Welcome to lecture 40. This is the last part of hypothesis testing. In this lecture, we will discuss about two important tests of variants such as ANOVA and chi square test ANOVA stands for analysis of variance, this is based on F distribution of probability is used to statistically draw conclusions where output Y is continuous and inputs x are discrete categories such as operators, machines, etc For example, manufacturer of a steel component for aircraft has to maintain its length as poor one 3.8 mm there had been variations and outputs. He would like to see the significant source of variation among his three operators named A, B and C. As you see, in the table, output data is of continuous type and input data is of discrete that is three operators There are two types of The wall basically one way ANOVA conducted with one factor for example, here we are considering only one factor that is operator and two way ANOVA conducted with two factors for example, operators and machines etc.
When we conduct an ANOVA we are basically analyzing the parameters of an alpha, such as, source of radiation, degree of freedom, which is nothing but the sample size minus one sum of squares Mean Square, F statistic and B value. Let me show you these parameters in Minitab. Download the corresponding Minitab file from this lecture and double click to open. Click Start a NOAA one way. One way analysis of variance dialog box will open type leaks in the response box and operator and the factor text boxes ensure confidence level is by default 95.0 and click OK. The ANOVA results is as shown on the screen.
We can see the sources of radiation sum of square SS mean square and F statistic etc at the top we can also see the p value there there is also significant difference in the performance of operators operator B's performance differ significantly from others. How do we interpret ANOVA for this purpose let us assume a null hypothesis for this ANOVA as there is no difference among the performance of operators The P value we got here it was 0.03. We should remember that when p is low, null must go. It means there is enough statistical evidence to reject null hypothesis. In other words, performance of one operator differs significantly from the others. Let us discuss one more test of variance the chi square test.
It can be used as a statistical method to infer whether two categories are dependent or not By test of independence, let us analyze this with an example. Similar to the previous one, we discussed for ANOVA. To check whether quality of product depends upon operators the difference between ANOVA and chi square is no one is used when y is continuous and x is discrete, whereas, chi square is used when both y's and x are discrete. Let us conduct the test with Minitab. Download the relevant file and double click to open click Start. tables, chi square test, two way table in worksheet, double click on good quality and bad quality respectively and click OK.
The chi square results are as shown on the screen. Let us note the p value how do we analyze chi square test? Let us again assume some null hypothesis such as quality of output does not depend on operator performance. What is the analysis result? We need to check the p value remember the formulae for rejecting the null hypothesis when p is low, null must go. What is P value here?
It's 0.637 this is more than our significance level of 0.05. That means, P is not low here. Hence, we cannot reject null hypothesis. In this case, the performance of operator is not affecting the quality of products. I am aware that you still have a lot of doubts at the back of your mind. Don't worry.
In fact, as A green belt professional, you only need to have a basic understanding about the hypothesis testing is covered in depth for black belt training course. That's all for this lecture. With this, we have completed section for the Analyze phase of Six Sigma In this section we have covered the subsections of B Okay, such as a exploratory data analysis be hypothesis testing. Now, let us proceed to the last section of this course with the next lecture. Thank you