Hello and welcome to the fourth lesson of the Certified Six Sigma Green Belt Course offered by Simplilearn. This lesson will cover the details of the analyze phase. Let us start with the objectives of this lesson in the next screen.

After completing this lesson you will be able to: Discuss multi-vari studies and the causes Explain correlation and its types Discus the various hypothesis tests Discuss the use and application of F-test, t-test, ANOVA and Chi-square (Pronounce as: khai square) Let us proceed to the first topic of this lesson in the following screen.

In this topic, we will discuss exploratory data analysis in detail. Let us learn about multi-vari studies in the following screen.

Multi-vari studies or multivariable studies are used to analyze variation in a process. Analyzing variation helps in investigating the stability of a process. More stable the process, less is the variation. Multi-vari studies also help in identifying areas to be investigated. Finally, they help in breaking down the variation into components to make the required improvements. Multi-vari studies classify variation sources into three major types: Positional, Cyclical (Pronounce as: sye-clical), and Temporal. Click each type to learn more. Positional variation occurs within a single piece or a product. Variation in pieces of a batch is also an example of positional variation. In positional variation, measurements at different locations of a piece would produce different values. Suppose a company is manufacturing a metal plate of thickness 1 inch and the plate thickness is different at many points, it is an example of positional variation. Some of the other examples can be pallet stacking in a truck, temperature gradient in an oven, variation observed from cavity-to-cavity within a mold, region of a country, and line on invoice. Cyclical variation occurs when measurement differs from piece to piece or product to product but over a short period of time. Time is important because certain measurements may change if a product, such as a hot metal sheet, is measured after a long interval. If the measurement at the same location in a piece varies with different pieces, it is an example of cyclical variation. Other examples of cyclical variations are batch to batch variation, lot-to-lot variation, and account activity week-to-week. Temporal variation occurs over a longer period of time, such as machine wear and tear, and changes in efficiency of an operator before and after lunch. Temporal variations may also be seasonal. If the range of positional variation in a piece is more in winter than in summer, it is an example of temporal variation. The variation may occur because of unfavorable working conditions in winter. Process drift, performance before and after breaks, seasonal and shift based differences, month-to-month closings, and quarterly returns can be examples of temporal variation.

We will learn about creating a multi-vari chart in this screen. The outcome of multi-vari studies is the multi-vari chart. It depicts the type of variation in the product and helps in identifying the root cause. There are five major steps involved in creating a Multi-Vari chart. They are Select Process and Characteristics, Decide Sample Size, Create a Tabulation Sheet, Plot the Chart, and Link the Observed Values. Click each step to learn more. The first step is to select the process and the relevant characteristics to be investigated. For example, selecting the process where the plate of 1 inch thickness is being manufactured. In this process, four equipment, numbered 1 to 4, produce the 1 inch plates. The characteristic to be measured is the thickness of the plate ranging from 0.95 inch to 1.05 inches. Any plate thickness outside this range is a defect. The second step is to decide the sample size and the frequency of data collection. In this example, the sample size is five pieces per equipment and the frequency of collecting data is every two hours starting from 8 in the morning to 2 in the afternoon. Then the tabulation sheet is created where the data will be recorded. So one should measure the thickness of the plate being produced by the four equipment at a data collection frequency of two hours. The third step is to create a tabulation sheet. In this example, the tabulation sheet with data records contains the columns with time, equipment number, and thickness as headers. The fourth step is to plot the chart. In this example, a chart can be plotted with time on the X axis and plate thickness on the Y axis. The last step is to link the observed values. In this example, the observed values can be linked by appropriate lines.

We will continue to learn about creating a multi-vari chart in this screen. The path to create a Multi-Vari chart in Minitab is by selecting Stat, then Quality Tools, followed by Multi-Vari Chart. The Multi-vari chart created from the data recorded is shown here. The upper specification limit of 1.05 inches and the lower specification limit of 0.95 inches has been marked by green lines. Data outside these lines are defects. The blue dots show the positional variation. The dots are the measurements of pieces in a batch of any single equipment. The black lines join the mean of the data recorded from the equipment. The mean of the data recorded from the products of equipment number 3 is much below the similar mean of other equipment. This shows that equipment number 3 is producing more defects than the other equipment. The red line is the mean of the data recorded at a particular time. The red line rises toward the right, which means the data points shift up after 12 pm. This may be because of the change in operator efficiency after a lunch break at 12 pm. Multi-vari chart helps us visually depict the variations and establish the root causes of variations. In the next screen we will learn about simple linear correlation.

Correlation means association between variables. Simple Linear Regression and Multiple Regression techniques are very important as they help in validating the association of X with Y. The Coefficient Correlation shows the strength of the relationship between Y and X. The dependent variable Y may depend on many independent variables, X. But correlation is used to find the behavior of Y as one of the X’s changes. Correlation helps us to predict the direction of movement in values in ‘Y’ when ‘X’ changes. Statistical significance of this movement is denoted by correlation coefficient r. it is also known as Pearson’s Coefficient of Correlation. In any correlation, the value of the correlation coefficient is always between -1 (Pronounce as minus one) and +1 (Pronounce as plus one). Positive value of r denotes the direction of movement in both variables is same. As X increases, Y also increases and vice versa. Negative value of ‘r’ denotes that the direction of movement in both variables is in inverse fashion. As X increases, Y decreases and as X decreases, Y increases. When value of r is zero, it means that there is no correlation between the two variables. Higher the absolute value of ‘r’, stronger the correlation between ‘Y’ & ‘X‘. Absolute value of a number is its value without the sign. +4 (pronounced plus-four) has an absolute value of 4 and -4 (pronounced minus-four) again has an absolute value of 4. An ‘r’ value of greater than plus 0.85 or lesser than minus 0.85 indicates a strong correlation. Hence r value of -0.95 shows a stronger correlation than the r value of -0.74. The next screen will elaborate on correlation with the help of an example and illustrations through scatter plots.

The four graphs on the screen are scatter plots displaying four different levels of correlation. Correlation measures the linear association between the dependent variable or output variable Y and one independent or input variable X. As can be deduced from the graphs, a definite pattern emerges as the absolute value of correlation coefficient r increases. It is easy to see a pattern in r value of 0.9 and above than to see a pattern in r value of 0.07. It is difficult to find a pattern below correlation coefficient of 0.5. Click the example button to know more. To understand how correlation helps, let us consider an example. A correlation test was performed on the scores of a set of students from their first grade, high school, and under graduation. The under graduation score was the dependent variable and first grade score the independent variable. The value of correlation coefficient r was calculated to be 0.29, and the correlation between under graduation scores and high school scores was 0.45. This means the high school scores have higher correlation compared to the first grade scores. This states the performance of students in high school is a better indicator of their performance in under graduation than their performance in the first grade. Although the correlation exits, as both the values of r are less than 0.85 it will be difficult to draw a straight line on the scatter plot.

In this screen, we will learn about regression. Although correlation gives the direction of movement of the dependent variable Y as independent variable X changes, it does not provide the extent of the movement of Y as X changes. This degree of movement can be calculated using regression. If a high percentage of variability in Y is explained by changes in X, one can use the model to write a transfer equation Y is equal to f(x) and use the same equation to predict future values of Y given X, and X given Y. The output of regression on Y and X is a transfer function equation that can predict values of Y for any other value of X. Transfer function is generally denoted by f and the equation is written as Y equals to f of X. Y can be regressed on one or more X’s simultaneously. Simple linear regression is for one X and multiple linear regression is for more than one X. The next screen will focus on key concepts of regression.

There are two key concepts of regression, transfer function to control Y and Vital X. Click each concept to learn more. The output of regression is a transfer function f of X. Although the transfer function f of X gives the degree of movement in Y as X changes, it is not the correct transfer function to control Y as there may be a low level of correlation between the two. The main thrust of regression is to discover whether a significant statistical relationship exists between ‘Y’ & a particular ‘X’, i.e. (pronounced as “that is”), by looking at p-values. Based on regression, one can infer the vital x and eliminate the unimportant Xs (pronounced exes). The Analyze Phase helps in understanding if there is statistical relevance between Y and X. If the relevance is established using metrics from Regression Analysis, one can move forward with the tests. The Simple Linear Regression or SLR should be used as a Statistical Validation tool in the beginning of this phase.

In this screen, we will understand the concept of simple linear regression. A simple linear regression equation is a fitted linear equation and is represented by the equation shown here. In this equation, Y is the dependent variable and X is the independent variable. A is the intercept of the fitted line on the Y axis, which is equal to the value of Y at X equal to zero. B is the regression coefficient or the slope of the line and C is the error in the regression model, which has a mean of zero. The next screen will focus on the least squares method in simple linear regression.

With reference to the error mentioned earlier, if correlation coefficient of Y and X is not equal to one, meaning the relation is not perfectly linear, there could be several lines that could fit in the scatter plot. Notice the two graphs displayed. For the same set of five points, two different types of lines are drawn and both of them have errors. Error refers to the points on the scatter plot that do not fall on the straight line drawn. The second graph shows the Y intercept. Statistical software like Minitab fits the line which has the least value of errors squared and added. As is clear from the graph, error is the distance of the point from the fitted line. Typically, the data lies off the line. In perfect linear relation, all points would lie on the line and error would be zero. The distance from the point to the line is the error distance used in the SSE calculations. Let’s understand SLR with the help of an example in the next screen.

Consider the following example. Suppose a farmer wishes to predict the relationship between the amount spent on fertilizers and the annual sales of his crops. He collects the data shown here for the last few years & determines his expected revenue if he spends $8 annually on fertilizer. He has targeted sales of $31 this year. The steps to perform Simple Linear regression in MS excel are as follows. Copy the data table on an Excel Worksheet. Select all the data from B1 to C6. This is assuming the Years table appears in cells A1 to A6. Click Insert, and choose the Plain Scatter Chart. It is titled Scatter with only Markers. The basic scatter chart will appear, as shown on the screen. Right click on the data points in the Scatter Chart and choose the option, “Add Trendline” Then choose the option, “Linear” and select the boxes titled, “Display R-Squared value” and “Display equation.” A linear line will appear which is called the Best fit line or the Least Squares line. To use the data for Regression analysis, the interpretation of the scatter chart is as follows: The R-Square value or the Coefficient of Determination conveys if the model is good and can be used. The R-Square value here is 0.3797. It means, 38% of variability in Y is explained by X. The remaining 62% variation is unexplained or due to residual factors. Other factors like rain amount and variability, sunshine, temperatures, seed type, and seed quality could be tested. The low value of R-Square statistically validates poor relationship between Y and X. Thus, the equation presented cannot be used for further analysis. In a similar situation, one should refer to the Cause and Effect Matrix and study the relationship between Y and a different X variable.

We will discuss multiple linear regression in this screen. If a new variable X2 is added to the R-square model, the impact of X1 and X2 on Y gets tested. This is known as Multiple Linear Regression. The value of R2 (pronounced as “R square”) changes due to the introduction of the new variable. The resulting value of R2, which can be used in cases of Multiple Regression, is known as R-Square Adjusted. The model can be used if R2 Adjusted value is greater than 70%. We will look at the key concepts in the next screen.

The key concepts in Multiple Linear Regression are as follows: The residuals or the differences between the actual value and the predicted value give an indication of how good the model is. If the errors or residuals are small and predictions use X’s (pronounced as X-s) that are within the range of the collected data, the predictions should be fine. The Sum of Squares Total can be calculated as follows: Sum of Squares Total or SST equals the Sum of Squares of Regression or SSR plus Sum of Squares of Error or SSE. To arrive at Sum of Squares of Regression SSR, use the formula: SSR equals Sum of Squares Total or SST minus Sum of Squares of Error or SSE. Since SSR is SSE subtracted from SST, value of SSE should be less than SST. R-squared is Sum of Squares of Regression or SSR divided by Sum of Squares Total or SST. Calculating SST and SSE helps in determining SSR, and R-square. To get a sense of the error in the fitted model, calculate the value of Y for a given data using the fitted line equation. To check for error, take two observations of Y at the same X. The most important thing to remember in regression analysis is that the obtained fitted line equation cannot be used to predict Y for values of X outside the data. For example, it would not be possible to predict the amount spent on fertilizers for a forecasted sales of $15 or $60. Both data points lie outside the data set on which regression analysis is performed. If Y is dependent on many X’s, then simple linear regression analysis can be used to prioritize X, but it requires running separate regressions on Y with each X. If an X does not explain variation in Y, then it should not be explored any further. These were the interpretations of the simple linear regression equation. In the next screen, we will learn that despite a relationship being established between two variables, the change in one may not cause a change in the other. Let us discuss the difference between correlation and causation in the following screen.

A regression equation denotes only a relationship between the Y and the X. This does not mean that a change in one variable will cause change in the other. If number of schools & incidents of crime in a city rise together, there may be a relationship, but no causation. The increase in both the factors could be due to a third factor, that is, population. In other words, both of them may be dependent variables to an independent variable. Consider the graphs shown on the screen. The graphs on the left show the relations between number of sneezes and incidents of death with respect to population. Both have a positive correlation. Finding a positive correlation between incidents of deaths and number of sneezes does not mean we assume sneezing is the cause of somebody’s death, despite the correlation being very strong, as depicted in the graph on the right. Let us proceed to the next topic of this lesson in the following screen.

In this topic, we will discuss hypothesis testing in detail. Let us learn about statistical and practical significance of hypothesis test in the following screen.

The differences between a variable and its hypothesized value may be statistically significant but may not be practical or economically meaningful. For example, based on a hypothesis test, Nutri Worldwide Inc wants to implement a trading strategy which is proven to provide statistically significant returns. However it does not guarantee trading on this strategy would result in economically meaningful positive returns. When the logical reasons are examined before implementation the returns are economically significant.. The returns may not be significant when statistically proven strategy is implemented directly. The returns may not be economically significant after accounting for taxes, transaction costs, and risks inherent in the strategy. Thus there should be a practical or economic significance study before implementing any statistically significant data. The next screen will briefly focus on the conceptual differences between a null and an alternate hypothesis.

The conceptual differences between a null and an alternate hypothesis are as follows: Assume the specification of the current process is itself the null hypothesis. Null hypothesis, denoted as the basic assumption for any activity or experiment, is represented as Ho (pronounced as H-O). Null hypothesis cannot be proved; it can only be rejected or disproved. It is important to note that if null hypothesis is rejected, alternative hypothesis must be right. For example, assuming that a movie is good, one plans to watch it. Therefore, the null hypothesis in this scenario will be “movie is good.” Alternate hypothesis or Ha (pronounced as H-A) challenges the null hypothesis or is the converse of the Null Hypothesis. In this scenario, alternate hypothesis will be “movie is not good.” In the following screen, we will discuss Type 1 and Type 2 error.

Rejecting a null hypothesis when it is true is called Type I error. It is also known as ‘Producer’s Risk.’ For example, the rejection of a product by the QA team when it is not defective will cause loss to the producer. Suppose, when a movie is good, it is reviewed to be ‘not good’, this reflects Type I error (pronounced as type one error). In this case, the null hypothesis is rejected when it is actually true. The two important points to be noted are: Significance level or alpha is the chance of committing a Type 1 error The value of Alpha is 0.05 or 5%. Accepting a null hypothesis when it is false is called Type II error (pronounced as type two error). It is also known as ‘Consumer’s Risk.’ For example, the acceptance of a defective product by the Quality Analyst of an organization will cause loss to consumer who buys it. Minimizing Type II error requires acceptance criteria to be very strict. Suppose, when a movie is not good, it is reviewed to be good, this reflects Type II Error. In this case, the alternate hypothesis is rejected when it was actually true. The two important points to be noted are: ? (pronounced as Beta) is the chance of committing a Type II Error. The value of ? is 0.2 or 20%. Any experiment should have as less ? value as possible. The next screen will cover the key points to remember about type I and type II error.

As you start dealing with the two types of errors, keep the following points in mind. The probability of making one type of error can be reduced when one is willing to accept a higher probability of making the other type of error. Suppose the management of a company producing pacemakers wants to ensure no defective pacemaker reaches the consumer. So, the quality assurance team makes stringent guidelines to inspect the pacemakers. This would invariably decrease the beta error or type two error. But this will also increase the chance that a non-defective pacemaker is declared defective by the quality assurance team. Thus alpha error or type one error increases. If all null hypotheses are accepted to avoid rejecting true null hypothesis, it will lead to type two error. Typically, alpha is set at 0.05, which means that the risk of committing a Type one error is 1 out of 20 experiments. In case of any product, the teams must decide what type of error should be less & set the value of alpha and beta accordingly. In the next screen, we will discuss the power of test.

The power of a hypothesis test or the power of test is the probability of correctly rejecting the null hypothesis when it is false. Power of a test is represented by 1 minus beta, which is also the type two error. The probability of not committing a type two error is called the power of a hypothesis test. The power of a test helps in improving the advantage of hypothesis testing. The higher the power of a test, the better it is for purposes of hypothesis testing. Given a choice of tests, the one with the highest power should be preferred. The only way to decrease the probability of a type two error, given the significance level or probability of type one error, is to increase the sample size. It is important to note that quality inspection is done on sample pieces and not on all the products. So, beta error is a function of the sample size. If the sample size is not appropriate, the defects in a product line could easily be missed out giving a wrong perception of the quality of the product. This will increase the type two error. To decrease this error, the quality assurance team has to increase the sample size. In hypothesis testing, alpha is called the significance level and one minus alpha is called the confidence level of the test. In the next screen, we will focus on the determinants of sample size for continuous data.

The sample size can be calculated by answering three simple questions. How much variation is present in the population? At what interval does the true population mean need to be estimated? And how much representation error is allowed in the sample? Continuous data is data which can be measured. The sample size for continuous data can be determined by the formula shown on the screen. We will learn about the standard sample size formula for continuous data in the next screen.

Representation error or alpha error is generally assumed to be 5% or 0.05. Hence, the expression one minus alpha divided by two amounts to 0.975 or 97.5%. Looking up the value of Z 97.5 from the Z table gives the value 1.96. The expression reduces to the one shown on screen. When alpha is 5%, Z is 1.96. To detect a change that is half the standard deviation, one needs to get at least 16 data points for the sample. Click the example tab to view an example of continuous data calculation using standard sample size formula. The population standard deviation for the time to resolve customer problems is 30 hours. What should be the size of a sample that can estimate the average problem resolution time within plus or minus 5 hours tolerance with 99% confidence? To know with 99% confidence that the time to resolve a customer problem ranges between 25 and 35 hours, the value of Z for 99.5 must be 2.575. A good result should fall outside the range of 0.5%, which is 1 in 200 trials. It is expected that 199 out of 200 trials will confirm a proper conclusion. The calculation gives a result of 238.70. One cannot have 0.70 of a sample, so one needs to round up to the nearest integer. If there are 239 samples, the significance level is greater than 0.01, which indicates the confidence is less than 99%. Using 239 reduces alpha and increases the confidence level. The rounded up value 239 means the expectations are being met for the confidence level of the test.

We will learn about the standard sample size formula for discrete data in this screen. Like continuous data, one can find out the sample size required while dealing with discrete population. If the average population proportion non-defective is p, then population standard deviation can be calculated by using the expression shown on the screen. The expression for sample size is present. It is important to note that in this expression, the interval or tolerance is in percentage. Click the example tab to view an example of discrete data calculation using standard sample size formula. The non-defective population proportion for pen manufacturing is 80%. What should be the sample size to draw a sample that can estimate the proportion of compliant pens within ± 5% with an alpha of 5%? Consider calculating the sample size for discrete data, for which the population proportion non defective is 80% and the tolerance limit is within plus or minus 5%. Substituting the values, it is found the sample size should be 246. In this example, to know if the population proportion for good pens is still within 75-85 % and to have 95 % confidence that the sample will allow a good conclusion, one needs to inspect 245.86 pens. 0.86 of a pen cannot be inspected, so the value is rounded up to maintain the confidence level. Inspecting 245 or fewer pens reduces the confidence level. This means the Z value would be lower than 1.96 and alpha would be greater than 0.05). Suppose one is willing to accept a greater range in the estimate, the proportion is within 20 percent of the past results and approximately within 1 standard deviation of the proportion. Delta changes to 0.20, and the number of needed samples is 15.4 ? 16 (Pronounce as: fifteen point four is approximately equal to sixteen).

This screen will focus on the hypothesis testing roadmap. Though the basic determinants of accepting or rejecting a hypothesis remain the same, various tests are used depending on the type of data. From the figure shown on the screen, you can conclude the type of test to be performed based on the kind of data and values available. For discrete data, if mean and standard deviation are both known, the Z-test is used, and if mean is known but standard deviation is unknown, the t-test is used. If the standard deviation is unknown and if the sample size is less than 30, it is preferable to use the t-test. If variance is known, one should go for chi squared test. If mean and standard deviation are known for a set of continuous data, it is recommended to go for the Z test. For mean comparison of two with standard deviation unknown, go for t-test and for mean comparison of many with standard deviation unknown, go for F-test. Also, if the variance is known for continuous data, go for f-test. The next few screens will discuss in detail the tests for mean, variance, and proportions. Let us understand hypothesis test for means theoretical through an example in the next screen.

The examples of hypothesis testing based on the types of data and values available are discussed here. The value of alpha can be assumed to be 5% or 0.05. Suppose you want to check for the average height of a population. North American males are selected as the population here. From the population, 117 men are gathered as the sample and the readings of their height are taken. The null hypothesis is that the average height of North American males is 165cm, and the alternate hypothesis is that the height is lesser or greater than 165cm. Consider the sample size n as 117 (for Z-test) and Sample size n as 25 (for t-test); Sample average or x bar is 164.5 cm Using the data given, let us calculate the z calc value and t calc value. The population height is 165 cm with a standard deviation of 5.2 cm and the average height of the sample group is 164.5 cm. The test for significant difference should be conducted. First let us compute z calc value using the formula given on the screen. Hence, the Z-calc is 1.04, which is less than 1.96 or t-critical. Therefore, the null hypothesis cannot be rejected. Since z0.05 = 1.96, the null hypothesis is not rejected at 5% level of significance. The statistical notation is shown on the screen. Thus a conclusion based on the sample collected is that the average height of North American males is 165 cm. If the population standard deviation is not known, a t-test is used. It is similar to the Z-test. Instead of using the population parameter or sigma, the sample statistic standard deviation or s is used. In this example, the s value is 5.0. Let us now compute t value using the formula given on the screen. The statistical notation to reject null hypothesis is shown on the screen. The t-critical value is 2.064 and we know the t-calc value is 0.5, which is less than 2.064. Therefore, the null hypothesis cannot be rejected at 5% level of significance. Thus a conclusion based on the sample collected is that the average height of North American males is 165 cm. The conclusion of not rejecting the null hypothesis is based on the assumption that the 25 males are randomly selected from all males in North America. Null and alternative hypotheses are same for both Z-test and t-test. In both the examples, the null hypothesis is not rejected. In the next screen, we will understand the hypothesis test for variance with an example.

32.0: In hypothesis test for variance, Chi square test is used. In the case of a Chi square test, the null and alternate hypotheses are defined and the values of Chi square critical and Chi square are calculated. To understand this concept with an example, click the button given on the screen. The null hypothesis is that the proportion of wins in Australia or abroad is independent of the country played against. The alternate hypothesis is that the proportion of wins in Australia or abroad is dependent on the country played against. Chi square critical is 6.251 and chi square calculated is 1.36 Since the calculated value is less than the critical value, the proportion of wins of the Australia hockey team is independent of the country played or place.

In this screen, we will discuss hypothesis test for proportions with an example. The hypothesis test on population proportion can be performed. To understand this with an example, click the button given on the screen. Let us perform hypothesis test on population proportion. The null hypothesis is that the proportion of smokers among males in a place named R is 0.10, represented as P zero. The alternative hypothesis is the proportion is different than 0.10. In notation, it is represented as null hypothesis is p equals p zero against alternative hypothesis is p different than p zero. A sample of 150 adult males are interviewed and it is found that 23 of them are smokers. Thus the sample proportion is 23 divided by 150, which is 0.153. Substituting this value in the expression of Z given on the screen gives the result of 1.80. You can reject the null hypothesis at level of significance alpha if z is greater than z alpha. For 5% level of confidence, the Z value should be 1.96. Since the calculated Z value is more than what is required for 5% level of confidence, the null hypothesis is rejected. Hence, it can be concluded that the proportion of smokers in R is greater than 0.10.

In this screen, we will focus on comparison of Means of two processes. Means of two processes are compared to understand whether the outcomes of the two processes are significantly different. This test is helpful in understanding whether a new process is better than an old process. This test can also determine whether the two samples belong to the same population or different populations. It is especially required for benchmarking to compare an existing process with another benchmarked process. Let us proceed to the next screen to learn about the paired-comparison hypothesis test for means.

The example of two mean t-test with unequal variances is discussed here. Null and alternate hypotheses are defined. The average heights of men in two different sets of people are compared to see if the means are significantly different. For this test, the sample sizes, means and variances are required to calculate the value of t. Two samples of sizes n1 of 125 and n2 of 110 are taken from the two populations. The mean value of sample size 1 is 167.3 and sample size 2 is 165.8. The standard deviation for sample sizes 1 and 2 are 4.2 and 5.0 respectively. Using the formula given on the screen, the t value is derived as 2.47. The null hypothesis is rejected if the calculated value of t is more than the required value of t. In other words, reject null hypothesis at level of significance ? if Computed t value is greater than t of DF,? divided by 2. With a t-test, we’re comparing two means and the population parameter (sigma) is unknown. Therefore, we’re pooling the sample standard deviations in order to calculate t. The variances are weighted by the number of data points in each sample group. Since t223 and 0.025 equals 1.96, the null hypothesis is rejected at 5% level of significance. The test used here is known as a Paired t-test, and is considered a very powerful test. In the next screen, we will look into the example of the paired-comparison hypothesis test for variance of f test.

It is important to understand the different types of tests, through an example. Susan is trying to compare the standard deviation of two companies. According to her, the earnings of Company A are more volatile than those of Company B. She has been obtaining earnings data for the past 31 years for Company A, and for the past 41 years for Company B. She finds that the sample standard deviation of Company A’s earnings is $4.40 and of Company B’s earnings is $3.90. Determine whether the earnings of Company A have a greater standard deviation than those of Company B at 5% level of significance. Click the button given on the screen to know the answer. Susan has the data of the earnings of the companies. Distributions rarely have the same spread or variance. When processes are improved, one of the strategies is to reduce the variation. It is important to be able to compare variances with each other. A null hypothesis would indicate no change has occurred. If it can be rejected and the variance is lower, one can claim success. The statistical notation for this example is given on the screen. Suppose one has to compare two sets of company data. Susan has looked at the earnings of two companies. She has been studying the effects of strategy, management styles, and leadership profiles on the earnings of these companies. There are significant differences in these KPIV’s; she wants to know if they have an effect on the variance in the earnings. She has sample data over several decades for each company. By the given data, it can be concluded that earnings of Company A have a greater standard deviation than those of Company B. In calculating the F-test statistic, and always put the greater variance in the numerator.

Let us look at the F-test example of hypothesis test for equality of variance in this screen. The degrees of freedom for company A and company B are 30 and 40 respectively. The critical value from F-table equals 1.74. The null hypothesis is rejected if the F-test statistic is greater than 1.74. The calculated value of F-test statistic is 1.273 and therefore at the 5% significance level, the null hypothesis cannot be rejected. The next screen will focus on hypothesis tests f-Test for Independent Groups.

A restaurant which wants to explore the recent overuse of avocados suspects there is a difference between two chefs and the number of avocados used to prepare the salads. The data shown in the table is the measure of avocados in ounces. The weight of avocado slices used in salads prepared by two different chefs is to determine if one chef is using more avocados than the other. Perhaps the restaurant’s expenditures on avocados is greater this month than the average of the past 12 months. This is assuming there is no change in avocado prices or the amount of avocados being used. Click the tab to learn to conduct an F-test in MS Excel. The F-Test is conducted in MS Excel through the following steps. Open MS Excel Click Data Click Data Analysis (Please follow the facilitator instruction on how to install Add-ins) Select F-Test Two Sample for Variances In Variable 1 Range, select the data set for Group A, and select data set for Group B in Variable 2 Range. Click Ok The screenshot of the F-Test window is also shown here.

In this screen, we will discuss the F-test assumptions. Before interpreting the f-test, the assumptions to be considered are: Null Hypothesis – There is no significant statistical difference between the variances of the two groups, thus concluding any variation could be because of chance. This is Common Cause of Variation. Alternate Hypothesis – There is a significant statistical difference between the variances of the two groups, thus concluding that variations could be because of assignable causes too. This is Special Cause of Variation. The following screen will focus on F-test interpretations.

The interpretations for the conducted f-test are: From the Excel result sheet, the p-value is 0.03. If p-value is low or below 0.05, the null must be rejected. Thus, null hypothesis with 97% confidence is rejected. Also, the fact that variation could only be due to Common Cause of Variation is rejected. It is inferred from the test that there could be Assignable Causes of Variation or Special Causes of Variation. Excel provides the descriptive statistics for each variable. It also gives the degrees of freedom for each. F is the calculated F statistic. F-critical is a reference number found in a statistics book table. P(F?f) (Pronounce as: p of f less than or equal to f) is the probability that F really is less than F-critical, or that the null hypothesis would be falsely rejected. Since the p-value is less than the alpha, the null hypothesis can be confidently rejected. Alongside conducting a hypothesis test, a meaningful conclusion from the test has been drawn. The following screen will focus on hypothesis tests t test for independent groups.

As discussed earlier, the table shows the measure of avocados in ounces and the significant difference in their means needs to be inspected. If a significant amount of difference is found, it can be concluded that there is a possibility of Special Cause of Variation. The next screen will demonstrate how to conduct the 2 sample t-test.

The 2-Sample Independent t-test inspects two groups of data for significant difference in their means. The idea is to conclude if there is a significant amount of difference. If there is a statistical evidence of variation, one can conclude a possibility of Special Cause of Variation. The steps for conducting a 2-Sample t-test are: Open MS Excel, Click Data and Click Data Analysis Select 2 Sample Independent t-test assuming unequal variances. In Variable 1 range, select the data set for Group A, and select the data set for Group B in Variable 2 range. Keep the “Hypothesized Mean Difference” as 0. Click Ok In the following screen, we will focus on 2-sample independent t-test assumptions.

The assumptions for a 2-Sample Independent t-test are: Null Hypothesis – There is no significant statistical difference between the means of the two groups, thus concluding any variation could be because of chance. This is Common Cause of Variation. Alternate Hypothesis – There is a significant statistical difference between the means of the two groups, thus concluding that variations could be because of assignable causes too. This is Special Cause of Variation. The null hypothesis states the mean of group A is equal to the mean of group B. The alternate hypothesis states the mean of group A is not equal to the mean of Group B. Note that alternate hypothesis tests two conditions, Mean of A < Mean of B and Mean of A > Mean of B. Thus a two-tailed probability needs to be used. Before we interpret the t-test results, let us compare the 2 tailed and 1 tailed probability in the next screen.

2-tailed probability and 1-tailed probability are used depending on the direction of the alternate hypothesis. If the alternate hypothesis tests more than one direction, either less or more, use a 2-tailed probability value from the test. Example: If Mean of A is not equal to Mean of B then it is 2-tailed probability. If the alternate hypothesis tests only one direction, use a 1-tailed probability value from the test. Example: If Mean of A is greater than Mean of B, then it is 1-tailed probability. In the next screen, let us look at the two-sample independent t test results and interpretations.

The Results are shown in the table on the screen. The inference is as the 2-tailed probability is being tested, the p-value of 2-tailed probability testing is 0.24, which is greater than 0.05. If p-value is greater than 0.05, the null hypothesis is not rejected. This means, one cannot reject the fact that there is no significant statistical difference between the two means. Similar to the F-test, Excel provides the descriptive statistics for each group or variable. The t Stat is shown. Excel also shows one-tailed or two tailed data. For the one-tailed test, the alpha is 0.05; the error is expected to be in one direction. For the two-tailed test, the error is alpha/2 or 0.025. In this example, t stat or t-calculated is less than either t-criticals; therefore the null hypothesis cannot be rejected. Thus it can be inferred that both the groups are statistically same. We will discuss the paired t-test in the next screen.

Paired t-test is another Hypothesis test from the family of t-tests. The following points will help in understanding the paired t-test in detail. The paired t-test is one of the most powerful tests from the t-test family. The Paired t-test is conducted before and after the process to be measured. For example, a group of students score X in CSSGB before taking the Training program. Post the training program, the scores are taken again. One needs to find out if there is a statistical difference between the two sets of scores. If there is a significant difference, the inference could be that the training was effective. It is important to note that the Paired t-test interpretation shows the effectiveness of the Improvement measures. This is the main reason why Paired t-tests are often used in the improve stage. We will proceed to the next screen that focuses on the analysis of variance or ANOVA, comparison for more than two means.

A t-test is used for 1-sample. 2-sample tests are used for comparing two means. To compare the means of more than two samples, use the ANOVA method. ANOVA stands for Analysis of Variance. ANOVA does not tell the better mean, it helps in understanding that all the sample means are not equal. The shortlisted samples based on ANOVA output can further be tested. One important aspect of ANOVA is it generalizes the t-test to include more than two samples. Performing multiple two-sample t-tests would increase the chance of committing a type one error. Hence, ANOVAs is useful in comparing two or more means. The next screen will help in understanding this concept through an example.

As an example, consider the takeaway food delivery time of three different outlets. Is there any evidence that the averages for the three outlets are not equal? In other words, can the delivery time be benchmarked on the outlet? The null hypothesis will assume that the three means are equal. If the null hypothesis is rejected, it would mean that there are at least two outlets that are different in their average delivery time. In minitab, one can perform ANOVA in one of the statistics packages. Ensure that the data of the table is stacked in two columns. In the main menu, go to stat, anova, and then one way. The left column of the table will have the outlets and the right column will have the time in minutes. This is similar to the table shown on the screen. In the One-way analysis of variance window, select the response as delivery time and factor as outlet, and click OK. The output of this process is shown here. Notice the P value which is much higher than 0.05. The steps to perform ANOVA in excel are as follows: After entering the data to a spreadsheet, select the ANOVA-single factor test from the Data Analysis “Toolpak”. Select the array for analysis, designate that the data is in columns and select an output range. Excel shows the descriptive statistics for each column in the top table. In the second table, the ANOVA analysis shows whether the variation is greater between the groups or within the groups. It shows the sum of squares or SS, degrees of freedom (df), mean of squares (MS or sum of squares divided by n-1 or variance), the F statistic (MS-between divided by MS-within), p-value, and F-critical (from a reference table). The F and p are calculated for the variation that occurs within each of the groups and between the groups. If the conditions in the groups are significant, it would be expected to see the “Between Groups” SS much higher and the MS slightly higher. Let us now interpret the minitab ANOVA results. Since the p-value is more than 0.05, the null hypothesis is accepted. This means there is no significant difference between means of delivery time for the three outlets. Based on the confidence intervals, it is found that the intervals overlap, which means there is little that separates the means of the three samples. In one-way ANOVA where there was only one factor to be benchmarked, that is, the outlet of delivery. If there are two such factors, you may use the two-way ANOVA.

In this screen, we will learn about chi-square distribution. The Chi square distribution is one of the most widely used probability distributions in inferential statistics. It is also known as hypothesis testing and the distribution is used in hypothesis tests. When used in hypothesis test, needs one sample for the test to be conducted. The Chi square distribution is also known as chi squared. It has k-1 (Pronounce as: k one) degrees of freedom and is the distribution of a sum of the squares of k independent standard normal random variables. Suppose, in a field for 9 players, Player 1 comes in and can choose amongst all 9 positions available. Player 2 can choose only amongst 8, and so on. After all the 8 players have chosen their positions, the last player gets to choose the last position left. The 8 players are free to choose in a playing field of 9. 8 is the Degree of Freedom for this example. Conventionally, Degree of freedom is n-1, where n is the sample size. For example, if w, x, y, and z are four random variables with standard normal distributions, the random variable f which is the sum of w Square, x square, y square, and z square, has a chi square distribution. The degrees of freedom of the distribution or the df equals the number of normally distributed variables used. In this case, df equals four. The formula for chi squared distribution is shown on the screen. It is important to note that F of o stands for an observed frequency and f of e stands for an expected frequency. The next screen will explain chi square test through an example.

Suppose the Australian hockey team wishes to analyze its wins at home and abroad against four different countries.. The data has two classifications and the table is also known as a two by four contingency table with two rows and four columns. The expected frequencies can be calculated assuming there is a relationship. Thus, expected frequency for each of the observed frequency is equal to product of row total and column total divided by overall total. One has to find out how to calculate the expected frequency. If the observed frequency is three wins against South Africa in Australia, then it would convert to total wins at home, which is twenty one divided by the total number of wins or thirty one, and the result is multiplied by five. The result is three point three nine. Similarly, the expected population parameters for all cases are found. In this step, all the information of the previous screen is combined and the table is populated. The estimated population parameters are calculated and added. The formula estimates the observed frequency to calculate the final chi square index which in this case is one point three six. It is important to note that there is a different chi square distribution for each of the different numbers of degrees of freedom. For the chi square distribution, the degrees of freedom are calculated as per the number of rows and columns in the contingency table. The equation for degrees of freedom should be noticed. The number of degrees of freedom is equal to three. Assuming an alpha of ten percent, the chi square distribution in the chi square table is noticed and a critical chi square index of six point two five one is arrived at. Chi square calculated value is one point three six. Both the values of the chi square index should be plotted. The critical chi square distribution divides the whole region into acceptance and rejection, while the calculated chi square distribution is based on data and conveys whether the data falls into an acceptance or rejection region. Therefore, as the calculated value is less than the critical value and falls in the acceptance region, the proportion of wins of the Aussie team at home or abroad has nothing to do with the opponent being played against.

Following is the quiz section to check your understanding of the lesson.

Let us summarize what we have learned in this lesson. Multi-vari studies are used to analyze variation in a process. Correlation means association between variables and Simple Linear Regression and Multiple Regression are the two main techniques. Hypothesis testing is conducted on different sets of data. The analysis of variance is used to compare means of more than two sample sets. A t-test is used for 1-sample & 2-sample tests are used for comparing two means. ANOVA is used to compare the means of more than two samples and the Chi-square distribution is used for probability distributions in inferential statistics.

With this, we have come to the end of this lesson. We will discuss Improve phase in the next lesson.

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