calculate Two Sample T Hypothesis Test (Equal Variance)

  • State the claim of the test and determine the null hypothesis and alternative hypothesis
  • Determine the level of significance
  • Calculate degrees of freedom
  • Find the critical value
  • Calculate the test statistics
  • Make a decision, the null hypothesis will be rejected if the test statistic falls in the rejection region: greater than the critical value for a right-tailed test, less than the negative critical value for a left-tailed test, or beyond either one for a two-tailed test
  • Finally, Interpret the decision in the context of the original claim.

Example of a Two Sample T Hypothesis Test (Equal Variance) in a DMAIC Project

Two Sample T test mostly performed in Analyze phase of DMAIC to evaluate the difference between two process means are really significant or due to random chance, this is basically used to validate the root cause(s) or Critical Xs (see the below example for more detail)

Two-tailed (Equal variance)

Example: Apple orchard farm owner wants to compare the two farms to see if there are any weight difference in the apples. From farm A, randomly collected 15 apples with an average weight of 86 gms, and the standard deviation is 7. From farm B, collected 10 apples with an average weight of 80 gms and standard deviation of 8. With a 95% confidence level, is there any difference in the farms?

  • Null Hypothesis (H0) : Mean apple weight of farm A is equal to farm B
  • Alternative Hypothesis (H1) : Mean apple weight of farm A is not equal to farm B
  • n1=15
  • n2=10
  • S12=49
  • S22 =64
  • X̅1 =86
  • X̅2 = 80

Significance level: α=0.05

Degrees of freedom df: 15+10-2= 23

Calculate critical value

Refer two tailed t table for 23 degrees of freedom

One and Two Sample T Tests
Two Tailed T Test

If the calculated t value is less than -2.069 or greater than 2.069, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value less than the critical value, hence failed to reject null hypothesis ( H0). So, there is no significant difference between mean weights of apples in farm A and farm B.

Two Sample T 2 Tailed Equal Variance template file

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Right-Tailed (Equal Variance)

Example: The thickness of ceramic tile of vendor A is slightly greater than the vendor 2. The researcher randomly collected samples from both vendors. Assuming the population variances are equal, at 95% confidence level, is there enough evidence to support the claim?

Two Sample T Hypothesis Tests
  • Null Hypothesis (H0) : Mean tile thickness of vendor A is equal to vendor B
  • Alternative Hypothesis (H1) : Mean tile thickness of vendor A is greater than vendor B
  • Significance level: α=0.05
  • n1=12
  • n2=12

Degrees of freedom df: n1+n2-2=12+12-2= 22

Calculate mean (X̅) of each sample

  • X̅1 = (x1 + x2 + ….+ xn) / n =18.7
  • X̅2 = (x1 + x2 + ….+ xn) / n =18.5

Calculate variance (S2) of each sample

  • S12 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1 =8.61
  • S22 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1=16.09

Calculate critical value

Refer one tailed t table for 22 degrees of freedom

If the calculated t value is greater than 1.717, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Two Sample T Hypothesis Tests

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is less than the critical value and it is not in rejection region, hence failed to reject null hypothesis ( H0). So, there is no significant difference between mean tile thickness of vendor A and vendor B.

Two Sample T Right Tailed Equal Variance template file

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Left-Tailed (Equal Variance)

Example: Samples of sales in similar shops in two different US states are taken for a new product with the following results. Is there any evidence that sales in state A is less than the state B. Assuming the population variances are equal, at 95% confidence level, is there enough evidence to support the claim?

  • Null Hypothesis (H0) : Mean sales in state A is equal to the state B
  • Alternative Hypothesis (H1) : Mean sales in state A is lesser than state B
  • Significance level: α=0.05
  • n1=12
  • n2=12

Degrees of freedom df: n1+n2-2=12+12-2= 22

Calculate mean (X̅) of each sample

  • X̅1 = (x1 + x2 + ….+ xn) / n =44.7
  • X̅2 = (x1 + x2 + ….+ xn) / n =42.8

Calculate variance (S2) of each sample

  • S12 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1 =53.3
  • S22 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1=28.2

Calculate critical value

Refer one tailed t table for 22 degrees of freedom

If the calculated t value is less than -1.717, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value (0.735) is greater than the critical value (-1.717) and it is not in the rejection region, hence we failed to reject the null hypothesis ( H0). So, there is not enough evidence that mean sales in state A is less than the state B.

Two Sample T Left Tailed Equal Variance template file

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Two Sample T Hypothesis Test (Equal Variance) Videos

2 SAMPLE T-TEST FOR MEANS (UNEQUAL VARIANCE)

Test for equality of two means.  This test is used to determine if the mean of one sample is not equal to the mean of another sample.

Two Sample T Hypothesis Test (Unequal Variance) formula

Two Sample T Hypothesis Tests
  • Where n1 and n2 are sample sizes
  • S12 & S22 are variances of sample 1 and sample 2
  • x̅1 and x̅2 are means of sample sizes

ASSUMPTIONS:

  • Observations (data):
    • Are Normally distributed.
      • Note: Although this method is relatively robust against non-normality the shape of distribution should at least be symmetric.
    • Within and between samples (groups) are independent of each other.
    • Are randomly “drawn”.
    • Are continuous.
      • Depending on some conditions (e.g. wide data range and field of applications), data that is not continuous (e.g. count) may still be used.
      • Not to use certain ratio type data (e.g. proportions and percentages).
  • Variances are unknown and estimated by sample variances (S2).
  • Variances are not equal, S12 ≠ S22

Steps to Calculate Two Sample T Hypothesis Test (Unequal Variance):

Step 1 – Calculate mean (X̅) of each sample

  •  
    X̅ = (x1 + x2 + ….+ xn) / n
  • Where, x1, x2, …xn are observations in each sample (“n1” = sample size in sample 1 and “n2” = sample size in sample 2)

Step 2 – Calculate variance (S2) of each sample

  • S2 = {(X̅- x1)2 + (
    X̅ – x2)2 + … +(X̅ – xn)2} / n-1

Step 3 – Calculate t value

Two Sample T Hypothesis Tests

Step 4 – Calculate degrees of freedom (DF)

Two Sample T Hypothesis Tests
  • Round to nearest integer

Step 5: Compare t-calc with critical value in the t-distribution table.

  • Look up the t-critical value in t-distribution table given calculated degrees of freedom in column α (significance level):
    • For 1 tailed test the t-critical is based on the given α
      • The commonly used T-critical value for a 1 tailed test is in the 0.05 α column
    • For 2 tailed test the t-critical is based on the given α divided by 2
      • The commonly used T-critical value for a 2 tailed test is in the 0.025 α column

INTERPRETATION:

  • Finally, interpret the decision in the context of the original claim.

Example of a Two Sample T Hypothesis Test (Unequal Variance) in a DMAIC Project

Two-tailed (Unequal Variance)

Example: Dogs are fed with two different diets. Test if the two diets differ significantly as regards their effect on their weight. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

  • Null Hypothesis (H0) : The mean dog weights are the same with Diet 1 and Diet 2
  • Alternative Hypothesis (H1) : The mean dog weights are not the same with Diet 1 and Diet 2
  • Significance level: α=0.05
  • n1=10
  • n2=10

Calculate mean (X̅) of each sample

  • X̅1 = (x1 + x2 + ….+ xn) / n =24.4
  • X̅2 = (x1 + x2 + ….+ xn) / n =20.00

Calculate variance (S2) of each sample

  • S12 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1 =25.6
  • S22 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1=56.22
  • S12/n1=2.56
  • S22/n2=5.62

Calculate critical value

Calculate the degrees of freedom

Refer two tailed t table for 16 degrees of freedom

If the calculated t value is less than -2.120 or greater than 2.120, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is less than the critical value and it is not in the rejection region, hence we failed to reject the null hypothesis ( H0). So, the mean dog weights are the same with Diet 1 and Diet 2.

Two Sample T 2 Tailed Unqual Variance template file

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Right-tailed (Unequal Variance)

Example: A researcher believes that students in school A have better computer knowledge than school B. Randomly computer marks were collected from the two schools for the same curriculum. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

  • Null Hypothesis (H0) : Mean students marks in school A is equal to the school B
  • Alternative Hypothesis (H1) : Mean students marks in school A is greater than the school B
  • Significance level: α=0.05
  • n1=19
  • n2=17

Calculate mean (X̅) of each sample

  • X̅1 = (x1 + x2 + ….+ xn) / n =26.19
  • X̅2 = (x1 + x2 + ….+ xn) / n =20.65

Calculate variance (S2) of each sample

  • S12 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1 =29.23
  • S22 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1=84.95
  • S12/n1=1.54
  • S22/n2=5.00

Calculate critical value

Calculate the degrees of freedom

Two Sample T Hypothesis Tests
Two Sample T Hypothesis Tests

Refer one tailed t table for 25 degrees of freedom

If the calculated t value is greater than 1.708, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Two Sample T Hypothesis Tests

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is greater than the critical value and it is in the rejection region, hence we reject the null hypothesis ( H0). So, the mean students marks in school A is greater than the school B.

Two Sample T Right Tailed Unequal Variance template file

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Left-tailed (Unequal Variance)

Example: A manufacturer believes that breaking strength of plastic parts of supplier A is less than the supplier B. Operator randomly checked the breaking strength of plastic parts of two suppliers. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

  • Null Hypothesis (H0) : Mean breaking strength of plastic parts of supplier A is equal to the supplier B
  • Alternative Hypothesis (H1) : Mean breaking strength of plastic parts of supplier A is less than supplier B
  • Significance level: α=0.05
  • n1=12
  • n2=12

Calculate mean (X̅) of each sample

  • X̅1 = (x1 + x2 + ….+ xn) / n =161.67
  • X̅2 = (x1 + x2 + ….+ xn) / n =158.08

Calculate variance (S2) of each sample

  • S12 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1 =10.97
  • S22 = {(X̅- x1)2 + (X̅ – x2)2 + … +(X̅ – xn)2} / n-1=25.72
  • S12/n1=0.91
  • S22/n2=2.14

Calculate critical value

Calculate the degrees of freedom

Two Sample T Hypothesis Tests

Refer one tailed t table for 19 degrees of freedom

If the calculated t value is less than -1.729, then reject the null hypothesis.

Test Statistic

Two Sample T Hypothesis Tests

Interpret the results:

Two Sample T Hypothesis Tests

Compare t calc to t critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is greater than the critical value and it is not in the rejection region, hence we failed to reject the null hypothesis ( H0). So, Mean breaking strength of plastic parts of supplier A is equal to the supplier B.

Two Sample T Left Tailed Unequal Variance template file

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Two Sample T Hypothesis Test (Unequal Variance) Videos

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1 Sample Wilcoxon Non Parametric Hypothesis Test

1 sample Wilcoxon non parametric hypothesis test is one of the popular non-parametric tests. One sample t-test is to compares the mean of the population to the known value (i.e more than, less than, or equal to a specific known value). The t-test always assumes that random data and the population standard deviation is unknown.

Wilcoxon Signed-Rank test is the equivalent non-parametric t-test, and this may be used when the dependent variable is not normally distributed. The Wilcoxon signed test is designed to test hypotheses about the location (median) of a population distribution.

History

The Wilcoxon signed test was first developed by Frank Wilcoxon, an American chemist in the year 1945, but popularized by Sidney Siegel in 1956. Wilcoxon Signed test can be used for a single sample, matched paired data (example before and after data), and also for unrelated samples ( it is almost similar to Mann Whitney U test).

Disambiguation

1 sample Wilcoxon non parametric hypothesis test is a rank-based test and it compares the standard value (theoretical value) with the hypothesized median. Usually, the t-test depends on the sample mean which is not so stable in heavy-tailed distribution; hence Wilcoxon test efficiency is high when compared to the t-test.

The name Sign test and Wilcoxon signed ranked test look similar and both are used for one sample & two samples, but Wilcoxon signed rank test is more powerful than the signed test.

Wilcoxon Signed-Rank Test for Paired Samples: This test is mainly an alternative to the t-test for paired samples i.e. if the requirements for the two paired t-tests are not satisfied then we can easily perform this test. It has three requirements all of which should be satisfied in order to perform this test. It has two methods: the exact one and the advanced one.

Hypothesis of 1 sample Wilcoxon Signed test

Left-tailed test:

  • Null Hypothesis H0: The population median (η) is greater than or equal to hypothesized median(η0)- η≥ η0
  • Alternative Hypothesis : H1: The population median(η) is less than the hypothesized median(η0): η<η0

Right-tailed test:

  • Null Hypothesis H0: The population median (η) is less than or equal to hypothesized median(η0)- η≤η0
  • Alternative Hypothesis : H1: The population median(η) is greater than the hypothesized median(η0)- η>η0

Two-tailed test:

  • Null Hypothesis H0: The population median (η) is equal to hypothesized median(η0)- η=η0
  • Alternative Hypothesis : H1: The population median(η) is not equal to the hypothesized median(η0)- η≠η0

Assumptions of the one sample Wilcoxon test

  • Differences between the data value and the hypothesized median are continuous
  • Data follows the symmetric distribution
  • Observations are mutually independent of each other
  • Measurement scale is at least an interval

Procedure to execute One Sample Wilcoxon Non Parametric Hypothesis Test

  • Identify the difference between each individual value and the median
  • If the difference between the individual value and the median is zero, ignore it.
  • Ignore the signs of the different values and assign the lowest rank to the smallest difference value. If the values have been tied, then consider the mean value.
  • Compute the sum of ranks of positive difference values, and negative difference values (W+ and W-)
  • If the values are (>20), the normal approximation would be
1 Sample Wilcoxon Non Parametric Hypothesis Test
1 Sample Wilcoxon Non Parametric Hypothesis Test

Where t is the rank of tied values

  • Calculate the z-value using
1 Sample Wilcoxon Non Parametric Hypothesis Test
  • Compare the test statistic, W, with the critical value in the tables; the null hypothesis can be rejected if W is less than or equal to the critical value.
  • Now, compare the test statics with critical values in the tables and make a decision, the null hypothesis will be rejected if the test statistic, W, is less than or equal to the critical value
  • Interpret the decision in the context of the original claim.

Example of One Sample Wilcoxon Test

In a law college, random samples of 10 students’ marks are noted below, is there evidence at the 5% confidence level to suggest that the median mark is greater than 67?

  • Null Hypothesis H0: The population median value ≤ 67 marks
  • Alternative Hypothesis : H1: The population median value > 67 marks

Ignoring the signs, rank the differences smallest rank =1

Separate the positive and negative ranks

  • Sum of + ranks =40
  • Sum of – ranks=15
  • Smallest value among 40 & 15 =15
  • Test statistics =15

Critical value =11

The null hypothesis will be rejected if the test statistic, W, is less than or equal to the critical value.

Since the test statistic value is greater than the critical value, hence we fail to reject the null hypothesis. There is no significant evidence that the median rank is greater than 67.

http://www.unm.edu/~marcusj/WilcoxonSR.pdf

Six Sigma Black Belt Certification 1 Sample Wilcoxon Test Questions:

Question 1:  Which of the following scenario the decision will impact the Wilcoxon test?

(A) Ties the values between the samples
(B) Ties of values never impact the decision
(C) Ties values within one sample
(D) Ties the values always impact the decision

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