## Fishing and Hypothesis Testing! Let’s assume that we are in the business of fish farming. We want to make sure that the mean weight of all the fishes in the pond must be 2 Kg before we sell them into the market. It is not possible to take out all of them (we don’t know how many fishes are there!) from the pond and measure their individual weight. So what we do is to take out a sample of say, 25 fishes and measure their weight. If the mean weight of the 25 fish is near to 2 Kg (more precisely, if it is between 2 ± 0.15 Kg), we assume that lot is ready to be sold in the market. This is hypothesis testing. We are trying to estimate the population parameter (mean weight of all fishes in the pond) based on the mean weight of the sample of 25 fishes with an acceptance criterion in our mind (between 2 ± 0.15 Kg).

In Inferential statistics we try to estimate the population parameter by studying a sample drawn from that population. It is not always possible to study the whole population (called as census). Take an example of rice being cooked in a restaurant. The entire lot of rice taken for the cooking may be considered as the population. Let’s further assume that there is a set protocol for cooking and after a predetermined cooking time, chef hypothesizes that the whole lot might have been cooked properly. In order to check his hypothesis, he takes out few grains of the cooked rice (sample) and then he test his hypothesis by subjecting the sample to some test (pressing them between the fingers). Finally based on the sample’s result, chef takes the decision whether the whole lot of rice is cooked or not. We have executed following steps in order to make an estimate about the degree of cooking of the entire lot of rice based on the small sample drawn from the pot.

• We select the population to be studied (the entire lot of rice under cooking)
• A cooking protocol is followed by the chef and he makes a hypothesis that the whole lot of rice (population) might have been cooked properly → trying to make an estimate about the population parameter
• Then we draw a sample from the pot → sample
• We have a criterion in our mind to say that the rice is overcooked or undercooked or properly cooked as per expectation → we set a threshold limit (confidence interval CI) within which we assume that rice is properly cooked. Less than that, it is undercooked and more than that, it is overcooked.
• We test the sample of the cooked rice by pressing them between our fingers → Test statistics
• The results of the test statistics is compared with the threshold limit set prior to conducting the experiments and based on this comparison, decision is taken whether the whole lot of rice is cooked properly or not → inference about the population parameter.

It must be somewhat clear from the above discussion that, we use hypothesis testing to challenge whether some claim about a population is true or not utilizing the sample information, for example,

• The mean height of all the students in the high school in a given state is 160 Cm.
• The mean salary of the fresh MBA graduates is \$65000
• The mean mileage of a particular brand of car is 15 Km/liter of the gasoline.

All of the above statement is some kind of population parameter that we hypothesized to be true. To test these hypothesis, we take a sample (say height of 100 students selected at random from the high school or salary data of 25 students selected at random from the MBA class) and subject it to some statistical tests called as test statistics (equivalent to pressing the rice between the fingers) to conclude the statement made about the population parameter is true or not.

But, before we go any further, it is important to understand that we are using a sample for estimating the population parameter and the size of the sample is too less (relative to the size of population). As a result, estimating population parameter based on the sample statistics would involve some uncertainty or the error. This is represented by following equation

Population Parameter = Sample statistics ± margin of error

The above equation gives an interval (because of ± sign) between which a population parameter is expected to be found. This interval is called as confidence interval (CI). This means every sample drawn from the population would give different Confidence Interval!!

Suppose we are trying to estimate the population mean (which is usually unknown) and we draw 100 samples from the population, all those 100 samples would give 100 different CI because all samples would have different sample mean and different “margin of error”. Now question arises “does all 100 CI thus obtained would contain the population parameter?” As stated earlier, because of the sampling error, we can never accurately estimate the population parameter, in other words, we understand that there will be some degree of error in estimating the population parameter based on the sample statistics. Hence, we should be wise enough to accept an inherent error rate prior to conducting any hypothesis testing. Let’s say that if I collect 100 samples from the population and obtained 100 different CI, then there are chances that 5 CI thus obtained might not contain the population parameter. This is called as error α or type-I error. This α represents the acceptable error or the level of significance and it has to be determined prior to conducting any hypothesis testing. Usually it is a management decision. For more detail see “Is it difficult for you to comprehend confidence interval?

Based on the above discussion, a 7-Step Process for the Hypothesis Testing is used (note: step-2 is described before step-1, this is done because it helps us in writing the hypothesis correctly)

Step 2: State the Alternate Hypothesis.

This is denoted by Ha and this is the real thing about the population that we want to test. In other words, Ha denotes what we want to prove.

For example:

• The mean height of all students in the high school is 160 Cm.
• Ha: μ ≠ 160 Cm
• The mean salary of the fresh MBA graduates is \$65000
• Ha: μ ≠ \$65000
• The mean mileage of a particular brand of car is greater than 15 Km/liter of the gasoline.
• Ha: μ > 15 Km/liter

Step 1: State the Null Hypothesis.

This is denoted by Ho. We state the null hypothesis as if we are extremely lazy persons and we don’t want to do any work! For example if new gasoline is claiming to have an average mileage of greater than 15Km/liter then my null hypothesis would be “it is less than or equal to 15 Km/liter” hence, by doing so, we would not take any pain in testing the new gasoline. We are happy with status quo!

So the null hypothesis in all of the above cases are

• The mean height of all students in the high school is 160 Cm.
• Ho: μ = 160 Cm
• The mean salary of the fresh MBA graduates is \$65000
• Ho: μ = \$65000
• The mean mileage of a particular brand of car is greater than 15 Km/liter of the gasoline.
• Ha: μ ≤ 15 Km/liter

Therefore, if you want me to work, first you make an effort to reject the null hypothesis!

Step 3: Set α

But, before we go any further, it is important to understand that we are using a sample for estimating the population parameter and the size of the sample is very less than then the size of the population. And because of this sampling error, estimating population parameter would contain some uncertainty or error. There is two types of error that can occur that we can make in hypothesis testing.

Following is the contingency table for the null hypothesis. We can make two errors, first rejecting the null hypothesis when it is true (α error) accepting the null hypothesis when it is false (β error). Hence, the acceptance limit for both the error is decided prior to hypothesis testing. Using z-transformation or t-test, we determine the critical value (threshold) corresponding to error α.

The level of significance α is the probability of rejecting the null hypothesis when it is true. This is like rejecting a good lot of material by mistake. Whereas the β is the called as type-II error and it is the probability of accepting the null hypothesis when it is false. This is like accepting a bad lot of material by mistake.

 Let’s understand null and alternate hypothesis graphically Following are the distribution of Ho and Ha with means μa & μb respectively and both having a variance of σ2. We also have an error term α, representing a threshold value on the distribution of Ho beyond which, we would fail to accept the Ho (in other words, we are would reject the Ho). Now the issue that is to be resolved is “how we can say that the two distributions represented by Ho and Ha are same or not” It is usually done by measuring the extent of overlap between the two distributions. This we do by measuring the distance between the mean of the two distributions (of course we need to consider the inherent variance in the system). There are statistical tools like z-test, t-tests, ANOVA etc. which helps us in concluding, whether the two distributions are significantly overlapping or not.

Step 4: Collect the Data

Step 5: Calculate a test statistic.

The test statistic is a numerical measure that is computed from the sample data which, is then compared with the critical value to determine whether or not the null hypothesis should be accepted. Another way of doing is to convert the test statistics to a probability value called as p-value, which is then compared with α, to conclude whether the hypothesis that was made about the population is to be accepted or rejected.

Also See “p-value, what the hell is it?”

Conceptualizing “Distribution” Will Help You in Understanding Your Problem in a Much Better Way

Is it difficult for you to comprehend confidence interval?

Step 6: Construct Acceptance / Rejection regions. The critical value is used as a benchmark or used as a threshold limit to determine whether the test statistic is too extreme to be consistent with the null hypothesis. Step 7: Based on steps 5 and 6, draw a conclusion about H0.

The decision, whether to accept or reject the null hypothesis is based on following criterion:

• If the absolute value of the test statistic exceeds the absolute value of the critical value in, the null hypothesis is rejected.
• Otherwise, the null hypothesis fails to be rejected (or simply Ho is accepted)
• Simplest way is to compare α and the p-value. If p-value is < α, reject the Ho.

Summary:

The null and alternative hypotheses are competing statements made about the population based on the sample. Either the null hypothesis (H0) is true or the alternative hypothesis (Ha) is true, but not both. Ideally the hypothesis testing procedure should lead to the acceptance of H0 when H0 is true and the rejection of H0 when Ha is true. Unfortunately, the correct conclusions are not always possible because hypothesis tests are based on sample information therefore, we must allow or we must have a provision for the possibility of type-I and type-II errors.

## “p-value” What the Hell is it? When ever we go to the supermarket, say to buy tomatoes, we go to the vegetable section and by merely looking at them, we make a hypothesis in our mind that all tomatoes must be of good or bad quality. What we are doing is, we are intuitively providing a qualitative limit on the quality and we can call it as theoretical limit. Now we go to the shelf and pick a sample of tomatoes and press then between our fingers to check the hardness, we can take it as the experimental value on hardness. If this experimental value is better than the theoretical value  we end up in buying the tomatoes. In business decisions when we want to compare two processes, we have a theoretical limits represented by α and a corresponding experimental value represented by p-value.  If p-value (experimental or observed value) is found to be less then the α (theoretical value), then two processes are different.

You must have understood the following “normal distribution” after having gone through so many blogs on this site. Let’s revise what we know about the normal curve If a process is stable, it will follow the bell shaped curve called as normal curve. It means that, if we plot all historical data obtained from a stable process – it will give a symmetrical curve as shown above. The distance from the mean (μ) in either direction is measured in the terms of σ. The σ represents the standard deviation (a measurement of variation) The main characteristic of the above curve is the proportion of the population captured in-between any two σ values. For example μ±2σ would contain 95% of the total population and μ±3σ would contain 99.73% of the total population.

The normal curve doesn’t touches the x-axis i.e. it extend from – to + . This information is very much important for understanding the p-value concept. The implication of this statement is that, there is always a possibility of finding an observation between – to + or in other words “even a stable process can give a product with specification anywhere between – to + ”. But, as you move away from the mean, the probability of finding an observation decreases, for example, the total probability of finding an observation beyond μ±2σ is only 5% (2.5% on either side of the normal curve). This probability decreases to 0.3% for an interval μ±3σ. Now, let’s understand this: I am manufacturing something and my process is quite robust and it follows the normal distribution. As per point number-2 (see above) there is always a possibility that the specification of the product can fall anywhere between – ∞ to +∞ . But, I can’t go to my customer and make this statement. The point that I want to make here is that, there has to be a THRESHOLD DISTANCE (control limits) from the mean (say μ ± xσ) as the acceptance criterion for the product and if specification falling beyond this threshold limit, would be rejected (will not be shipped to the customer). In other words, if my process is giving me the products with sampling distribution of mean beyond the threshold limits, then I will assume that my process has deviated from the SOP (standard operating procedure) due to some assignable cause and now the current process is different from the earlier process! Or simply there are two processes that are running in the plant.

Generally μ±3σ is taken as the threshold limit. This threshold is represented by alpha (α) or the % acceptable error. In present case (μ ± 3σ), α = 0.3% or 0.003. From the above point, it is clear that, as long as the process is giving me the sampling distribution of mean within μ±3σ, we would say that the products are coming from the same process. If the sampling distribution of mean of a batch of the manufactured product is falling beyond μ±3σ then, it would represent a different process.

Till now we have defined a theoretical threshold limit called as alpha (α). Now consider two sampling distribution of mean of two processes described below In case-3, we can confidently say that the two processes are different as there is a minimum overlap of two sampling distribution of means. But, what about case-2 and case-1? In these two cases, taking decision would be difficult because there is significant overlap of two distributions! (At least appears to be). In these circumstances, we need a statistical tool to access whether the overlap is significant or not, in other words a tool is required to ascertain that the sampling distribution of mean of two processes are significantly apart to say that the two processes are different. In order to do that we need to collect some data from both the processes and then subject them to some statistical tests (z-test, t-test, F-test etc.) to check whether the difference between the mean of two processes is significant or not. This significance obtained by collecting a samples from both the population followed by a statistical analysis, the result is obtained in the form of a probability term called as p-value. The point to be noted here is that, the p-value is generated from some statistical test (equivalent to an experiment value).

We can say that the α is the theoretical threshold limit and the p-value is the experimentally generated threshold limit and if the p-value is less than or equal to the theoretical threshold limit α then we would say two processes are really different.

When we say the p-value < α, it means that the sampling distribution of mean of the new process is significantly different from the existing process.

To summarize

1. In general a = 0.05, there is only 5% chance that two processes are same.
2. If p-value (experimental or observed value) is < α (theoretical value), then new process is different.

More details would be covered in hypothesis testing

## Conceptualizing “Distribution” Will Help You in Understanding Your Problem in a Much Better Way Abstract: You will be surprised that we all are aware of this concept of distribution and are using it intuitively, all the time! Don’t believe me? Let me ask you a simple question, to which income class do you belong? Let’s assume that your answer is middle income class. On what basis did you made this statement? Probably in your mind you have following distribution of income groups and based on this image in your mind, you are telling your position is towards the left side or towards the middle income group on this distribution. Figure-1: How we are making use of distribution in our daily life, intuitively
 Note: This article gives a conceptual view of the tools that we use in inferential statistics. Here we are not explanting the concept of sampling or  the sampling distribution. Instead we are using distribution of individual values and assuming them to be normally distributed (which is not always the case) in order to explain the concept and also using it for the illustration purpose. We advise readers to read something on “sampling and sampling distribution” immediately after reading this article for better clarity as we are giving oversimplified version of the same in the present article. Don’t miss the “Central limit theorem”.

Introduction to the Concept of Distribution

When we say that my child is not good at studies, you are drawing a distribution of all students in your mind and implicitly trying to tell the position of your child towards the left of that distribution. Whenever we talk of adjectives like rich, poor, tall, handsome, beautiful, intelligent, cost of living etc., we subconsciously, associate a distribution to those adjective and we just try to pinpoint the position of a given subject onto this distribution.

What we are dealing here is called as inferential statistics because, it helps in drawing inferences about the population based on a sample data. This is just opposite of probability as shown below. Figure-2: Difference between probability & statistics

This inferential statistics empower us to take a decision based on the small sample drawn from a population.

Why, it is so difficult to take decisions or what causes this difficulty?

This is because we are dealing with samples instead of population. Let’s assume, we are making a batch of one million tablets (population) of Lipitor and before releasing this batch in to the market, we want to make sure that each tablet must be having Lipitor content of 98-102%. Can we analyze all one million tablets? Absolutely not! What we actually do is to analyze, say 100 tablets (Sample) selected at random from one million tablets and based on the results, we accept or reject the whole lot of one million tablets (we usually use z-test or t-test for taking decisions)

BUT, there is a catch. Since we are working with small samples, there is always a chances of taking a wrong decision because the sample thus selected may not be homogeneous enough to represent the entire population (sampling error). This error is denoted by alpha (α) and is decided by the management prior to performing any study i.e. we are accepting an error of α. It means that there is a probability of α that we are accepting a failed batch of Lipitor. Since α is a theoretical threshold limit then, it must be vetted by some experimental probability value. This experimental or the observed probability value is called as p-value (see blog on p-value).

Another aspect of the above discussion arises if we draw two or more samples (of 100 tablets each) and try to analyze them. Let me make it more complicated for you. You are the analyst and I come to you with three samples and want to know from you, whether all these three samples are coming from a single batch (or belong to the same parent population) or not? Point I want to emphasize is that, even though multiple samples are withdrawn from the same population but they would seldom be exactly the same because of the sampling error. The concept is described in following figure-3. This type of decision where sample size ≥ 3, is taken by ANOVA. Figure-3: The distribution overlap and the decision making (or inferential statistics)

We have seen earlier that α is the theoretical probability or a threshold limit beyond which we assume that the process is no longer the same. This theoretical limit is then tested by collecting a dataset followed by performing some statistical tests (t-test, z-test etc.) to obtain an experimental or observed probability value or the p-value and if, this p-value is found to be less than α, we say that samples are coming from two different populations. This concept is represented below Figure-4: The relationship between p-value and the alpha value for taking statistical decisions.

Let’s remember the above diagram and try to visualize some more situations that we face every day, where we are supposed to take decisions. But before we do that, one important point, we must identify the target population correctly otherwise whole exercise would be a futile one.

For example

As a high end apparel store, I am interested in the monthly expenditure of females, but wait a second, shouldn’t we specify what kind of females? Yes, we require to study the females of following two categories

The employed and the self-employed females (great! at least we have identified the population categories to be compared). Now next dilemma is whether to consider the females of all age groups or the females below certain age? As my store is more interested in young professionals hence I would compare the above two groups of females but with an age restriction of less than or equal to thirty years. Figure-5: Identifying the right population for study is important

Another important point, in order to compare two (using z-test or t-test) or more samples (using ANOVA), we also require information about the mean and standard deviation of the samples, before we can tell whether they are coming from same or different parent population.

For example, the mean monthly expenditure on apparel by a sample of 30 employed females is \$1500 and the mean expenditure by 30 self-employed females be \$1510. Immediately we will try to compare these two means and conclude that two means are almost the same. In back of our mind we are assuming that even though means are different but there will some variation in the data and if, we consider this variation then this difference is not significant. Remember! We have made some kind of distribution in our mind before making this statement. (statistically we do it by two sample t-test) Figure-6: Significant Overlap between distributions indication no difference between them

What if, the mean expenditure by self-employed females be \$1525, then we can say it’s not a big difference to be significant (again we are assuming that there will be a variability in the data). What if, the mean expenditure by self-employed females is \$1600, in this case we are certain that the difference is significance. In all three cases discussed above, it is assumed that variance remained constant. Figure-7: Insignificant Overlap between distributions indication that there is a difference between them

In real life, whenever we encounter two samples, we are tempted to compare the mean directly for taking decisions. But, in doing so, we forget to consider the standard deviation (variation) that is there in the data of two samples. If we consider the standard deviation and then if we find that there is no significant overlap between the distribution of the monthly expenditure by employed females and the distribution of the monthly expenditure by self-employed, then we can conclude that the expenditure behavior of the two groups are different (see figure-6 & 7 above).

Some other situations that could be understood by drawing the distribution. It will help us in comprehending the situation in a much better way.

Women workforce are protesting that there is a gender biasness in the pay scale in your company, is it so?

Once again, be careful about selecting the population for the study! We should only compare males/females of same designation or with same work experience. Let’s take the designation (males & females at manager and senior manager level) as a criterion for the comparison. Since, we have identified the population, we can now select some random samples from both genders belonging to manager and senior manager level. We can have two situations, either the two distribution overlaps or do not overlap. If there is a significant overlap (p-value > α) then there is no difference in salary based on the gender. On the other hand, if two distribution are far apart (p-value < α), then there is a gender bias. Figure-8: Intuitive scenarios for taking decision, based on the degree of distribution overlap

Our new gasoline formula gives a better mileage than the other types of gasoline available in the market, should we start selling it?

This problem can be visualized by following diagram. But be careful! While measuring mileage, make sure you are taking same kind of car and testing them on the same road and running them for the same number of kilometers at a same constant speed! Since number of samples ≥ 3, use ANOVA. Figure-9: Understanding the gasoline efficiency using distribution

New filament increase the life time of a bulb by 10%, should we commercialize it?

For this problem, let’s produce two sets of bulbs, first set with the old filament and second set with the new filament. This is followed by testing the samples from each group for their lifespan, what we are expecting is represented below Figure-10: Understanding the filament efficiency using distribution

A new catalysts developed by R&D team can increase the yield of the process by 5%, should we scale-up the process?

Here we need to establish whether the 5% increase in yield is really higher or not. Can this case be represented by case-1 or by case-2 in above diagrams?

The efficacy of a new drug is 30% better than of the existing drug in the market, is it so?

The soap manufacturing plant finds that some of the soap are weighing 55 gm. instead of 50-53 gm t(he target weight)., should he reset the process for corrective actions?

A new production process claims to reduce the manufacturing time by 4 hrs, should we invest in this new process?

The students of ABC management school are offered better salary than that of the XYZ School, is it so? Colleges advertise like that!

Let’s have a look how the data is usually manipulated here. In order to promote a brand, companies usually distort the distribution when they compare their products with the other brands. Figure-11: Misuse of statistics

ABC College or any other company promoting their brands would take samples from the upper band of their distribution and then they compare it with the distribution of the XYZ College or with other available brands. This gives a feeling that ABC College or a given brand in question is better than others. Alternatively, you can take competitor’s samples from the lower end of their distribution for comparison for getting the feel good factor about your brand!

Yield of a process has decreased from 90% to 87%, should we take it as a six sigma project?

Again, we need to establish whether the decrease of 3% yield is really significant or not. Can this case be represented by case-1 or by case-2 in above diagrams?

If we look at the situations described in points 4-8 above, we are forced to think “what is the minimum separation required between the mean of two sample, to tell whether there is significant overlap or not” Figure-12: What should be the minimum separation between distribution?

This is usually done in following steps (this will be dealt separately in next blog on hypothesis testing)

1. Hypothesis Testing
1. Null and alternate hypothesis
2. Decide α
2. Test statistics
1. Use appropriate statistical test to estimate p-value like Z-test, t-test, F-test etc.
3. Compare p-value and α
4. Take decision based on whether p-value is < or > α

Concept of distribution and the hypothesis testing

Let’s see how the above concept of distribution helps in understanding the hypothesis testing. In hypothesis testing we make two statement about the same population based on the sample. These two statement are known as “null” and “alternate” hypothesis.

Null Hypothesis (H0): Mean mileage from a liter of new gasoline ≤ 20 Km (first distribution)

Alternate Hypothesis (Ha): Mean mileage from a liter of new gasoline > 20 Km (second distribution)

The above two statement can be represented by following two distribution Figure-13: Distributions of null and alternate hypothesis

Now, if H0 is true i.e. new gasoline is no better than the existing one then, we would expect two distributions to overlap significantly (p-value > a) Figure-14: Pictorial view of the condition when null hypothesis is true

On the other hand if H0 is false or Ha is true (new gasoline is really better than the existing one) then these two distribution will be far from each other or there would no significant overlap of the two distributions (p-value < a) Figure-15: The pictorial view of the case when null hypothesis is not true

Above discussion can be extended to understand ANOVA, Regression analysis etc.

Summary

This article tries to give a pictorial view to a given statistical problem, we can call it as “The Tale of Two Distributions”.

Any business problem that requires decision making can be visualized in the form of a overlapping or a non-overlapping distributions. This will give a pictorial view of the problem to the management and would be easy for comprehending the problem.

Another point that is important here is the exercise if identifying the right target population i.e. we must make sure that an apple is compared to an apple!

Going forward, this understanding will help you in understanding hypothesis testing in upcoming blog.

## 7QC Tools: Why do we Require to Plot X-bar and R-charts Simultaneously Abstract: The main purpose of the control charts is to monitor the health of the process and this is done by monitoring both, accuracy and the precision of the process. The control charts is a tool that helps us in doing so by plotting following two control charts simultaneously for accuracy and precision. Control chart for mean (for accuracy of the process) Control chart of variability (for Precision of the process) E.g. X-bar  and R chart (also called averages and range chart) and X-bar  and s chart

The Accuracy and the precision

We all must be aware of the following diagram that explains the concept of precision and accuracy in that analytical development.

Case-1:

If you are hitting the target all the time at the bull’s eye is called as  accuracy and if all your shots are concentrated at the same point then it is called as Precision. Figure-1: Accuracy and precision

Case-2:

You are off the target (inaccurate) all the time but your shots are concentrated at the same point i.e. there is not much variation (Precision)

Case-2:

It is an interesting case. Your shots are scattered around the bull’s eye but, on an average your shots are on the target (Accuracy), this is because of the average effect. But your shots are wide spread around the center (Imprecision).

Case-4

In this case all your shots are off target and precision is also lost.

Before we could correlate the above concept with the manufacturing process, we must have a look at the following diagram that explains the characteristics of a given manufacturing process. Figure-2: Precision and Accuracy of a manufacturing process

The distance between the average of the process control limits and the target value (average of the specification limits) represents the accuracy of the process or how much the process mean is deviating from the target value.

Whereas the spread of the process i.e. the difference between LCL and UCL of the process represents the precision of the process or how much variation is there in the process.

Having understood the above two diagrams, it would be interesting to visualize the control chart patterns in all of the four cases discussed above. But, before that let’s have a look at the effect of time on a given process i.e. what happens to the process with respect to the time?

As the process continue to run, there will be wear and tear of machines, change of operators etc. and because of that there will be shift and drift in the process as represented by four scenarios described in the following diagram. Figure-3: Process behavior in a long run

A shift in the process mean from the target value is the loss of accuracy and change in the process control limits is the loss of precision. A process shift of ±1.5σ is acceptable in the long run.

If we combine figure-1 and figure-3, we get the figure-4, which enable us to comprehend the control charts in a much better way. This gives picture of the manufacturing process in the form of control charts in four scenarios discussed above. Figure-4: Control chart pattern in case of precision and accuracy issue

Above discussion is useful in understanding the reasons behind the importance of the control charts.

1. Most processes don’t run under statistical control for long time. There are drifts and shift in the process with respect to the time, hence process needs adjustment at regular interval.
2. Process deviation is caused by assignable and common factors/causes. Hence a monitoring tool is required to identify the assignable causes. This tool is called as control charts
3. These control charts helps in determining whether the abnormality in the process is due to assignable causes or due to common causes
4. It enables timely detection of abnormality prompt us to take timely corrective action
5. It provides an online test of hypothesis that the process is under control
1. Helps in taking decision whether to interfere with process or not.
1. H0: Process is under control (common causes)
2. Ha: Process is out of control (assignable causes) 6.  Helps in continuous improvement: Figure-5: Control Charts provide an opportunity for continuous improvement

## Is it Difficult for you to Comprehend the Concept of Confidence Interval? Try this out Abstract:

I found it very difficult to comprehend the concept of sampling, sampling distribution of mean and the confidence interval. These concepts plays a very important role in inferential statistics, which is a integral part of six sigma tools. This is an attempt to simplify the concept of confidence interval or simply CI.

In our practical life we need to take the decisions about the population based on the analysis of a sample drawn from that sample. Example, a batch of one million tablets is to be qualified by QA based on a sample of say, 50 tablets. The confidence interval from the samples, enables us to have an interval estimate that may contain the population parameter with some degree of error α.

If we assume that the population average is a golden fish that we want to catch from a pond using 100 types of nets of different sizes (equivalent to CI). Then 95 of those nets (out of 100 nets) would capture the golden fish if we consider an error of 5%.

Introduction to confidence interval

Let’s assume that we are in the fishing business and we have our own farm where we are raising fish right from the egg stage to the mature fish. Traditionally, we will incubate the eggs for 10-15 days to get the larvae and then that would be transferred to the juvenile tank where, they would be fed and monitored for another 3 months. Once they are 3 months old we would again transfer all fish from the juvenile tank to a larger tank where they would be on different diet for another 3 months. After this all fish would be sold to the wholesaler.

Now problem is that we would like to sell only those fish which are having weight around 900-950 Gm. In order to maximize the profit per fish. Any fish less than that would be a loss to us as, I would we giving away more number of fish for a given order from the wholesaler. You can say that let’s weigh each and every fish and sell only those between 925-950 Gm. Yes, that can be a solution but, imagine the efforts required to take out every fish, weighing them and making arrangement to keep them in some other tank. This is very difficult as, at a given time we are having 1000-1500 fishes in the tank. Is there a solution by which we can estimate the average weight of all the fishes in the tank?

Yes, Statistics does that. But how?

Statistically it can be done by drawing a random sample and calculating the average of the sample and then trying to estimate the average of the population. But, we need to understand a very important point here, there are chances that the sample thus collected might not be representing the entire population of fish (this is called as sampling) hence, there would be some margin of error in calculating the population average.

Therefore, the average weight of all fish in the pond can be expressed as

= average weight of the sample ± margin of error

Or simply

Average population parameter = sample’s average ± margin of error

Sample Average: This average will vary from sample to sample but, as the sample size increases and then the average of the sample will be closer to the population average.

Margin of Error: This margin of error is calculated statistically. Right now it is sufficient to know that the margin of error is directly proportional to the standard deviation of the sample and inversely proportional to the sample size. It means that if we want to have a narrow interval for the estimation of the population average then, we need to take a larger sample with small variation. This term is generally known as standard error

There is also a statistical constant involved in the calculation of margin of error, this is called as critical t-value. This critical t-value depends on the presumed error α and the degree of freedom.

What is this error α?

This error or risk and is denoted by α. Let’s assume, in above fish example, we draw 100 samples of 5 fish each and calculate all 100 interval using equation-1. We are assuming that all these interval would contain the population average, but there are chances that some intervals might not contain the population average. This because we are working with samples and sampling error bound to happen. So we assume an error or a risk α that, out of 100 interval (calculated from 100 sample) there will be α number of intervals that would not contain the population average. This α is denoted in % or in the probability terms. For example α = 0.05 mean that there is 5% chances or 0.05 probability that 5 out of 100 intervals thus calculated might not contain the population average. This α, is decided prior to starting any experimentation using samples. It’s purely a business decision based on the risk appetite of the company. We can work with α = 0.05 or 0.1 or 0.15. Generally we work with 0.05.

Once we have defined α, we can now discuss t-critical.

The t-critical is the threshold value (like z-value, discussed earlier) on the t-distribution beyond which process is no longer the same i.e. if the observations are falling in the region < tcritical (in below figure), then we would say that the samples are coming from the same parent population otherwise it is coming from the different sample. This t-value is characterized by two parameter α and degree of freedom (df = number of observations-1 or simply n-1) and its value can be obtained from the t-distribution table. When to take a or α/2?

As we have considered the total acceptable error of 5%, now based on the scenario there are chances that the interval calculated might miss the population parameter on either side of the interval. Hence the total error is distributed at both end of the interval equally. For example, say the interval calculated is 915 to 935 Gm. Now, the chances are there that the actual population mean might be less than 915 or more than 935. So if the total error we started with is 5% then, 2.5% is distributed at both the end. This is a case of two tailed test and error on single side is represented by α/2. If this has been a one tail test, then there is no need to do this and the error is represented by α. Now, we are ready estimate the population parameter

Average population parameter = sample’s average ± margin of error Now we can see that, if we want to have a narrower interval then we need to decrease the term “margin of error” and for that, we have to increase the sample size or decrease the σ. Since controlling σ is not in our hand (it is the characteristics of the random sample), we can increase the sample size in order to reach closer to the population parameter.

Interval calculated above is called as CONFIDENCE INTERVAL or simply CI.

Concept of CI obtained from a group of samples is illustrated below Example:

We have no idea about the average weight of all fishes in the pond and we also don’t have any idea about the standard deviation of the weights of all fishes in the tank. In that case we have to estimate the average weight of all fishes in the pond based on the sample’s average and its standard deviation.

Let’s take out the first sample of five fish from the pond and calculate its CI. Methodology to be applied Average weight of five fish from first sample = 928 Gm

Standard deviation of five fish from first sample = 27.97

Now calculate tα/2,df

As α/2 = 0.025 and df = 5-1 = 4

Therefore from t-distribution table Margin of error Confidence Interval of the first sample Inference from the above CI

From the first sample, we got a CI of 892 to 964, it means that the average weight of all fishes in the pond is between 892 to 964 Gm. But, still we can’t pin point the exact average of all fishes in the pond!

Further, if we draw 99 more samples of 5 fishes each and calculate the corresponding CI then we will find that 95 CI out of 100 CI would contain the population average. Only 5 CI would not contain the population average. But, still we can’t pin point the exact average of all fishes in the pond!

Let’s draw some more samples and calculate their CI How we can use this concept in production:

Suppose we made a lot of a product (be a million tablets, bulbs etc.) and QA need to qualify that batch. What he does is to take a random sample and calculates its CI. If this CI contains the population mean (specification), he would pass the lot. Have a look at the following blogs for application part.

Related Blog for the utility of CI

How to provide a realistic range for a CQAs during product development to avoid unwanted OOS-1.

How to provide a realistic range for a CQAs during product development to avoid Unwanted OOS-2 Case Study

But always remember this

If we assume that the population average is a golden fish that we want to catch from a pond using 100 types of nets of different sizes (equivalent to CI). Then 95 of those nets out of 100 nets) would capture the golden fish if we consider an error of 5%.

###### Common Misconception about CI

The biggest mistake we make while interpreting the confidence intervals is that we think CI represents the percentage of the data from a given sample that falls between two limits. For example, in above example, the first CI was found to be 893-963 Gms. People would make a mistake of assuming that there is 95% chance that the mean of all fishes would fall within this range. This is incorrect!

Following books gives an excellent presentation of confidence interval, sampling distribution of mean through cartoons

## Why Standard Normal Distribution Table is so important? ###### Abstract

Since we are entering the technical/statistical part of the subject hence, it would be better for us to understand the concept first

For many business decisions, we need to calculate the likelihood or probability of an event to occur. Histograms along with relative frequency of a dataset can be used to some extent.. But for every problem we come across we need to draw the histogram and relative frequency to find the probability using area under the curve (AUC).

In order to overcome this limitation a standard normal distribution or Z-distribution or Gaussian distribution was developed and the AUC or probability between any two points on this distribution is well documented in the statistical tables or can be easily found by using excel sheet.

But in order to use standard normal distribution table, we need to convert the parent dataset (irrespective of the unit of measurement) into standard normal distribution using Z-transformation. Once it is done, we can look into the standard normal distribution table to calculate the probabilities. From my experience, I found the books belonging to category “statistics for business & economics” are much better for understanding the 6sigma concepts rather than a pure statistical book. Try any of these books as a reference guide.

Introduction

Let’s understand by this example

A company is trying to make a job description for the manager level position and most important criterion was the years of experience a person should possess. They collected a sample of ten manager from their company, data is tabulated below along with its histogram. As a HR person, I want to know the mean years of experience of a manager and the various probabilities as discussed below

Average experience = 3.9 years

What is the probability that X ≤ 4 years?

What is the probability that X ≤ 5 years?

What is the probability that 3 < X ≤ 5 years?

In order to calculate the above probabilities, we need to calculate the relative frequency and cumulative frequency Now we can answer above questions

What is the probability that X ≤ 4 years? = 0.7 (see cumulative frequency)

What is the probability that X ≤ 5 years? = 0.9

What is the probability that 3 < X ≤ 5 years? = (probability X ≤ 5) – (probability X < 3) = 0.9-0.3 = 0.6 i.e. 60% of the managers have experience between 3 to 5 years.

Area under the curve (AUC) as a measure of probability:

Width of a bar in the histogram = 1 unit

Height of the bar = frequency of the class

Area under the curve for a given bar = 1x frequency of the class

Total area under the curve (AUC) = total area under all bars = 1×1+1×2+1×4+1×2+1×1 = 10

Total area under the curve for class 3 < x ≤ 5 = (AUC of 3rd class + AUC of 4th class) /total AUC = (4+2)/10 = 0.6 = probability of finding x between 3 and 5 (excluding 3)

Now, what about the probability of (3.2 < x ≤ 4.3) =? It will be difficult to calculate by this method, as requires the use of calculus.

Yes, we can use calculus for calculating various probabilities or AUC for this problem. Are we going to do this whole exercise again and again for each and every problem we come across?

With God’s grace, our ancestors gave us the solution in the form of Z-distribution or Standard normal distribution or Gaussian distribution, where the AUC between any two points is already documented.

This Standard normal distribution or Gaussian distribution is widely used in the scientific measurements and for drawing statistical inferences. This normal curve is shown by a perfectly symmetrical and bell shaped curve.

The Standard normal probability distribution has following characteristics

1. The normal curve is defined by two parameters, µ = 0 and σ = 1. They determine the location and shape of the normal distribution.
2. The highest point on the normal curve is at the mean which is also the median and mode.
3. The normal distribution is symmetrical and tails of the curve extend to infinity i.e. it never touches the x-axis.
4. Probabilities of the normal random variable are given by the AUC. The total AUC for normal distribution is 1. The AUC to the right of the mean = AUC to the left of mean = 0.5.
5. Percentage of observations within a given interval around the mean in a standard normal distribution is shown below The AUC for standard normal distribution have been calculated for all given value of p ≥ z and are available in tables that can be used for calculating probabilities. Note: be careful whenever you are using this table as some table give area for ≤ z and some gives area between two z-values.

Let’s try to calculate some of the probabilities using above table

Problem-1:

Probability p(z ≥ 1.25). This problem is depicted below Look for z = 1.2 in vertical column and then look for z = 0.05 for second decimal place in horizontal row of the z-table, p(z ≤ -1.25) = 0.8944

Note! The z-distribution table given above give the cumulative probability for p(z ≤ 1.25), but here we want p(z ≥ 1.25). Since total probability or AUC = 1, p(z ≥ 1.25) will be given by 1- p(z ≤ 1.25)

Therefore

p(z ≥ 1.25) = 1- p(z ≤ -1.25) = 1-0.8944 = 0.1056

Problem-2:

Probability p(z ≤ -1.25). This problem is depicted below Note! Since above z-distribution table doesn’t contain -1.25 but the p(z ≤ -1.25) = p(z ≥ 1.25) as standard normal curve is symmetrical.

Therefore

Probability p(z ≤ -1.25) = 0.1056

Problem-3:

Probability p(-1.25 ≤ z ≤ 1.25). This problem is depicted below For the obvious reasons, this can be calculated by subtracting the AUC of yellow region from one.

p(-1.25 ≤ z ≤ 1.25) = 1- {p(z ≤ -1.25) + p(z ≥ 1.25)} = 1 – (2 x 0.1056) = 0.7888

From the above discussion, we learnt that a standard normal distribution table (which is readily available) could be used for calculating the probabilities.

Now comes the real problem! Somehow I have to convert my original dataset into the standard normal distribution, so that calculating any probabilities becomes easy. In simple words, my original dataset has a mean of 3.9 years with σ = 1.37 years and we need to convert it into the standard normal distribution with a mean of 0 and σ = 1.

The formula for converting any normal random variable x with mean µ and standard deviation σ to the standard normal distribution is by z-transformation and the value so obtained is called as z-score. Note that the numerator in the above equation = distance of a data point from the mean. The distance so obtained is divided by σ, giving distance of a data point from the mean in terms of σ i.e. now we can say that a particular data is 1.25σ away from the mean. Now the data becomes unit less! Let’s do it for the above example discussed earlier Note: Z-distribution table is used only in the cases where number of observations ≥ 30. Here we are using it to demonstrate the concept. Actually we should be using t-distribution in this case.

We can say that the managers with 4 years of experience are 0.073σ away from the mean and on the right hand side. Whereas the managers with 3 years of experience are -0.657σ away from the mean on left hand side.

Now, if you look at the distribution of the Z-scores, it resembles the standard normal distribution with mean = 0 and standard deviation =1.

But, still one question need to be answered. What is the advantage of converting a given data set into standard normal distribution?

There are three advantages, first being, it enables us to calculate the probability between any two points instantaneously. Secondly, once you convert your original data into standard normal distribution, you are ending in a unit less distribution (both numerator & denominator in Z-transformation formula has same units)! Hence, it makes possible to compare an orange with an apple. For example, I wish to compare the variation in the salary of the employees with the variation in their years of experience. Since, salary and experience has different unit of measurements, it is not possible to compare them but, once both distributions are converted to standard normal distribution, we can compare them (now both are unit less).

Third advantage is that, while solving problems, we needn’t to convert everything to z-scores as explained by following example

Historical 100 batches from the plant has given a mean yield of 88% with a standard deviation of 2.1. Now I want to know the various probabilities

Probability of batches having yield between 85% and 90%

Step-1: Transform the yield (x) data into z-scores

What we are looking for is the probability of yield between 85 and 90% i.e. p(85 ≤ x ≤ 90)  Step-2: Always draw rough the standard normal curve and preempt what area one is interested in Step-3: Use the Z-distribution table for calculating probabilities.

The Z-distribution table given above can be used in following way to calculate p(-1.43 ≤ z ≤ 0.95)

Diagrammatically, p(-1.43 ≤ z ≤ 0.95) = p(z ≤ 0.95) – p(z ≤ -1.43), is represented below p(-1.43 ≤ z ≤ 0.95) = p(z ≤ 0.95) – p(z ≤ -1.43)= 0.83-0.076 = 0.75

75% of the batches or there is a probability of 0.75 that the yield will be between 85 and 90%.

It can also be interpreted as “probability of getting a sample mean between 85 and 90 given that population mean is 88% with standard deviation of 2.1”.

Probability of yield ≥ 90%

What we are looking for is the probability of yield ≥ 90% i.e. p(x ≥ 90) = p(z ≥ 0.95) Diagrammatically, p(-1.43 ≤ z ≤ 0.95) = p(z ≤ 0.95) – p(z ≤ -1.43), is represented below p(x ≥ 90) = p(z ≥ 0.95) = 1-p(z ≤ 0.95) = 1- 0.076 = 0.17, there is only 17% probability of getting yield ≥ 90%

Probability of yield between ≤ 90%

This is very easy, just subtract p(x ≥ 90) from 1

Therefore,

p(x ≤ 90) = 1- p(x ≥ 90) = 1- 0.17 = 0.83 or 83% of the batches would be having yield ≤ 90%.

Now let’s work the problem in reverse way, I want to know the yield corresponding to the probability of ≥ 0.85.

Graphically it can be represented as Since the table that we are using gives the probability value ≤ z value hence, first we need to find the z-value corresponding to the probability of 0.85. Let’s look into the z-distribution table and find the probability close to 0.85 The probability of 0.8508 correspond to the z-value of 1.04

Now we have z-value of 1.04 and we need find corresponding x-value (i.e. yield) using the Z-transformation formula  Solving for x

x = 90.18

Therefore, there is 0.85 probability of getting yield ≤ 90.18% (as z-distribution table we are using give probability for ≤ z) hence, there is only 0.15 probability that yield would be greater than 90.18%.

Above problem can be represented by following diagram Exercise:

The historical data shows that the average time taken to complete the BB exam is 135 minutes with a standard deviation of 15 minutes.

Fins the probability that

1. Exam is completed in less than 140 minutes
2. Exam is completed between 135 and 145 minutes
3. Exam takes more than 150 minutes

Summary:

This articles shows the limitations of histogram and relative frequency methods in calculating probabilities, as for every problem we need to draw them. To overcome this challenge, a standardized method of using standard normal distribution is adopted where, the AUC between any two points on the curve gives the corresponding probability can easily be calculated using excel sheet or by using z-distribution table. The only thing we need to do is to convert the given data into standard normal distribution using Z-transformation. This also enables us to compare two unrelated things as the Z-distribution is a unit less with mean = 0 and standard deviation = 1. If the population standard deviation is known, we can use z-distribution otherwise we have to work with sample’s standard deviation and we have to use Student’s t-distribution.

## ANOVA by Prof. Hunter We are excited about the quality of videos available on youtube, on almost every topic. Look at this video on ANOVA by none other than Prof. Hunter himself.  These video was shot in 1966 in black & white but experience the contents.

ANOVA-1

ANOVA-2

## Six Sigma Video lectures from IIT Karagpur by Prof. Bagachi This is a collection of 40 excellent video lectures from Prof. Bagachi which would be helpful in understanding the concepts.

6sigma-Video Lectures by Prof. Bagachi