Ted Hessing

Confidence Intervals

Posted by Ted Hessing

When we use a sample group to gain insight into an entire population, whether we’re talking about people or a product built in a factory, we risk the sample group not completely reflecting the whole population. Therefore, we need confidence intervals.

Further, we use confidence intervals to calculate a degree of certainty that the sample group accurately represents the entire population from which they were drawn. Another way of thinking of it is that if you drew the same-sized sample group hundreds of times and performed the same measurements, a certain percentage of confidence intervals in those sample groups will contain the population means.

A confidence interval is a range of values. So, you can have a percentage of certainty that the population’s mean lies within that range in any given sample from that population.

This page covers every kind of confidence interval you’ll need for your exam. Jump straight to the one you need:

New to the concept of estimating a population from a sample? Start with Point and Interval Estimation first, then come back here for the worked examples of each type.

Confidence Interval vs. Confidence Level

A confidence interval is a range of values that probably contain the population mean.

However, a confidence level is a percentage of certainty that, in any given sample, that confidence interval will contain the population mean.

Confidence Interval vs. Prediction Interval

Prediction intervals are used to calculate the next probable data point. They tell you about the distribution of data values, whereas confidence intervals tell you about a probable population mean.

Estimating Population Mean and Standard Deviation

In most cases, we won’t know the mean and standard deviation for an entire population. However, we can calculate the mean and standard deviation for each of our samples. So we can use the sample figures to estimate those of the whole population. This is what we call a point estimate.

Once we find the point estimate, we also need to know how accurate it is. For instance, the Central Limit Theorem says that in a large enough sample group (30 or more), the sample’s mean and standard deviation will be close to those of the population.

Margin of Error

The margin of error is the maximum expected difference between the actual population parameter and a sample estimate of the parameter. In other words, it is the range of values above and below sample statistics.

Margin of error widely used in surveys tells the degree of uncertainty that the survey results might have.

Margin of error = 1/√n

Where n= sample size

The margin of error is the product of the critical value and the standard error in the confidence interval. If the margin of error increases, the confidence level increases. Similarly, if the margin of error decreases, the confidence level also decreases.

For example, to estimate the confidence interval, we may use X̅±E

E is the margin of error.

Confidence interval = X̅±Zα/2*σ/√n

  • Where σ = population standard deviation
  • Zα/2 = normal distribution value for the desired confidence level
  • n = sample size
  • X̅ = sample mean

Z Scores

This page uses a cumulative z table: each value is the area to the left of z, so a 95% confidence level (0.025 in each tail) means finding 0.975 in the table, which gives 1.96. If your handbook’s table shows the area from the mean (0 to z), look for 0.475 instead; if it shows the upper tail, look for 0.025.

A Z score is the number of standard deviations between a data point and its mean. Thus, you can use a Z-score table to find the corresponding Z-score for common confidence levels or calculate the α value using this formula:

α = 1 - confidence level

If your desired confidence level is 95%, then your calculation would look like this:

α = (1 - 0.95) 
α/2 = (1 - 0.95) / 2 = 0.025
Zα/2 = Z0.025 =1.96

Common confidence intervals and corresponding Z scores

Desired Confidence Interval Z Score
90% 1.645
95%1.96
99%2.576

Confidence Interval Question Using Z-Score

Question: We conduct a random survey of 500 newly-enrolled university students. We know that the standard deviation for university enrollment age is eight years. The mean age of our sample is 24. Calculate the confidence interval for all first-year university students with a 99% confidence level to 3 decimal places.

Calculation: The first step is to consult a Z-score table. A 99% confidence level requires a Z-score of 2.576.

Zα/2 = 2.576, X̅ = 24, σ = 8 and n = 500

Margin of error E = Zα/2*σ/√n = 2.576*8/√800 = 0.922

Confidence interval = X̅±Zα/2*σ/√n = 24±0.922

Answer: Hence, the confidence interval for the age of first-year university students is 23.07 to 24.922, with a 99% confidence level.

Confidence Interval Question Using T-Score

Question: A factory produces tennis balls. A sample of 19 balls is taken from one factory production day. The mean weight of the sample balls is 58.2g. The standard deviation for the sample balls is 0.4g. Calculate the confidence interval with a confidence level of 95%.

Calculation: The sample size is too small to use a Z-score. Instead, use a T-score, which uses a t-distribution. Finding a confidence interval for a mean is a two-tailed test.

You’ll need an alpha score. To calculate it, use this simple equation:

α = (100% - confidence level%)
α = (100% - 95%) 
α = 5%

You also need the degrees of freedom (df), which is the number of samples minus one. Or in equation form:

df = n - 1

df for this question is 18.

So, use a T-table to look up the T-score needed for a two-tailed test with an α of 5% and a df of 18: the answer is 2.101.

tα/2=2.101, X̅=58.2, σ=0.4 and n=19

Margin of error E =tα/2*σ/√n = 2.101*0.4/√19 = 0.192

Confidence interval = X̅±tα/2*σ/√n = 58.2±0.192

Answer: Hence, a 95% confidence interval for tennis balls produced in the factory is 58.008 to 58.392g.

Calculating the Confidence Interval Comparing Two Population Means

Using samples from each population, you can also use confidence intervals to compare two population means. Use this method to compare two different manufacturing methods or to look for differences in two groups of people (for example, smokers and non-smokers). You could also use it to decide whether or not it will be acceptable to pool your two population samples into one larger sample.

The confidence interval for comparing two means is a range of values in which the difference between those two means might lie.

We use a similar equation to the one we use to calculate a population mean. Instead of looking for x, we’re looking for the difference between means.

Z = ((x̄1 - x̄2) - (x1 - x2)) / √((s12 / n1) + (s22 / n2))

Or, in a slightly easier-to-read format:

Confidence Intervals

Example

High blood pressure has been causally linked to smoking tobacco products. To test this, you want to compare systolic blood pressure between smokers and nonsmokers. You’ll use a confidence level of 95%.

  • You have 45 smokers and 56 non-smokers, with similar variance (age, gender, health levels) in each group.
  • In the sample group of smokers, the mean systolic rate is 138.
  • In the sample group of nonsmokers, the mean systolic rate is 135.
  • The standard deviation for smokers is 16.5.
  • The standard deviation for nonsmokers is 14.9.
  • The Z-score you need is 1.96.
n1 = 45
n2 = 56
Z = 1.96 (found by looking up 95% confidence level on the chart.)
s1 = 16.5
s2 = 14.9
x̄1 = 138
x̄2 = 135

Confidence Interval=(x̄1 - x̄2)±Z*√((s12 /n1 )+(s22 /n2)

Firstly, plug the numbers into the equation, so our confidence interval, with a confidence level of 95%, is::

Confidence Interval=(x̄1 - x̄2)±Z*√((s12 /n1 )+(s22 /n2)
= (138 - 135) ± 1.96* √((16.52 / 45) + (14.92 / 56))
= (3)±1.96×√(10.014)
= (3)±1.96×3.16
= (3)±6.19
Confidence Interval≈(3−6.19,3+6.19)
≈ (−3.19,9.19)

To find the margin of error, use:
margin of error = Z * √((s12 / n1) + (s22 / n2))
margin of error = 1.96 * √((16.52 / 45) + (14.92 / 56)) 
margin of error = 1.96 * √((272.25 / 45) + (222.01 / 56))  
margin of error = 1.96 *  √(6.05 + 3.9645) 
margin of error = 6.2025 

Therefore, the margin of error for the difference in means between smokers and nonsmokers, with a 95% confidence level, is 6.19. This means that we are 95% confident that the true difference in means falls within ±6.19 units of the sample difference.

Additional Confidence Intervals Videos

Confidence intervals for Variation

The point estimate for σ is s. s2 is the most unbiased estimate of σ2.

Use the chi-square distribution to construct a confidence interval for the variance and standard deviation

If the random variable x has the normal distribution, the distribution of:

Confidence Intervals

For sample size n > 1

Confidence intervals for variation:

Where n= sample size

s2 = point estimate of variance:

Example: XYZ pharmaceutical company randomly selected 25 samples of flu medicines. The sample variance is 6 milligrams. For instance, assume the weights are normally distributed and construct a 99% confidence interval for the population variance.

n=25

Degrees of freedom = n -1 = 24

s2 = 6

chi square α/2 = 45.55

chi square 1-α/2 = 9.886

Confidence Intervals

Confidence intervals for variation:

Confidence Intervals

Hence, with 99% confidence, you can say that the population variance is between 3.161 and 14.56 milligrams.

Confidence intervals for Proportion

Central theorem says that with larger samples, every sample proportion will have a normal distribution. For larger sample sizes, sample size times proportion (np) and n (1-p) great than equal to 5, the normal distribution can be used to calculate the confidence interval for a proportion.

The confidence intervals for proportion is:

Confidence Intervals

Where:

n=sample size

p= population proportion estimate

Zα/2 = appropriate confidence level from Z table

Example:  In a bulb manufacturing unit, 24 defective bulbs were identified in a sample size of 400 bulbs. Calculate the 95% confidence interval for the proportion.

n=400

P = 24/400 = 0.06

1-p = 0.94

Zα/2 = 1.96 (95% confidence)

0.06±0.0232

0.036≤p≤0.083

So, with 95% confidence, the population proportion is between 0.036 and 0.083.

What is the Difference Between Control Limits and Confidence Intervals?

These are two entirely different concepts. One is used in the Analysis of a process, and the other in the Control of a process.

Further, Control limits depend on your population or sample’s distribution. They can be defined as the average + or – 3 standard deviations. “Control limits are obtained based on the nature of the distribution of data that you collect; if a process is in control doesn’t mean that your process is stable. Hence control limits give the limits at that instant.”

These are different from Specification limits which generally show up on control charts. They are assigned by businesses for what is viable for them. A colleague once described them as if the process goes above this level, we all must update our resumes. So, if it goes below this other limit, don’t worry about updating resumes; no one will ever hire us again!

Confidence Interval

Confidence intervals are a device of statistics for when you do not have perfect knowledge of all of the data. For example, imagine you are trying to infer the chance of some event happening by sampling from a population. Let’s take US voting polls. CNN can’t possibly get all of the voting data, but they can sample the population through exit polls. From that sample, they can predict the winner. So, the question is, how sure and certain are they of how accurate their answer is? Are they 90% certain? Are they 95% certain? 99.5%?  It all depends on the confidence level required. Thus, you get answers like, “We are projecting that candidate X has an 80% chance of winning with a 95% level of confidence.”

“While confidence level signifies how confident you are that the population lies within the range one has specified and this range is nothing but the confidence interval, and this has nothing to do with the control limits as control limits keeps changing if a process is not in stable though its in control.”

Also, see:

How to Calculate a Sample Size Given Standard Deviation, Confidence Interval, and Margin of Error

Other Confidence Intervals Problems

Also, you can find some more confidence interval problems, with links to worked answers, here: Finding the Sample Size Needed for a Confidence Interval for a Single Population Mean.

ASQ Six Sigma Black Belt Exam Confidence Intervals Questions

Question: Which of the following describes the 95% confidence interval of a 20% absentee rate in a department with 30 people?

(A) 6% to 34%
(B) 8% to 32%
(C) 13% to 27%
(D) 17% to 23%

Answer:

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Ted Hessing

Z Scores (Z Value) & Z Table

Posted by Ted Hessing

Z scores (Z value) is the number of standard deviations a score or a value (x) is away from the mean. In other words, the Z-score measures the dispersion of data. Technically, a Z-score tells you how many standard deviations value (x) are below or above the population mean (µ). If the Z value is positive, it indicates that the value or score (x) is above the mean. Similarly, if the Z value is negative, it means the value (x) is below the mean.

Z-scores are also how you answer a common question: what is the probability of a defect? The further a data point sits from a specification limit, the higher its Z-score (also called the process sigma), and the higher the process sigma, the less defective the process. This is actually where “Six Sigma” gets its name: a Six Sigma process has six standard deviations (6σ) between the process mean and the nearest specification limit.

To find that probability yourself, calculate Z = (x – µ) / σ, where x is your sample data point, µ is the process mean, and σ is the standard deviation. Once you have that Z-score, the Z-table below converts it directly into a defect probability.

What is a Standard Normal Distribution?

A Normal Standard Distribution curve is a symmetric distribution where the area under the normal curve is 1 or 100%. The standard normal distribution is a type of special normal distribution with a mean (µ) of 0 and a standard deviation of 1.

A standard normal distribution always has a mean of zero and has intervals that increase by 1. Each number on the horizontal line corresponds to the z-score. Hence, use Z Scores to transform a given standard distribution into something that is easy to calculate probabilities on as it can determine the likelihood of some event happening. 

Z Scores (Z Value) & Z Table & Z Transformations

Any normal distribution with any value of mean (µ) and a sigma can be transformed into the standard normal distribution, where the mean of zero and a standard deviation of 1. This is also called standardization.

A Z-score tells how much standard deviation a value or score is from the mean (µ). For example, if a Z-score is negative 3 means the value (x) is 3 standard deviations left of the mean. Similarly, if the Z-score is positive 2.5 means the value (x) is 2.5 standard deviations to the right of the mean (µ).

This is a common transformation, so there is a reference chart that allows us to look up values. Those values correlate to the value under the normal distribution curve. In other words, what’s the chance of an event happening? We use the Z table to find the percent chance.

How to Interpret the Z-score Table

This page uses a cumulative z table: each value is the area to the left of z (z = 1.00 gives 0.8413). If your handbook’s table shows the area from the mean (0 to z), add 0.5 to its value for positive z to match; if it shows the upper tail, subtract its value from 1.

Most importantly, the Z-score helps to calculate how much area that specific Z-score is associated with. A Z-score table is also known as a standard normal table used to find the exact area. The Z-score table tells the total quantity of area contained on the left side of any score or value (x).

Z Scores (Z Value) & Z Table & Z Transformations

Z Score Table Download

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There are two methods to read the Z-table:

Case 1: Use the Z-table to see the area under the value (x)

The first column in the Z-table top row corresponds to the Z-values and all the numbers in the middle correspond to the areas.

For example, a Z-score of -1.53 has an area of 0.0630 to the left of it. In other words, p(Z<-1.53) = 0.0630.

The standard normal table is also used to determine the area to the right of any Z-value by subtracting the area on the left from 1. 1-AreaLeft = Arearight

For example, a Z-score of 0.83 has an area of 0.7967 to the left of it. So, the Area to the right is 1 – 0.7967 = 0.2033.

Z Scores (Z Value) & Z Table & Z Transformations

Case 2: Use the Z-table to see what that score is associated with a specific area.

P(o<=Z<=x)

  • Pick the right Z row by reading down the right column.
  • Read across the top to find the decimal space.
  • Finally, find the intersection and multiply by 100.

For example, the Value of Z corresponds to an area of 0.9750 to the left of it is 1.96.

What if the Z Score is off the Chart?

Z score tables sometimes only go up to 3. But depending on the spread of the population, z scores could go on for a while. A Z score of 3 refers to 3 standard deviations. That would mean that more than 99% of the population was covered by the z score. There’s not a lot left, but there is some. Use Excel to find the actual value if your table doesn’t go that high.

How do you Calculate a Z Score?

How to Create Z score in Excel

Using the Z score, find the percentage by using the formula: 1-NORMSDIST(Z), where Z is your calculated Z Score.

How to Calculate a Z Score by Hand

There are 2 different situations you need to be aware of when calculating a z score:

  • Z score for a sample
  • Z score for a population

While the z-score equations look very similar, remember that calculating the standard deviation of a population is different than the way you calculate the standard deviation of a sample.

The formula for transforming a score or observation x from any normal distribution to a standard normal score is :

Calculating a Z Score (generic)
Calculating a Z Score (generic)

Calculating a Z Score for a Population

z_pop

Instead of memorizing this formula, let’s see what it actually does

Experiment 1: What does a z-score actually measure?

Set μ = 0 and σ = 1.
Now move x to:

  • 1
  • 2
  • 3

Watch how far x moves from the mean.

👉 What you should observe:

  • z = distance in standard deviations
  • spacing is consistent

💡 Insight:

A z-score is not a value. It’s a distance.

Experiment 2: Change the Mean (μ)

Keep x fixed (e.g., x = 2).
Now move μ left and right.

👉 Observe:

  • The entire curve shifts
  • z-score changes even though x didn’t move

💡 Insight:

Z-score is relative, not absolute.

Experiment 3: Change the Spread (σ)

Keep x and μ fixed.
Increase σ from 1 → 4.

👉 Observe:

  • Curve becomes flatter and wider
  • z-score gets smaller
  • shaded probability changes

💡 Insight:

The same distance means different things depending on variability.

Experiment 4: Same x, Different σ → Different Probability

Try:

  • x = 2, μ = 0, σ = 1
  • x = 2, μ = 0, σ = 3

👉 Observe:

  • Same raw value
  • very different probabilities

💡 Insight:

Probability depends on distribution, not just the value.

Experiment 5: Connect to Z-Table Thinking

Adjust x until z ≈ 1.00, 2.00, 3.00

👉 Observe:

  • ~84%, ~97.5%, ~99.9%

💡 Insight:

This is what the Z-table is giving you.

Z Score for a Sample
z_sample

How many parts in a population will be longer or greater than some number?

Z score examples using standard deviation

Example 1: Longer than / Greater than

Hospital stays for admitted patients at a certain hospital are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to last for longer than 300 hours?

  • x=300
  • x̅ = 200
  • s=75

Z= x- x̅/s =(300-200)/75= 100/75= 1.33

Z score from the table for 1.33 = 0.9082

Z Scores (Z Value) & Z Table & Z Transformations

Since, we are looking for longer, solution is P(x>300) = P(Z>1.33) = 1- P(Z<1.33)= 1-0.9082 = 0.0918 = 9.18%

Z Score Positive Template

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Example 1.1

A batch of batteries with an average of 60v and a Standard Deviation of 4v. If 9 batteries are selected at random, what is the probability that the total voltage of the batteries is greater than 530?

Answer

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Example 2: Less than

Hospital stays, for admitted patients at a certain hospital, are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to last less than 75 hours?

  • x=75
  • x̅ = 200
  • s=75

Z= x- x̅/s =(75-200)/75= -125/75= -1.667

Z score from the table for -1.667 = 0.0475

Z Scores (Z Value) & Z Table & Z Transformations

Since we are looking for less than, the solution is = 4.75%

Z Score Negative Template

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Example 2.2

What is the probability that the average voltage is less than 62?

Answer

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Example 3: Both less than AND greater than. (Percentage outside the range)

The mean inside diameter of a sample of 200 washers produced by a machine is 0.502 inches and the standard deviation is 0.005 inches. The purpose for which these washers are intended allow a maximum tolerance in the diameter of 0.496 to 0.508 inches, otherwise, the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.

Less than

  • x=0.496
  • x̅ = 0.502
  • s=0.005

Z= x- x̅/s =(0.496-0.502)/0.005= -1.2

Z score from the table for -1.2 = 0.1151

Since we are looking for less than, solution for lower bound = 11.51%

More than

  • x=0.508
  • x̅ = 0.502
  • s=0.005

Z= x- x̅/s =(0.508-0.502)/0.005= 1.2

Z_score from the table for 1.2 = 0.8849

Since we are looking for more than that, the solution is 1-0.8849 = 0.1151 = 11.51%

So, the total is 23.02%

Example 4: Both upper and lower bound. (Percentage Inside the range.)

The weights of 500 American men were taken and the sample mean was found to be 194 pounds with a standard deviation of 11.2 pounds. What percentages of men have weights between 175 and 225 pounds?

P(X<175)

  • x=175
  • x̅ = 194
  • s=11.2

Z= x- x̅/s =(175-194)/11.2= -1.6964

Z_score from the table for -1.6964= 0.0455

P(X<175) = 4.55%

P(X<225)

  • x=225
  • x̅ =194
  • s=11.2

Z= x- x̅/s =(225-194)/11.2= 2.7678

Z score from the table for 2.7678 = 0.9971=99.71%

Since we are looking for weights between 175 and 225, P(175<x<225) = 99.71%-4.55% = 95.16%

Z Scores (Z Value) & Z Table & Z Transformations

 Additional Notes:

What do you do when the sample size is less than 20?

Great question! You'd apply student t-scores.

Six Sigma Green Belt Z Score Questions

Question: This formula Z = (X - μ)/σ is used to calculate a Z score that, with the appropriate table, can tell a Belt what ____________________________________.

A) Ratio the area under the curve to the total population

B) Number of Standard Deviations between X and μ

C) The Median of the sample population is

D) Proportion of the data is between X and μ

Answer: 

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Question: A battery manufacturer was considering changing suppliers for a particular part. The purchasing manager required that the average cost of the part be less than or equal to $32 in order to stay within budget. A sample of the 32 initial deliveries had a Mean of the new product upgrade price of $28 with an estimated Standard Deviation of $3. Based on the data provided, the Z value for the data assuming a Normal Distribution is?

A) 0.67

B) 1.33

C) 2.67

D) 4.33

Answer:

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Ted Hessing

Chi-Square Goodness of Fit Test

Posted by Ted Hessing
Goodness of fit tests
Photo by darkday

A chi-square goodness-of-fit (GOF) test checks how well an observed categorical data distribution fits an expected distribution. In other words, it tests whether the frequencies observed in different categories match some hypothesized frequencies. This is useful when you have one categorical variable and a theory or historical data about what the distribution of that variable should be. For example, you might want to verify if defect occurrences follow a Poisson distribution, or if customer preferences are evenly distributed among choices. The chi-square GOF test is a non-parametric test (no assumption of normality; it deals directly with category counts).

When to use Chi-Square Goodness of Fit?

A chi square goodness-of-fit test can be conducted when there is one categorical variable with more than two levels. If there are exactly two categories, then a one proportion z test may be conducted.

Reference.

The chi-square goodness-of-fit test requires 2 assumptions2,3:

1. Independent observations;

2. For 2 categories, each expected frequency EiEi must be at least 5.

For 3+ categories, each EiEi must be at least 1 and no more than 20% of all EiEi may be smaller than 5.

Reference

When to use: Use a GOF test in Six Sigma during Measure or Analyze when you need to validate distributional assumptions. For instance, if you assume defects per hour follow a Poisson distribution, a GOF test can confirm or refute that. Or if historical data suggest a certain proportional breakdown of defect causes, you can test if current data still fits that pattern. This helps ensure you’re using the correct statistical models and can reveal changes in process behavior.

Hypotheses: Typically, for a goodness-of-fit test:

  • H₀ (null hypothesis): The observed distribution fits the expected distribution. (No significant difference between observed and expected frequencies.)
  • H₁ (alternative hypothesis): The observed distribution does not fit the expected distribution (at least one category’s frequency is significantly different from expected).

In plain terms, H₀ assumes any differences between observed and expected counts are just due to random chance, while H₁ asserts that the differences are too large to be random, indicating the distribution has changed or the assumption was wrong.

Assumptions: Each observation falls into one of the categories, and you have a sufficiently large sample such that expected counts in each category aren’t too low. A common rule of thumb is that expected counts should be at least 5 in each category (or if that’s not possible, no more than 20% of categories have expected < 5)sixsigmastudyguide.com. If expected frequencies are very small, the chi-square approximation may not be accurate (you might need to combine categories or use an exact test).

Steps to Perform a Chi-Square Goodness-of-Fit Test

  1. Define the hypotheses:
    • Null (H₀): The observed category proportions are equal to the expected proportions (no difference between observed and expected distribution).
    • Alternative (H₁): The observed proportions differ from the expected (at least one category is not in line with the expected distribution).
  2. Choose a significance level (α): Commonly α = 0.05 (5% significance) is used, corresponding to 95% confidence.
  3. Collect data and calculate expected frequencies: Determine the expected count for each category under H₀. Often this is done by multiplying the total sample size by the expected percentage for each category. Expected values might be given by a theoretical distribution (e.g., a probability model or historical proportions). Ensure the total of expected counts equals the total observations.
  4. Compute the chi-square statistic:
    χ2=∑i=1k(Oi−Ei)2Ei,displaystyle chi^2 = sum_{i=1}^{k} frac{(O_i – E_i)^2}{E_i},χ2=i=1∑k​Ei​(Oi​−Ei​)2​,
    where O<sub>i</sub> is the observed frequency for category i and E<sub>i</sub> is the expected frequencysixsigmastudyguide.comsixsigmastudyguide.com. Each term $(O – E)^2/E$ measures the squared difference relative to expectation. Sum these across all k categories to get χ².
  5. Determine degrees of freedom: For goodness-of-fit, df = (number of categories – 1) minus any parameters estimated from the data. If the expected proportions are completely specified in advance, df = k – 1. (For example, if you have 4 categories, df = 3.) If you estimated parameters (like the mean of a Poisson from the data), subtract those as well.
  6. Find the critical value: Because this test is right-tailed (we care about large χ²), find the critical χ² value for the chosen α and the calculated degrees of freedom. This value comes from the chi-square distribution table. For instance, at α = 0.05 and df = 3, the critical value is 7.815 (meaning that 5% of the time, a random χ²(3) would exceed 7.815). If using p-value approach, compute p (the area to the right of your χ² statistic).
  7. Make a conclusion: Compare your χ² statistic to the critical value. If χ²_stat > χ²_critical (or p-value < α), reject H₀. This means the differences between observed and expected are statistically significant: the data do not fit the expected distribution. If χ²_stat is smaller than critical (p > α), you fail to reject H₀, meaning any differences are small enough to be explained by chance, and the distribution fits as assumed.

Goodness-of-Fit Example: Did Our Customer Distribution Change?

To illustrate, consider a realistic Six Sigma scenario: A loyalty program manager wants to know if the age distribution of the company’s priority customers has changed over the last decade. Ten years ago, the company categorized its priority customers by age and found the distribution to be: 15% were under 25, 40% were 25-40 years old, 25% were 40-60, and 20% were over 60. Now, in the Analyze phase of a project, the manager takes a random sample of 500 current priority customers and observes their age breakdown. We can use a chi-square goodness-of-fit test to see if the current sample’s distribution differs from the 2010 baseline.

Expected age distribution of priority customers in 2010 (historical baseline). Each percentage represents the proportion of customers in that age group (SixSigmaStudyGuide.com).

From the 2010 data (expected distribution):

  • Under 25: 15%
  • 25-40: 40%
  • 40-60: 25%
  • Over 60: 20%

If we expect those proportions to still hold, the expected frequencies out of 500 customers would be:
Under 25: 0.15 * 500 = 75
25-40: 0.40 * 500 = 200
40-60: 0.25 * 500 = 125
Over 60: 0.20 * 500 = 100

Observed age distribution of a 2020 sample of 500 priority customers (SixSigmaStudyGuide.com). These are the actual counts in each age group from the sample.

The observed frequencies in the 2020 sample of 500 are:
Under 25: 75
25-40: 220
40-60: 150
Over 60: 55

Let’s run the chi-square GOF test at 95% confidence (α = 0.05). First, state hypotheses:

  • H₀: The 2020 age distribution matches the 2010 distribution (no change).
  • H₁: The 2020 age distribution differs from 2010 (at least one age group has a different proportion).

We already have expected (E) and observed (O) counts. Now calculate the χ² statistic:

  • Under 25: O=75, E=75 ⇒ (75-75)²/75 = 0
  • 25-40: O=220, E=200 ⇒ (220-200)²/200 = (20)²/200 = 400/200 = 2.0
  • 40-60: O=150, E=125 ⇒ (150-125)²/125 = 625/125 = 5.0
  • Over 60: O=55, E=100 ⇒ (55-100)²/100 = (-45)²/100 = 2025/100 = 20.25

χ²_stat = 0 + 2.0 + 5.0 + 20.25 = 27.25 (approximately).

Degrees of freedom = 4 categories – 1 = 3. For df = 3 at α = 0.05, the chi-square critical value is 7.815sixsigmastudyguide.com. Our test statistic (27.25) is much larger than 7.815, falling deep in the right tail beyond the cutoff. This means the probability of seeing a χ² that large under H₀ is extremely low. We therefore reject H₀.

Conclusion: The sample data provide strong evidence that the age distribution of priority customers in 2020 is different from the 2010 distribution. Specifically, the chi-square contributions suggest the biggest differences are in the older age groups (the “Over 60” category had far fewer people than expected, contributing heavily to χ²). In Six Sigma terms, if this was a KPI or a customer demographic critical to quality, we’d note that the customer base has shifted significantly over the decadesixsigmastudyguide.com.

(Notice that this GOF test was one-tailed (right-tail) because only a large χ² (indicating a big deviation) would lead us to reject H₀. Small deviations wouldn’t).

Chi square goodness of fit test specifically tells how well a categorical (nominal or ordinal) sample distribution fit into a hypothetical distribution. In other words, it is the test used to check if the sample data is consistent with a hypothesized distribution of the population. The goodness of fit test is used to determine the observed sample distribution matches or fits the expected values; hence we use the term goodness of fit.

The goodness of fit is a non-parametric test because it does not rely on estimates of a population parameter like mean or variance to make an inference on the characteristics of the population.

Structure of a Chi-Square Goodness of Fit Test

The goodness of fit tests is structured in cells; therefore, the observed frequency goes in each cell. Furthermore, the distribution you are trying to match would have a theoretical frequency. Then, the chi-square is summed across all cells.

Use the data values structured into cells, explicitly requiring a calculated chi-square test statistic. The unknown distribution is tested; likewise, the Degrees of Freedom vary according to the distribution.

GOF Distribution  | Degrees of Freedom

Normal | # cells – 3

Poisson | # cells – 2

Binomial | # cells – 2

Uniform | # cells – 1

Steps to perform Chi-Square goodness of fit

Step1:

Firstly, define the null hypothesis and alternative hypothesis

  • Null hypothesis (H0): There is no difference between the observed value and the expected value
  • Alternative hypothesis (H1): There  is a significant difference between the observed value and the expected value

Step 2:

Secondly, specify the level of significance.

Step 3:

Thirdly, compute the χ2 statistic.

  • O is the observed value
  • E is the expected value

Step 4:

Fourthly, calculate the degree of freedom:

The degrees of freedom in chi-square test depends on the sample distribution

Step 5:

Then, find the critical value based on degrees of freedom.

Step 6:

Finally, draw the statistical conclusion:

If the test statistic value is greater than the critical value, reject the null hypothesis. Hence, we can conclude that there is a significant difference between the observed value and the expected value.

Chi-Square goodness of fit test Example 1: Did the Distribution Change?

Ten years ago, US airlines categorized the priority customers (those who completed 10,000 miles traveled in a year) based specifically on age:

Similarly, in 2020, 500 priority passengers were sampled, and below are the results:

At a 95% confidence level, would you conclude that the population distribution of priority customers changed in the last 10 years?

  • Null hypothesis (H0): The sample data meet the expected distribution.
  • Alternative hypothesis (H1): The sample data does not meet the expected distribution.

Level of significance: α=0.05:

Degrees of freedom = number of categories (n)= 4

n-1 =3

Chi-square critical value for 3 degrees of freedom =7.815

The test statistic value is greater than the critical value; hence, we can reject the null hypothesis.

So, we can conclude that the priority customers in 2020 are different than those expected based on the 2010 population.

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Chi-Square Normality Test

Many statistical techniques (regression, ANOVA, t-tests, etc.) rely on the assumption that data is normally distributed. Hence, the chi-square goodness of fit test is one of the good options to check whether the data follows a normal distribution.

Furthermore, the chi-square goodness of fit test is an alternative to the Anderson-Darling test, Kolmogorov-Smirnov (K-S) test, and Shapiro-Wilk test to test the normality.

Similar to other statistical hypothesis tests, the chi-square goodness of fit test also needs to compute the test statistics and find the critical value for a given degree of freedom and confidence level. If the chi-square value is greater than the critical value, reject the null hypothesis.

How to conduct Chi-Square goodness fit normality test

  • Determine the null hypothesis, i.e. data is sampled from a normal distribution and alternative hypothesis.
  • Compute the sample mean as well as the standard deviation.
  • Then, define the confidence level.
  • Bin the data: Determine non-overlapping bins, and then count values in each bin
  • Furthermore, find the cumulative probability for each category endpoint.
  • After that, compute the probability that a randomly selected value would go into each category.
  • Find expected observations for each bin, which is the product of the probability of observation would fall in the bin compared to the sample size (n).
  • Likewise, compute the chi-square statistic. χ2 = Σ [ ( Observed frequency – Exp frequency) 2 / Exp frequency].
  • Find the degrees of freedom based specifically on the number of categories or bins. The degrees of freedom for the chi-square goodness of fit test is always the number of categories-1-2(two estimated parameters, mean and standard deviation) = k-3 (note: valid if mean and standard deviation is not given).
  • Find the critical value based on degrees of freedom.
  • Finally, draw the statistical conclusion: If the test statistic value is greater than the critical value, reject the null hypothesis.

Chi-Square normality test example

Example: 28 students’ weights (in kgs) are collected in ABC school. Test whether the data is normally distributed, given that the confidence level is 95%.

  • H0: Students’ weights in an ABC school follow a normal distribution
  • H1: Students’ weights in an ABC school do not follow a normal distribution
Firstly, compute the sample mean and standard deviation.
  • Select all the data and enter” =Average(B2:H5)” for mean and “ Stdev(B2:H5)” for standard deviation
  • Confidence level = 95%
Secondly, bin or categories the data and count the values
Thirdly, find the cumulative probability for each category endpoint.
  • In cell H11 type NORMDIST (20, L2, L4, TRUE),
  • Similarly H12 =  NORMDIST (25, L2, L4, TRUE),
  • H13 =  NORMDIST (30, L2, L4, TRUE),
  • H14=  NORMDIST (35, L2, L4, TRUE),
  • H15 =  NORMDIST (∞, L2, L4, TRUE),
Fourthly, compute the probability that a randomly selected value would go in each category.
  • For I 11 cell, ie less than 20 copy the same value ie 0.118431892
  • In cell I 12, i.e. values between 25 to 20= H12-H11 = 0.350912244-0.118431892 = 0.232480352
  • Similarly for I13 = H13-H12=0.310801442
  • I14 = H14-H13 = 0.226512236
  • I15 = H15-H14 = 0.111773954
Then, find expected observations in each bin, which is the product of the probability of observation would fall in the bin to the sample size (n) =28.
  • J11 = I11*$G$16 = 3.316092984
  • Similarly, J12=I12*$G$16=6.509449848
  • J13=I13*$G$16=8.702440383
  • J14=I14*$G$16=6.342346084
  • J15= I15*$G$16=3.129670701
Finally, find the chi-square statistic. χ2 = Σ [ ( Observed frequency – Exp frequency) 2 / Exp frequency]
  • K11 = (G11-J11)2/J11 = 0.522331777
  • Similarly, K12 =  (G12 – J12)2 / J12 = 1.871731198
  • K13=(G13-J13)2/J13=0.056699325
  • K14=(G14-J14)2/J14 = 0.865071869
  • K15=(G15-J15)2/J15 = 0.242029645
  • Then, find the sum of chi-square values = K16= sum(K11:K15) =3.55786
  • Following that logic, the Degrees of freedom = No of categories -3 =5-3 =2
  • Consequently, the Critical value =5.991

Conclusion: The chi-square value of 3.557 is less than the critical value of 5.991. In hypothesis testing, a critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis. If the chi-square calculated was smaller than the critical value, then the data did fit the model, therefore, failed to reject the null hypothesis. So, students’ weights in ABC school follow a normal distribution.

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Chi-Square Normality Testing

Ted Hessing

Dispersion

Posted by Ted Hessing

Dispersion is the degree of variation in the data. In other words, the extent of the spread of values from the mean. Items in a data set tend to differ from each other and the mean. So, dispersion measures the extent to which different items tend to disperse away from the central tendency.

The measure of central tendency gives the central value around which all the values spread along with the central value, but that does not give the correct picture of the variability of the data.

The better method to find the spread or scatter of the data is the measure of dispersion. It attempts to quantify the variability or spread of the data. The measure of dispersion tells how the data spreads or scatters around the center value. Dispersion is nothing but “distance from the average”.

Why Measure of Dispersion Is Important

For Example, Set-1 household income data is $70K, $60K, and $50K respectively; the average is $60K. Set-2 household income is $20K, $50K, and $100K; the average is $60K.

Dispersion

Dispersion indicates the extent of uniformity. Uniformity and degree of variation are inversely proportional. In other words, more uniformity, less variation. From the above example, though the average salary of the two sets is the same ($60K), the first set of household data indicates more uniformity. Similarly, the second set of data shows more variation around the center.

If the means of two or more series are the same, do not think of them as similar because their other characteristics (dispersion, skewness, kurtosis), may differ. Hence, the method of dispersion helps to find the correct variation of the data.

Objective of Dispersion

  • To determine the reliability of an average
  • Compare two or more data sets with regard to the variability
  • Facilitate computations of other statistical measures
  • To indicate the level of uniformity of variables

Different types of Measures of Dispersion

There are several measures of dispersion, the most common being are

Range

Range is the difference between the maximum and the minimum value.

Example: The age of the randomly sampled audience in a theatre is 55, 16, 23, 65, 45, 34, 28, 37, 58, and 24. Find the Range.

Arrange the values in ascending order: 16, 23, 24, 28, 34, 37, 45, 55, 58, 65

Range = Maximum-Minimum = 65 -16 = 49

  • The range can be used with ordinal or interval ratio variables but cannot be used with a nominal scale.
  • The disadvantage of Range: it completely depends upon the extreme values. Thus, the range is affected by the outliers.

Variance

Variance measures the dispersion of a set of data points around their mean value.

Population variance, denoted by sigma squared, is equal to the sum of squared differences between the observed values and the population mean, divided by the total number of observations.

Sample variance is denoted by s squared and is equal to the sum of squared differences between observed sample values and the sample mean, divided by the number of sample observations minus 1.

Standard Deviation

Standard Deviation is the most popular measure of dispersion. The symbol for the measurement of dispersion in a population is denoted by the Greek letter sigma σ. Similarly, the sample standard deviation denoted by s (or sd) is a point estimate for the population standard deviation / the dispersion statistic for samples.

Dispersion

Standard deviation is the positive square root of the arithmetic mean of the squares of the deviations of all the observations from their mean.

Unlike range, the standard deviation takes each data point into account while calculating the dispersion.

The lower the standard deviation, the lower the process variability. In other words, data points are closer to the mean or average value.

Example: Determine the standard deviation of the weights (in lbs) of 12 samples.

n =12 and n -1 =11

Average of 12 samples X̅= ΣX/n = 1848 / 12 = 154

Compute the deviation i.e (X-X̅) and then square each deviation (X-X̅)2

Dispersion

Sample standard deviation s= √Σ(X-X̅)2 / n-1 = √3702 / 11 = 18.3

s is the standard deviation of the sample (18.3) which is used as an estimate for the population from which the sample was taken.

Coefficient of Variation

The Coefficient of Variation is the standard deviation relative to the mean. In other words, the Coefficient of Variation is equal to the standard deviation divided by the mean, and it is expressed as a percentage. It is also known as relative standard deviation. See the dedicated Coefficient of Variation page for the full formula walkthrough, worked examples, and how to interpret it.

COV = s/ X̅ * (100%) or COV = σ / µ * 100%

Example: From the above example s = 18.3 and X̅ = 154

Coefficient of Variation COV = s/ X̅ * (100%) = 18.3 / 154 * 100 = 11.9%

Inter Quartile range

Inter quartile range is one of the measures of dispersion. It is defined as the difference between the first and third quartiles.

IR = Q3 – Q1

Example: Find the Inter Quartile range of 10 students’ marks in a science subject.

56, 78, 35, 89, 92, 52, 66, 72, 84, 96

  • Arrange the values in ascending order
  • 35, 52, 56, 66, 72, 78, 84, 89, 92, 96
  • n=10
  • Calculate Q1 = ¼ (n+1) = 11 / 4 = 2.75th term = 52 + (0.75) * (56 – 52) = 55
  • Calculate Q3 = ¾ (n+1) = 3 * 11 / 4 = 8.25th term = 89 + (0.25) * (92 – 89) = 89.75

Inter Quartile range = Q3-Q1 = 89.75 – 55 = 34.75

Kurtosis

Kurtosis is a statistical measure to determine whether the data are heavy-tailed or light-tailed relative to a normal distribution.  In other words, Kurtosis is a measure of the thickness of the tails of a distribution.

Kurtosis is a measure of the peakedness of a distribution. The larger the kurtosis, the more peaked the distribution. The standard normal distribution has a kurtosis of 3.

There are three types of Kurtosis

  • Leptokurtic: It is a curve having a high peak than the normal curve. Too much concentration of the items near the center. Leptokurtic distribution has positive kurtosis values. Most financial returns are Leptokurtic.
  • Platykurtic: It is a curve having a low peak (flat) than the normal curve. There is less concentration of items near the center. A platykurtic distribution shows a negative excess kurtosis.
  • Mesokurtic: It is a curve having a normal peak or the normal curve. There is an equal distribution of items around the center values. Mesokurtic has zero value kurtosis value. In other words, Mesokurtic is the same as the normal distribution.

 What is Excess Kurtosis?

It is a metric that compares the kurtosis of a normal distribution to the kurtosis of a distribution. The normal distribution kurtosis value is 3.

Hence, Excess Kurtosis = Kurtosis – 3.

We need six points (strata, units of measure) under the upper control limit of the range in order to have sufficient measurement discrimination. This is due to rounding. Too much rounding causes a loss of information about dispersion in a sample.

http://www.six-sigma-material.com/Measures-of-Dispersion.html (Measures of Dispersion)

Important Videos on Dispersion

Ted Hessing

Box Plots (aka Box and Whisker Plots)

Posted by Ted Hessing

What is a Box and Whisker Plot?

Box plots are graphical tools that let you easily visually compare variations between the data sets evaluated. It also makes comparisons quickly and easily. In other words, Box plots are used in statistics to graphically display various parameters at a glance.

Graphical analysis is one of the best ways to analyze the problems in Six Sigma projects. It is an effective way to visualize data patterns and provides key insights into the data. Different graphical tools, such as Run charts, Scatter diagrams, Histograms, and Pareto Chart depict different data features such as trend, frequency, dispersion, and shape of the distribution. Box plot, a.k.a Box and Whisker plot, is a pictorial representation of continuous data. Box plots show the max, min, median, interquartile range Q1, Q3, and outlier.

Box Plot
From Microsoft

Rectangles are drawn to show the second and the third quartiles. And there is a vertical line that shows the median value. Both the upper and the lower quartiles are displayed as horizontal lines on both sides of the rectangle. It has the name box and whisker because the lines extend vertically from the boxes. These lines indicated the variability outside the upper quartiles and lower quartiles.

  1. Median.
  2. Maximum.
  3. Minimum
  4. First quartile (0-25%)
  5. Third quartile (75%)

Why Box Plots

Box plots are very effective and easy to read. Graphically it shows the variation between multiple variables and the variations within the ranges.

It is also called a box-and-whisker plot or “five number summary.” In other words, quartiles, the median, and the highest and lowest values. It shows how the data scattered within those ranges will help for effective decision-making.

When to use Box plot?

  • Box plots compare multiple data sets from independent sources that may have some relationship.
  • To compare the process changes before and after the improvements
  • Box plots help to display various parameters at a glance.
  • To read the median, interquartile range, and outliers in the data set
  • When the data is in metric scale levels such as age, temperature, etc.

What Can Box Plots Tell Us?

  • Middle values: The box indicates the range in which the middle 50% of all data lies
  • Median: The blue horizontal line indicates the median of the group.
  • 1st Quartile and 3rd Quartile: The lower end of the box is the 1st quartile, and the upper end is the 3rd quartile.
    • Useful in visualizing most of the spread of variation.
    • Ex. Stock 1 typically has the least variation.
  • Skewness: By looking at a box plot you can tell if your data distribution is skewed if the line inside the box is not centered.
    • Ex. Stock 4’s median is not centered. Thus this data is skewed.
  • Extreme values: The vertical lines (whiskers) extend to the most extreme data points that are still within 1.5 x IQR of the box (Q1 – 1.5 x IQR and Q3 + 1.5 x IQR). If there are no outliers, the whiskers end at the max and min values.
    • Ex. Stock 5 tends towards larger values.
  • Data points beyond the whiskers (more than 1.5 x IQR outside the box) are plotted as outliers with dots, small circles, or stars.
  • Some of the box plots also make use of an additional character to represent the mean of the sample data.
  • Some of the box and whisker plots make use of a cross-hatch that is placed on the whiskers. All the box plots will definitely have whiskers.
Box plots
Box plots

How to Make a Box and Whisker Plot?

Now, coming to the creation of box and whisker plots, you will need to follow the points given below:

  • Gather the data: Get the data for which you need to create plots.
  • Organize this data: Order the data by sorting them numerically, in ascending or increasing order.
  • Calculate Medians: Find out the Median of this data range. This Median will divide the data into halves. And to find out the quartiles, compute the Medians of the 2 halves. If the data set has an odd set of numbers, the Median would have an equal amount of numbers on both sides. And, if the data set has an even set of numbers, the 2 numbers in the middle would have to be added up and averaged.
  • Calculate 1st and 3rd quartiles: This can be found out by finding out the Median of each of the halves.
  • Draw plot lines: Now, you will need to draw plot lines and mark the Median and quartiles.

Box Plot Example

Example: Draw a Box plot for the following 15 points

23,28,44,72,66,54,89,91,26,24,59,74,81,36,77

Step 1: Organize the data: Arrange the data in ascending order

23,24,26,28,36,44,54,59,66,72,74,77,81,89,91

Step 2: Calculate median: 23,24,26,28,36,44,54,59,66,72,74,77,81,89,91

59 is the median value

Step 3: Find 1st Quartile: find the median of the lowest half of the data set

23,24,26,28,36,44,54

28 is the median value

Step 4: Find the 3rd Quartile: find the median of the upper half of the data set

66,72,74,77,81,89,91

77 is the median value

Step 5: Draw a 5-number summary: Draw plot lines and mark the Lowest, Highest points, Median, and quartiles.

Uses of Box and Whisker Plot

There are various uses for this plot. Given below are some of the main features and advantages:

  • Easy to use: It is a convenient way to depict the numerical data groups in a graphical manner, especially with the help of quartiles.
  • No Assumptions: These display the variations in samples without making any kind of assumptions on the statistical distributions.
  • Skewness and dispersion: The box plats are not parametric. The spacing between various parts of the box shows the scenes and dispersion in the data.
  • Various statistical values: Other than the points, these plots also help estimate range, mid-range, mid-hinge, inter-quartile ranges, standard deviations, and tri-mean. These plots help to find the Median, Maximum, Minimum, 1st Quartile (25%), 3rd Quartile (75%). So, one can easily find out even the 2nd, 9th, 25th, 50th, 75th, 98th, and 99th percentile, along with the Median.
  • Uniformity: These plots are quite uniform. The box’s top and bottom are the first and third quartiles. The bands inside the box are the second quartile or the Median.
  • Easier comparisons: It also makes the comparisons between the data sets easier and quicker.
  • Equal spacing: If there is a normal distribution of data, all the marks on the plot will be equally spaced.

How to Make Box and Whisker Plots in Excel?

The emergence of Excel has made life easier for all analysts and statisticians. It is so much easier to present data in various forms on Excel. There are pivot tables, bar graphs, and box and whisker plots, which help represent the data in a more understandable format. This article will give you detailed information about how to make a box and whisker plot. But, before we start with that, we need to understand the basic question: “What is a box and whisker plot?” and how does that help us?

Given below are the steps for making these plots in Excel:

  1. Type the sample data in the worksheet.
  2. Select the cells that have data. Now, click on Chart on the Insert menu.
  3. On the tab, Standard types, click on Stock under the Chart Type. Then, click on the fourth chart from there.
  4. Now, click on the Next tab and click on Data Series in Rows under Data Range.
  5. Then on the tab Legend, uncheck the Show Legend check box.
  6. On the tab Axes, uncheck the Value (Y) Axis under the Secondary Axis and then click on Finish.

This way, you would have created your box and whisker plots. Hope you have learned how to make box and whisker plots and could do them on your own.

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