ASQ Six Sigma Black Belt Practice Exam (50% off online class)

hello my name is John Lee and I'm the president of Alpha training and consulting and I have a true passion for preparing students for asq certification exams I love preparing students for asq certification exams but today we're going to go over to practice exam questions for the Six Sigma Black Belt exam and we're taking those questions right out of the certified Six Sigma Black Belt primer which comes from Kali Council of Indiana and they gave us permission to use these questions so we greatly appreciate it thank you Quality Council of Indiana we so much like their primers that we actually made them part of our class so what you do is you take all our lectures all our exams and then when you're finished with that you review the primer and we go over all the end of chapter questions with you in a video like this one now if you're interested in that class at all that prep class just go to WWE SQ CSS BB for certified six sigma black belt again that's a excuse CSS BB comm set website dedicated to nothing but the asq certification called six sigma black belt I think I have over ten videos embedded in there and answers any question you may have concerning the black belt certification alright let's get started on those questions alright here we are back to question six and I'll read that again increasing performance in a Lean Six Sigma corporation meaning one and a half Sigma shift from three sigma to for Sigma would reduce defects per million by a factor of one alright in the table you'll need in order to solve this problem as in the appendix and then my revision it's page three and it's Six Sigma failure rates and with a one and a half Sigma process shift or with no process now you can do it on the Z table but let me read the question again here increasing performances in a Lean Six Sigma corporation as soon as it says Lean Six Sigma corporation you know they're going to take into account the one and a half Sigma shift and that's why you'll use this table right here you'll be using the one with the shift not with no shift now remember you could also do this by I going to your z-table and there at 3 Sigma 4 Sigma take 3 Sigma subtract 1 and 1/2 from it and then which would give us one and a half Sigma's that you'd look up in the Z table and 4 Sigma and look up 2 and 1/2 a Z of 2 and 1/2 would give you the same answer as this table here but this table makes a little quicker and of course we like to go fast so let's look at 3 Sigma here and it tells us 3 Sigma with the 1 and 1/2 Sigma shift is 60 6810 point six three parts per million so I'm going to write that down 66 thousand eight hundred and ten point six three parts per million okay that's four three Sigma now we'll go to four sigma column here there it is six thousand two hundred nine point seven six thousand two hundred nine point seven parts per million and so now let's go to the white board and calculate this out and get the correct answer alright so here we are at the white board to get the factor we take the sixty six thousand eight hundred ten point six three parts per million and remember that was for three Sigma divided by six thousand two hundred nine point seven parts per million remember this was for the four Sigma process so let's divide that out and see what our factor of improvement is here we go sixty-six thousand eight hundred and ten point six three divided by six thousand two hundred and nine point seven equals and so we're a factor of ten point seven six now he asked you just kind of round things off a lot of times so this may just the correct answer maybe ten but let's go back to our question sheet and see if we can find the correct answer alright here we are back at question six and I'll read that again increasing formitz in a Lean Six Sigma corporation meaning 1/2 Sigma shift from 3 Sigma 2 4 Sigma would reduce defects per million by a factor of what there it is it was ten point seven something there was an 11 I'd pick it but there's not ten is the closest thing to it so I'm going to go with C and that's question six is C so that is correct all right here we are question 16 from an upper management perspective what has been the principal motivating factor in embracing Lean Six Sigma well upper management what are they what is the biggest reason that upper management loses their job it's because they do not meet their financial goals so they have to make that happen they have to create profit so it's all going to be about I see this right here bottom line results whenever you see upper management you should think bottom line results they have to make those numbers what they need to be to keep everyone happy or they won't have a job so a bottom line results I'm sure is the correct answer market share growth now if you grow market share but you don't increase profits you're still in trouble defect reductions I mean that's all good but this is a subcategory to bottom line results all these could be subcategories to a so a is the most general because all these could be correct customer focus we know ASG likes customer focus but if it doesn't create bottom line results upper management's in trouble so just remember if it says upper management you should think they have to make money here we are at question seven a Six Sigma project requires twenty three thousand dollars of initial investment and the training costs of six thousand unless we have thirty thousand invested thus far spread over a six month period okay so it's gonna take six months to finish this project sounds like the project is expected to save the company three thousand dollars per month starting in the third month so we have three months where we can't start collecting savings okay so already it's taken three months starting in the third month ignoring interest taxes what is the payback period okay in other words how long is it going to take me to get my $29,000 back if I'm paying $3,000 a month so let's go to the whiteboard and work this one out alright here we are at the whiteboard I wrote down the cost that we're going to invest we're gonna invest twenty three thousand upfront six thousand dollars for training so we add those together and get twenty nine thousand dollars then we take twenty nine thousand dollars and how long is it going to take us if we're charging if we're getting back three thousand dollars per month so we pull out our trusty calculator here and go twenty nine thousand dollars divided by three thousand dollars equals nine point six six seven months okay and it took how long before we could implement this because payback starts the minute you start investing money so day one and we had to go three months so you have to add three months to nine months and they're gonna probably round it two whole numbers they did if I remember correctly so this is either going to be nine plus three which will be twelve or nine plus or ten if you round it to ten and plus the three months so it's either gonna be twelve or thirteen months let's go back to the question and see if that answer exists all right here we are and let's read that again a Six Sigma project requires twenty three thousand dollars of initial investment and training costs of six thousand spread over a six-month period and then it says the project is expected to save three thousand per month starting in the third month so we calculated nine or ten months so a lot of people are gonna pick nine or ten months depending on how you round it but it's incorrect why because the payback period starts the moment you start investing money into this and they start investing money three months before so it's either going to be twelve or thirteen when you add the three months in depending on which way around thirteen months is not an option but twelve months is so 3.7 is D all right here we are question 14 the term metrics most frequently refers to what okay a unit of measurement could be it is a measurement so I'm not going to throw a out be the metric system no net present value maybe if you meet your metrics you'll have a good net present value but it doesn't align to that statement so B and C are out there the best one so far is a an evaluation method yes you give people metrics and see if they can reach those metrics if they do they're doing a good job if they exceed them they're doing a great job if they don't meet them they're not doing such a good job so is an evaluation method so it's either a or D but I feel D best aligns with the intent of this question so 314 is D okay team briefing presentations to senior management should include which of the following considerations okay what do we know about senior management they don't have a lot of time so when we do work with senior management we're supposed to do it very efficiently we talked about that in the test taking skills modules so again team briefing presentations to senior management should include which of the following considerations since time of senior management is Val you address all potential details the first part is okay since the time of senior management is valuable but it should say you know just address the main points it says all details that's not good so a is out of their be every member of the team should have a speaking briefing role yeah if that works but I'm not going to do it just so everyone can get a pat on the back I'm in front of senior management the most important thing is that I communicate efficiently effectively and get out of there because they don't have a lot of time okay okay identify the problem of the proposed action desired identify the problem in the proposed I like that one I really like that one you're just getting right down to the point and getting out of their handout should require author's explanations not really I think senior management just come in tell me what I need to know and leave because I have a lot of other things to do and so I think it's see on this 14.5 is see okay it is excessive conflict within a work team okay I know this was in our example questions that we used in test-taking strategies remember asq does not like conflict period has a negative effect on team members and should be avoided I already know aids the correct one has a positive effect no most often results in win-win Oh engages shine member participation now it usually does just the opposite is that so this one no doubt is a 513 which of the following techniques has proven useful in translating customer needs into product design features well what's the object objective of qfd quality function deployment is to convert customer wants needs and desires into product and service attributes well in so many words that is the definition of quality function deployment so I already know the correct answer is going to be the quality function deployment you should remember that they're going to test you on it every time the objective of quality function deployment is to convert customer wants needs and desires into product and service attributes so remember that one and because you're bound to see it on the certification exam so that wants D what is the danger in bunching ideas immediately in two closely related categories when using an affinity diagram okay remember affinity diagram we don't have a lot of information yet so it's going to be fairly subjective but let's see what we have here the wrong category might be selected I like this might be as a non absolute so that has something going for it and could you choose the wrong can I guess I don't know but it's easy to change it if you do so but let's see what else I do like the non absolute thought patterns could be biased yes I like there it is again so we have two non absolutes so which one do I like better the wrong category might be selected or thought third patterns could be biased now they're both possibilities no doubt but if the wrong category selected so what you just changed the heading on it it's not a big deal so B is a bigger deal thought patterns could be biased remember I said this was subjective which means bias plays a part in it and could and probably does to some extent so B went out on a arrows might be aimed in the wrong direction okay whatever I know that you need arrows but if you did if they're aimed in the wrong direction it's like a so what changes the direction it's not a big deal so B definitely wins between a and C the problem resolution can be overlooked are we getting a little ahead of ourselves with the affinity diagram when we talked about when we talked about prom resolutions and we're not even close to resolving the problem so this being subjective and biased in there I would go with B 514 is a B yes it is when performing calculations on sample data a continuous relative frequency graph called a histogram results I don't know it depends on what kind of sample data so now I don't think a is correct besides when I do perform calculations calculations doesn't necessarily mean graphical I guess it could but there's just too many holes in a a is out of there rounding the data has no effect on the mean and standard deviation yeah you can round things incorrectly for one thing so that could be a problem I would throw out a though I keep B before I throw out a or I mean I keep B so I would throw out a so that C coding the data has no effect on the mean and standard deviation now it has an effect I think I'm gonna go to the white board and go over this one a little more because I don't think it's a or B it really and so I think it's going to be C or D coding and rounding effect both the mean and the standard deviation okay well let's go to the white board and discuss question two alright here we are that white board notice I have three numbers here 301 302 303 and when they have a test question this is usually what I do when it's concerning coded data and I already know from past experience one two and three gives me a standard deviation of one we performed that in some of the modules we'll do it again here but here's my data I coat it to just be one two and three and my average becomes two and my standard deviation is 1 so it so now let's do the non coded data my average is 302 so if I plug one two and three I get an average of two if I plug in 301 302 303 the non coded data I get 302 so did it change the average absolutely okay then I take the standard deviation of the non-coated I get one I take the standard deviation of the coded I get one so in this case standard deviation didn't change but the average did so let's go back and look at the question again see if we can find the right answer all right here we are back at question two we know it's either C or D coded data has no effect on the mean and standard deviation well in our example it did change the mean so I threw out C coded coding and rounding oh I didn't see this first coding and rounding effect both the mean and the standard deviation okay we already discuss rounding can most certainly have an impact and so even on the standard deviation coding we know has a difference on the mean but didn't have an impact on the standard deviation but rounding will impact both of them so that's going to be the correct answer must be d 6.2 is d yes it is the calibration of measuring instruments is necessary to maintain accuracy yes that's true to reduce biases to increase accuracy remember that they test on that a lot I'll say it again decreasing bias increases accuracy so how does calibration affect precision it does not then consider to be independent of one another the precision increases over the working range of the instrument no I mean it may be true on some instrument out there but he didn't tell me any information about the instrument a cannot be the correct answer the precision decreases over the working range now calibration has a minimal impact on precision that's correct si is correct the precision will vary over the working range of the instrument now it has to be see six point seven is seen yes it is it's going to seven point seven the reported CPK for a process with an average of 28 and a spread of 10 units okay what does spread mean okay here's another one spread means is another words way of saying we plus minus three sigma okay so you have to remember the spread we also had another term here a minute ago the natural limits or something like that means the same as spread but you're more likely to hear the word spread on the certification exam means plus minus three Sigma's and upper in the lower specification limits of 35 and 15 respectively would be so they're looking for the CPK well you can calculate CP upper and CP lower but this is a time test so you don't want to calculate both if you don't have to how do you know which one's going to be the CP CP upper CP lower is going to be the CPK well it's wherever the average is closer to then there's the upper spec limit how far away are they from 28 to 35 35 minus 28 is 7 okay so we're seven units away from the upper spec limit the averages 28 and 15 28 minus 15 is 1525 it's gonna be 13 so there's 13 units from 22 the average to the lower spec limit there's only 7 to the upper spec limit so seek the upper is going to be CPK now some of you will get that some of you will not it's okay we're going to go to the whiteboard and draw some pictures and work this out all right here we are at the whiteboard and I've drawn out the lower spec limit in the upper spec limit that was given in the problem and I just found that middle right in between there was 25 now the average of the distribution is 28 I wrote 25 in here so I can get an idea of which side 28 lies on and of course 20 what you have to ask yourself this 28 between 25 and 15 or 28 between 25 and 35 of course that's between 25 and 35 so I'm going to draw it out here this will be the average of my distribution just 28 and then I always draw the distribution just so I don't forget what I'm doing and which one is the average closer to the upper spec limit or the lower spec limit it's closer to the upper spec limit so if you wanted to we could get this answer remember CPK is the lower of CP upper and CP lower I already know the lower the CPK is gonna be CP upper but you may not know that so we'll do both just in case and so here is the formula CP upper equals upper spec limit minus the average divided by three standard deviations now the question said the spread of the distribution is what was it ten units okay what does the spread mean it means plus minus three so the width of this whole distribution here plus minus three is ten well three Sigma is half of that so this is going to not be ten three Sigma's 5 because plus minus 3 Sigma is 10 so just 3 Sigma be half of that will make 3 Sigma equal to 5 what am I worried about here more than anything else you have to understand that the spread means plus minus 3 Sigma and the calculation for CP uh prints me lower three Sigma's so it's only half of the ten is going to give us five so hopefully you're okay with that now let's put in the upper spec limit is 35 minus the average which is 28 and that is going to equal let's get out our calculator there we go 35-28 equals divided by five equals one point four so the CP upper is going to be one point four there we have it now I already know that CPK because this is a shorter distance than over here but let's say you don't feel comfortable with that so you just want to do the calculation so that's okay doesn't take that long CP lower equals average minus the lower spec limit divided by 3 Sigma now we remember we know 3 Sigma is equal to 5 because the spread will stand half of thats 5 and then the average is 28 put that in there 28 minus 15 and let's get our calculator out and we'll calculate CP lower 28 minus 15 equals divided by five equals two point six so CP lower equals 2.6 now remember what CPK is that's what they asked us to calculate CPK is the lesser of CP upper and CP lower so the lesser is 1.4 so that will be the correct answer to this question let's go back to the Blues page and make sure that answer is available all right here we are back at the blue page and notice we do have the answer B 1.4 so that is the correct answer for question 7 question 22 and it says process capability analysis is often defined as so it's asking for a definition of process capability what did I tell you in an earlier module when they defined process capability process capability must have one associated with it spec limits it always will it has to a member the formulas upper spec limit minus average divided by 3 Sigma so it has to have spec limits associated with it and it's also a probability so probability spec limits are the two key words when you define process capability well let's look through here I don't see any spec limits there and a I don't see any spec limits and B specification limits and C D specifications there it is so I know already it's either C or D so I'm not gonna read the other ones the variability allowed by the specification limits No so what capability analysis does gave Billy analysis calculates the probability that not what's allowed you could but it's probably not the correct answer let's see if these better the determination that the process can meet product specifications yes determination is a probability so there you have it probability specifications this one is no doubt d hopefully that helps you you're going to get one of these questions on the certification exam the probability of a train arriving on time and leaving on time is 0.8 the probability of the same train arriving on time is 0.8 for the probability of this train leaving on time is 0.8 6 given the train arrived on time given that the train arrived on time what is the probability that will leave on time all right let's go to the whiteboard and work on this one all right welcome to the whiteboard let's go ahead and solve this problem I've written down all the probabilities the probability of a train arriving and leaving on times point 8 probability of it arriving on times point 8 for probability of it leaving on time is 0.8 6 well this is called conditional probabilities and this is the formula we'll use probability of a and that vertical line means given probability of a probability of a taking place given B has already taken place you need to get good at saying that the probability of a given B has already taken place equals the probability of a and B taking place divided by the probability of B well let's get started then the probability of a and B probability of arriving and leaving so two things happen they only gave us one probability what two things happen and that gives us point eight so we know that's going to go in the numerator 0.8 so that's easy enough now the question is we have two more probabilities left over which one goes in the denominator well let's read this again and I'm asking you it's probability of B the earlier event or the later event look at it again probability of a what's the probability of a taking place given B has already taken place B is already taking place so B is the earlier probability which takes place earlier the probability of arriving or the probability of leaving it's the probability of arriving is the earlier events 0.84 will go in the denominator and that's how you solve this problem the difficult part is you don't know what goes in the denominator well you do now the denominator is the earlier of the two events so now we can get our calculator out and clear it of course and go 0.8 divided by 0.8 for and it gives us a ninety five point two percent chance of what there's a 95 there's a point nine two five two percent chance that the train will leave on time given it arrived on time and that is the correct answer let's go back to the blue sheets all right here we are back at question 37 remember 36 we calculated to be point nine five and so D is the correct answer they're three a random sample size in is to be taken from a large population having a standard deviation of one the sample size is to be determined so that there will be a 5% risk probability of exceeding 0.01 in other words five percent alpha risk a point one tolerance error and using the sample mean to estimate mu which of the following values is nearest the required sample size okay this is a very common test question and it has to do at that formula for sample sizes so we're gonna go to the white board and work this one out all right here we are at the white board here's the formula you need to know if this is not in your formula package please write it in there and it says in or sample size equals Z squared times Sigma squared divided by what I call Delta squared I think in the primer they call it e or error but this is the practical significance and that is point one and let me read that again because a random sample size of n is to be taken from a large population having a standard deviation of one right there the sample size to be determined so that there will be a five percent alpha risk probability of exceeding the point one tolerance error okay so basically they're saying this is the practical significance and the point one here it doesn't say it's greater than or less than it just says it wants to know if it's the difference of 0.1 which means plus/minus that means it's two-tailed so when you look up the 5% alpha risk for Z you have to look up the two and a half percent don't forget that it's probably one of the big reasons people miss this question and on every asq exam I've ever taken and I've taken a lot of them it's always with this formula they always use the two tail so 1.96 all right and now once you understand that it's just a matter of putting it in the formula which I've done here and plugging away in your calculator so let's do that and it's going to be 1.96 squared times 1 squared really article you didn't really need to do that give me equals then divided by 0.1 squared don't forget to square everything that'll get you in trouble quickly and that gives me a sample size of 384 point 1 6 now on these types of calculations they'll usually round down 384 but let's go to the blue pages and see what the options are all right here we are back at the blue pages and notice for the last question the closest one there is a 385 that is the correct answer all right let's go to question 11 determine whether the following two types of Rockets have significantly different variances at the 5% level assume that the larger variance goes in the numerator okay this is good to know because it tells us that it's a right tailed test on a right tailed F test the larger number goes in the numerator and if it's a left tailed test they do the inverse okay so that's one thing you have to understand and most of them on the certification exam are right tailed tests to be honest with you I've never had a left tailed test on a certification exam on the F statistic doesn't mean it can't happen but that's the case with this one another thing you need to know here and let me ask you so you can look is this a one tail or a two-tailed test determine whether the following two types of rockets have significantly different variances different different is a keyword for a two-tailed test if it would have said greater than or less than it would have been a one tailed test but this is a two-tailed test so you have to take that five percent alpha risk level you have to divide it by two and so you have to look up the critical statistic in the two and a half percent alpha risk table so that is probably the number one reason people miss this problem all right and it gives us rocket a rocket be rocket a has 61 readings and there's it's is this a standard deviation of variance remember what variance is Sigma squared equals variance okay did they give us variance or standard deviations it's very important because F equals center deviation 1 squared divided by standard deviation 2 squared or variance 1 over variance 2 because standard deviation squared is variance well they gave us variance this 1347 miles squared how do I know it's variance because the units are squared which means the number was also so these are variances that's another reason students missed this question is their square something that's already been squared that'll get you in trouble okay what we need to do then is go to the whiteboard and work this out so let's go to the whiteboard all right here we are back at the whiteboard and notice I've written the formulas out standard deviation B squared equals variance B remember they gave us variance so we don't have to square it again so this is variance of B is 2237 since it's the larger number it will go in the numerator on a right-tailed F test if it was a left-tail that would be the inverse of this the larger one would go in the denominator but again the most common one is a right tailed test so there it is variance be very it's a there's the sample sizes we'll need that to look up our critical statistic because remember degrees of freedom for this is n minus 1 okay let's plug that into the formula so f is going to equal the larger variance 2237 divided by the smaller one 1347 so let's pull out the calculator here and let's calculate this it's going to be two thousand two hundred and thirty seven divided by 1347 equals and that is one point six six is the correct answer there one point six six now we just need to look up the critical statistic now remember it said it want to know if there's a difference so it's a two-tailed test as a result I have to go to the two and a half percent alpha risk table remember that's unique to the F table every table has its own alpha risk so we have to go the two-and-a-half percent alpha risk table and let's look at the numbers again the enumerator has thirty-one and the denominator has 16 remember degrees of freedom for this is n minus 1 so we had 31 in the numerator so 31 minus 1 is 30 and in the denominator we had 61 so 61 minus 1 is 60 so where this row intersects this column that's our critical statistic so F critical is 1 point 8 2 all right so here we are back at the whiteboard notice our F critical is one point eight two and our calculator is one point six six well since our critical is greater than our calculated we do not have enough evidence to reject the null hypothesis in other words there is not they are not statistically significantly different at the 5 percent alpha risk level let's go back to the blue pages all right here we are back at question eleven on the blue pages remember we did not find a statistically significant difference significant difference no no significant difference because if calculated is less than the F tabulated that is true it must be B let's read the other ones significant difference because F calculated is greater than F table no and so it's going to be B is the correct answer for 811 all right welcome to question 24 here if a sample size of 16 yields an average of 12 well 16 isn't 30 so we're going to have to use the T statistic here for small so if a sample size of 16 yields an average of 12 and a standard deviation of 3 estimate the 95% confidence interval for the population well we're all whenever we do a confidence interval whether we use T or Z it's always going to be about predicting the population assume a normal distribution okay let's go to the whiteboard and work on this problem all right here we are at the whiteboard and I'm just usually on these you don't calculate the upper confidence interval and the lower because usually you only need to do one to get the correct answer and since it's time test that's what you want to do and so we're just going this first start off with the upper confidence interval if we need the lower we'll calculate it in a little bit okay so the formula is x-bar plus T why did I use T instead of Z because the sample size was 16 less than 30 you use T so there it is times the standard deviation over the square root of n remember when they asked for confidence intervals you're taking a confidence interval of the average if it's a single event they have another name for it it's called estimate interval or something like that okay so I wrote down what we know we know X bar to be 12 we know standard deviation to be 3 we know the sample size to be 16 and we know that degrees of freedom is 16 minus 1 n minus 1 which gives us 15 so let's go to the T table and look up the T statistic now it will be a two-tailed and let's look at this real quick because I can't remember yeah it's too tail and it's 5 percent so we're going to look up a two and a half percent alpha risk for any confidence interval that's always two-tailed so let's go to our t-table and look this up so here we are at the tea table there's our degrees of freedom of 15 and we want a two and a half percent and that's two point one three one let's go back to the whiteboard all right here we are back at the whiteboard and so let's get started we have all our information we just need to plug the numbers in now so let's get to work on that let's replace the X bar with twelve and our T we know to be two point one three one and then our standard deviation is three and we're going to divide that by the square root of n which is the square root of 16 all right now let's calculate the tee time standard deviation first calculator out here and I'm going to go 3/16 square root equals 0.75 0.75 times T two point one three one equals and let me write that down so it makes a little easier for us okay and that was my answer from doing that math now let's add it to the average and that will give us our upper confidence interval which is thirteen point five nine eight so I'll write that down now that's our upper confidence interval if we need the lower one we'll just have to subtract from that but like I say honest time to exam usually I only give one because that's usually all you need let's go back to the blue pages and see if we can get this correct all right here we are back at the blue pages and notice we had thirteen point five nine eight and the closest one to that is thirteen point six so there's no other repeat if there was two thirteen point six is we'd have to go back and calculate the lower one but no need here we know the correct answer is a let's move on to the next question and it's question 25 in order to test whether the average output of one machine is the same or greater than another machine so this is a one tailed test a sample of ten pieces was taken from each the calculated T value turned out to be one point seven six seven okay so they already calculated it for us using a one tailed test with a significance level of 5% one concludes that okay this seems like a pretty easy one so let's go to our T table and figure out what our T critical is alright so here we are at the whiteboard and in one equal ten in two equalled ten they had ten of each in this study and for this type of problem we go in 1 plus n 2 minus two is the one sometimes this is an estimate this is actually for pooled standard deviation but we use it throughout the asq exam for these type of test questions because the other degrees of freedom was that skyscraper formula we don't want to use on a timed test nor does asq expect you to use that one on the test they expect you to use this so 10 plus 10 minus 2 is 18 so we have 18 degrees of freedom and let me see what else we need to know here because I forgot a one tailed test it says so we're gonna look up five percent alpha risk at 18 degrees of freedom so let's go to our T table so here we are our T table and we have 18 degrees of freedom and this was a one tailed test so T critical is going to be one point seven three four so you'll want to write that down I'm going to write it down so I don't forget it one point seven three four is the critical statistic now let's go back and compare that to the calculate alright here we are back at the blue page and there it is one point six seven and our critical was one point seven three four so our calculated is greater than our critical so we have enough evidence to reject the null hypothesis so the obtained T ratio does not fall within the critical region no it does fall in the critical region critical region is the region where you reject it if the value is in the shaded region you know reject it the critical value is what borders that so the border would be one point six seven there the border would be one point seven thirty four and our calculated is greater than that so it's in the critical region so it does not fall in no it does fall within the critical region there was no significant difference between the means no there most certainly is the null hypothesis was rejected that is the correct answer and C is the correct answer for question twenty-five 837 the current process produces 50 units per shift a new process yielded 52 units per shift for 16 straight shifts that's pretty good but that's not much of a difference 52 52 with a standard deviation of 4 units per shifts what is the level of confidence that the process has changed okay so we need to do a inference study here and what kind of in fear'd study is it going to be what is the level of confidence that the process is changed okay let's go to the whiteboard and work this one out all right here we aren't the whiteboard and this is a TN feared study we have two averages and a standard deviation of sample size of 16 shifts if you recall so this is the formula we will use and on average we had 52 at the new process 50 with the old process the standard deviation was 4 and it was for 16 straight shifts so 16 is the sample size well if we go 52 minus 50 that's 2 4 divided by the square root of 16 square root of 16 is 4 so 4 over 4 is 1 so we get down here and I put X bar there that should be T so T is 2 over 1 equals 2 now that's not the question it asks what was our confidence and it wanted to know the difference well difference means a two-tailed test so we have to keep that into a camp also so as I said this is a little bit of a different problem in that we have to be some term in the confidence here and we remember we had 16 in the degrees of freedom for that formula is n minus 1 so 16 minus 1 is 15 so this is the row we'll be working with and we're looking for something between two so we start out here to point 0.25 eight point six nine one point eight six six two is right between these two numbers which just happens to be five percent alpha risk and two-and-a-half percent alpha risk however remember this was to tail so when you go out of the table this means five percent and 10 percent so it's between five and ten percent so remember that as we go back to the blue pages it's between five and ten percent alpha risk all right here we are back to the blue pages and notice we have they didn't do it in alpha they did it in confidence because it said level of confidence changed so remember it was five to ten percent well I would made 95 10 would be 90 so the correct answer is B now if you would have solved this as a one-tailed problem you would have ended up with that as the correct answer how did I know it was a two-tailed problem because what is the level of confidence that the process has changed is there a difference doesn't say greater than or less than that's how I knew it was two-tailed and then I once I looked it up in the tea table that's a one tail so I have to multiply that by two when I bring it out of the table and that gives me my 90 and 95 so that's kind of a tricky one I mean imagine many of you would miss that one so be careful look for those two tails and one tailed test you cut it in half the alpha risk when it's two tails going into the table when you're bringing it out of the table you have to multiply it by two all right 46 it is desirable to reduce the variation in a process okay very good the current variance is known to be six okay notice they just gave you a standard deviation they didn't give you the number of sample size they did they're basically giving you the population variance notice they also said variation not standard deviation so those are things you have to be careful of on these types of test questions the current variance is known to be 6 the new process yielded a standard deviation of 2 well that's from 6 to 2 sounds pretty good for 25 trials what is the chi-squared calculated okay this isn't too bad let's go to the whiteboard and work this out but first I want you to know they asked for the chi-squared so they told you this is chi-squared how do we know it's chrisquared well you have two statistics that you can use to determine a difference in standard deviation to competing statistics F and chi-squared well F has to have two sample sizes because it has a degrees of freedom across the first row on degrees of freedom you need two degrees of freedom to use the table if that you only have one degree of freedom you can only use chi-squared well they gave us the current variance is known to be 6 okay they gave you didn't how many trials so you they did on this other one so you only have one set one pot one degree of freedom to work with well you have to use the chi-square table in that case that's how I knew this was kind when they gave me a variance they may give you a standard deviation but they gave me a variance without any how many numbers they used to calculate that that forces it to chi-squared let's go to the whiteboard and work this one all right welcome to the whiteboard where we're working on question 46 this kind of a tricky question to be honest with you because the test question I don't know if I read it right the first time but I read it again and realized that they gave us one standard deviation from the sample they did on the new process and gave us a standard deviation on the population value it was a variance so with that being said here's the formula chi squared equals n minus 1 times sample standard deviation squared that's the new standard deviation that they're comparing the old one to and divided by Sigma squared now they already gave us variance for the denominator it's very tricky but they gave a standard deviation for the numerator so I wrote all that down here and then I put it in the equation sample size is 25 minus 1 times standard deviation squared they gave a standard deviation so we have to square it however in the denominator at Sigma squared and they gave us variance so it's already been squared so you can see they're being a little bit tricky here and guess what they're going to do the same thing on the certification exam so you have to read those questions very carefully especially when it comes to variance and standard deviation some a lot of people want to put 2 or 6 as a standard deviation they would have squared it again that'll give you the wrong answer every time so again be very careful when you're reading these questions and differentiate between standard deviation and variance because they'll try to trick you on those all right and you've studied too hard to miss to fail the exam based on such a silly tricky test questions alright let's go get on calculator and get this figured out so there we go 25 minus 1 which we already knew to be 24 times 2 squared times 2 squared equals so the value of the numerator is 96 divided by 6 remember it's already been squared they gave us variance equals 16 so the correct answer is 16 let's go back to the blue pages and make sure that that is an option all right here we are back at the blue pages question 46 I want to read through this again so you can pick out the tricks is desirable to reduce the variation in the process so if this is an invert study which it ends up being it's either gonna be F test or chi-squared test chi-squared parametric test the current variance is known to be 6 aha and all they gave you is the value of the variance which is Sigma squared remember is 6 there's no sample size or anything so we can't use the if this is a chi squared study now and we know it a new process yielded a standard deviation of 2 for 25 trials then we have one sample size so we can go 25 minus 1 is 24 degrees of freedom and we can use the chi-square table we can't use the F table why because it demands two sets of degrees of freedom a set of two degrees of freedom one for the earlier process one for the later process so this is definitely chi-squared we already did the calculation we know that C is the correct answer all right here we are at question 22 the advantage of using the modern designed method of experimentation do e rather than the classical what was the classical one at a time well one at a time if you have 10 variables you have to change one at a time and you have what 10 runs I guess but you'll never pick up on interactions that's the big problem so the advantage of using the modern design method of experimentation rather than the classical is that everything is held constant except for the factor under investigation now that is the classical kind and they're asking for the advantage that's not an advantage experimental error is recognized but need not be stated in quantitative terms yes it needs to be stated in quantitative terms it's okay to take risk but you need to quantify risk wherever possible fewer terms and measurements are needed for valid and useful information this is true and it's also a very common test question on an asq exam but C is correct the sequence of measurement is often assumed to have no effect no that's not true you should do em essays before you do do E's to make sure you don't have a measurement problem so 922 is no doubt C the repeated trials in a designed experiment allow for what the repeated trials in a designed experiment allow for what first-order modelling well if I remember correctly first-order modelling means the main effects now you may want to look that up to make sure but that doesn't fit this intent determination of experimental error yes you must have repeats to experiment to calculate experimental error because experimental error is based on what standard hair or standard deviation and you can't do that with one number so you have to do replicates to get the the experimental error so B I think is probably the correct one nested experimentation I think that's one everything's in the same column but I'm not if I remember correctly it doesn't fit the intent of this the resolution of main effects yeah that would definitely help with resolution I would imagine so I have a problem here somewhat of the hair splitter B determination of experimental error or the resolution of main effects okay and I can see it going either way but you may get some resolution you will get resolution of main effects without repeatability but you can definitely improve resolution but you cannot do experimental error without repeated trials so 941 is B here we are at question 53 plaque at Berman experimental designs are called screening designs this is true a screening design can be defined as an experiment with interactions among the main effects no placket vermin is incapable of picking up on interactions as our most other screening designs if not all the use of a non geometric experimental design okay I believe it's still a geometric experimental design and identification of the key input factors okay a screening design can be defined as an identification of key input factors okay it will study key input factors I don't like any of these all that well to be honest with you a fractional factorial experiment it is fractional but fractional is less you'd throw away less runs usually so there's a there's a difference between fractional and screaming half fractional screening is highly fractionalized so I think on this one it is C but it's worded kind of funny so let's look that up 53 is it seen and yes it is a process is in control with a P bar of 0.1 zero and a sample size of 100 I'm assuming sample size is constant the three Sigma limits of the NP control chart are okay you're going to get these types of questions so you'll need to go and look up the control limit formulas and I want you to do that maybe put this on pause make sure you can find those because if you can't you're gonna be in trouble in the test in the meantime I'm going to go over to the whiteboard and write those formulas out and get ready to solve this problem but make sure you know where to find these let's go to the whiteboard all right here we are at the whiteboard notice I wrote the formula out this is going to be a timed test so I'm not going to calculate the upper control limit and the lower control limit if I don't have to I could you know approximate double the time so I'm just gonna do one see if I can get the correct answer I always use the upper control limit because I get to add and subtract it just makes a little easier on my calculator moves okay so the the formula M P bar plus three times the square root of M P bar times 1 minus P bar make sure you can find that in your formula sheet if you if it's not there write it in there remember we want all the formulas in one place so you don't have to look throughout the the primer and P bar was a given of 0.1 0 or 10% and sample size equals a hundred it is constant because they asked us to calculate the N P chart and P chart demands constant sample size P chart does not if you'll recall okay what is now we have to calculate n P bar while N P bar is a formula n times P bar okay so n is 100 P bar is zero point 1 0 I multiply both of those to get in P bar that's 10 that what does that mean it means on average I get 10 bad parts per subgroup all right now let's put those numbers into our formula so NP bar like I said is 10 so there's my 10 for n P bar plus 3 times the square root of n P bar there's my 10 again times 1 minus P bar which is 1 minus 0.1 0 and now let's just pull our calculator out and start crunching numbers so 1 minus 0.1 0 of course is 0.9 times 10 equals 9 square root equals 3 so this whole square root thing is equal to 3 times 3 equals 9 plus 10 equals 19 of course you'll probably L do a lot of that in your head and do it much quicker than we just did and that's good but let's go back to the blue pages and see if this is one of the options so here we are back at the blue pages and notice there is a 19 I believe that's what we calculated it is and that is the correct answer right here we are at question 12 and X bar and our chart was free for an operation using 20 samples with five pieces in each sample that's good remember they'll test you on that sometime you need to have at least 20 subgroups before you calculate the upper and lower control them and it should be over another reasonable amount of time to where the average has time to fluctuate and everything and so I've always good thus far x-bar was found to be thirty three point six and our bar was six point two zero and when they start throwing out numbers it means you're going to have to do some calculations I was surprised that there's not numbers down here for answers but let's read on during production a sample of five was taken and the pieces measured these numbers at the time this sample was taken both the average and range were within control limits neither the average nor range were then control limits etc etc so what we're going to have to do is we're gonna have to calculate the upper and lower control limit on both the average and range chart calculate this sample and range from this sample sample average and sample range and see how it relates to those control limits well this is going to demand that we go to the white board let's go to the white board and work this out all right here we are at the white board notice I wrote everything out we're going to use an x-bar chart that was a given in the problem they gave us X double bar of 33.6 they gave us our bar of 6.20 we need to calculate the upper and lower control limit and the upper control limit is X double bar plus eight two times our bar I wrote these down but make sure you can find those formulas and it can find them quickly and you feel good about it confidently okay and lower control limit equals X double bar minus a two times our bar so we and I looked up a two from the constant table for SPC it's 0.577 and once we have that I put everything in here X double bar 33.6 and they may just give you a verage in the question you use what they give you we assume it's X double bar 33.6 plus a2 I looked up in the table of five point five seven seven our bar of 6.20 now the only difference between upper and lower control limit is instead of adding a two times our bar you're subtracting a two times our bar remember eight two times our bar equals three Sigma 3 Sigma of the averages if they gave it they could give you different numbers here to where you don't have everything you need they just give you Sigma well and they give you a sample size then you go 3 times Sigma over the square root of n is going to give you the same thing they do that sometimes most people are trained on these old formulas but the new ones are more accurate what's the new one X double bar plus 3 times Sigma of individuals divided by the square root of n ok but in this case they gave us everything we need for this traditional formula that you are created so let's go ahead and punch those numbers in and see what we get for an upper and lower control limit the 6.20 times 0.577 three point five seven so when you do these where you have to calculate the upper and the lower control limit always write down the value of 3 Sigma so you don't have to calculate it twice so this is three point five seven seven which is what equal to three Sigma's right there three Sigma's of what averages are single events averages eight two times our bar is four Sigma of averages okay now we just simply add that plus thirty three point six equals thirty seven point one seven seven four so the upper control limit equals thirty seven point one seven seven four oh and I erased I erased my three sigma thing there that I told you to write down but that's okay I'll plug it in again but remember on the time test you don't want to plug it in again so you want to keep that written down which you'll have to in the test anyway so I'm gonna go point five seven seven times six point two zero equals three point five seven seven four there it is I couldn't remember it at first but that sounds right three point five seven seven four and then I'm going to subtract that from the average so thirty three point six minus three point five seven seven four three point five seven seven four equals basically 30 thirty point zero two two six and I have students ask me a lot well what do I round that round to on these asq exams and I've overdone it here usually a look and say okay there's two decimal points I'm going to go out three and so I went out for but if you go out and if you use that rule you're going to be just fine on the SQ exam all right now that we know the upper and lower control limit for the averages chart let's go ahead and go to the range chart and calculate the upper and lower control limit for that so here we are the formulas for upper control limit of the range is d4 times R Bar lower control limit is d3 times R Bar I looked it up d4 remember you should be able to find these equations easily and confidently before you go take the test I looked up d4 is 2.1 1/4 d3 for a sample size of 5 how did I get sample size of 5 they gave me 5 numbers in that sample group and that's how I knew it was five d3 equals zero when a sample size 5 remember it will always equal zero until I believe the sample size gets above seven I have it out right here let me see d3 yes at seven it starts taking upon itself a number other than zero that's the d3 value there so be aware of that it's not always zero just if it's less than seven sample size okay with that being said they already gave me the range earlier I wrote that down earlier and so upper control limit of the range equals my d4 value times my R bar which is six point two zero right there so let's go ahead and plug that in two point one one four times six point two zero r bar equals thirteen point one oh six eight okay that's the upper control limit I should write that right there okay and then of course the lower control limits multiplied by zero and so it's going to be zero that's an easy one alright now we know our upper and lower values of our upper and lower control limits now we need to go back to that subgroup calculate the average calculate the range and then see how that lies within these control limits all right so I wrote down those numbers that they gave us and I calculated the average and the range high minus low there and then I look at that and say okay where does the average lie on the control limits does it lie within or outside well thirty five point eight lies right in there it's a little bit above the average which would be thirty three point five something but it lies within the control limits but eighteen range of eighteen lies what above the upper control limit so the range shows special cause average does not let's go back to the blue pages and see if that if we can find the correct answer alright here we are back to the blue pages and let's see read these to see which one is the correct answer both the average and range are within the control limits no the average was the range was not neither the average nor the range within control limits no the average was within control limits only the average was outside control limits no so it must be d only the range was outside the control limits this is true so the correct answer for question 12 is D all right here we are at question 6 the most effective and efficient method of solving quality problems for a product is to concentrate areas and would concentrate efforts in which area well this is an easy one we've been teaching this from day one in the class and it's going to be a design and you'll be asked that question on every asq exam so hopefully you can get that one right probably on multiple occasions and a production quality improvement well all the improvement could be part of design but axiomatic design so it's kind of a hair splitter but no quality improvement in design if it would say quality improvement in design then we would have a real hair splitter but this one is no doubt a let's move on to the next question 7 question 7 the principal purpose of robust design techniques is to do what to make things robust make the product less sensitive to noise effects yes that's true so it's probably a use the tools of experimental design now there's although it is a big deal to create robust design using that but the ultimate objective is to create less sensitive products reduce the sources of variation yeah not necessarily you create the design to where it can handle the variation of the environment improve manufacturing quality now the whole thing about robustness is make the product less sensitive to noise effects so 11/7 is going to be 8 congratulations you have completed this video as you can see I have a lot of experience in asq certification exams I passed most the certification exams myself if you have any questions accomplish our comments please contact me at alpha TC comm again that's alpha TC comm for alpha training and consulting and I look forward to hearing from you once you get on the website go to contact us send me a message I'll get back with you as soon as possible thank you and have a great goodbye

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Keywords: ASQ CSSBB online training, ASQ SIX SIGMA BLACK BELT test online, ASQ Certified Six Sigma Black Belt certification exam, ASQ Certified Six Sigma Black Belt question bank, ASQ Certified Six Sigma Black Belt exam questions, ASQ CSSBB practice exam, ASQ Certified Six Sigma Black Belt practice questions, ASQ Certified Six Sigma Black Belt practice test, Six Sigma Black Belting question bank, Six Sigma Black Belt practice exam, Six Sigma Black Belting practice questions

Id: jNnHp6AZP6I

Length: 72min 41sec (4361 seconds)

Published: Thu Oct 03 2019