Need Help Analyzing Variance with Minitab
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 This topic has 12 replies, 6 voices, and was last updated 1 year, 4 months ago by Chris Seider.

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June 13, 2019 at 8:57 pm #239784
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Hello,
I’m new here and looking for some support from some expierienced six sigma mentors. I am currently working on a project and having trouble analyzing some data. This is my first project using minitab and any guidance would be greatly appreciated. My main question is regarding measurement system analysis and r&r gauge. So here I go.
I am trying perform a r&r gauge of the data below. The material I am measuring is bulk metal which is being ground to a specific particle size. I have no load scales so I am trying to develop a measurement system to track throughput, this will be my primary metric. I will use this metric along with runtime to hopefully justify any improvements in my system during my project.
We use a front end loader to load bulk metal into the system, we run several hundred tons of material a week. I want to analyze the weight of material in each scoop using the loader in hopes of using the # of scoops per shift to determine the input of material into the system. Using this and the runtime I will have my run rate.
So I have 6 team members, 3 on each of two shifts. Using large truck scales I measured the weight of 5 different scoops of material from each of the 6 team members.
My question is can you please help me use the r&r gauge or some other tool in minitab to analyze the variance between the samples? I know I have variation between team members and shifts and I want to see this statiscally. An example of my data is below.
Shift 1:
Team member 1 – scoop weights(tons)= 1.3, 1.35, 1.4, 1, 1.15
Team member 2 – scoop weights(tons)= 1.3, 1.4, 1.4, 1.05 1.2
Team member 1 – scoop weights(tons)= 0.9, 1.2, 1.4, 1, 1.3Shift 2:
Team member 4 – scoop weights(tons)= 2.1 1.9, 1.4, 1.85, 1.8
Team member 5 – scoop weights(tons)= 2.05, 1.9, 1.7, 2, 1.95
Team member 6 – scoop weights(tons)= 1.95, 2, 1.85, 2, 1.75Thanks for reading, your Feedback is much appreciated.
Brandon
0June 13, 2019 at 9:03 pm #239787
Katie BarryKeymaster@KatieBarry Include @KatieBarry in your post and this person will
be notified via email.@minitab — Are you free to weigh in?
0June 13, 2019 at 9:43 pm #239789
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Just a visual impression of your data says you have a statistically significant difference in mean scoop size between first and second shift. If we check it formally and pretend the measures within each team member can be treated as independent measures then what you have is 1.22 vs 1.88 (which, given the limit on significant digits reduces to 1.2 vs 1.9) with corresponding standard deviations of .168 and .172 (which rounds to .17 and .17). There is a significant difference in the means (P <.0001) and there is no significant difference (P = .92) with respect to shift variability.
One thing you should consider. The above assumes the measures within each team member are independent. If the measures you have reported are random draws from a single shift work for each team member then it is probably reasonable to treat the measures as independent within a given member. On the other hand, if the measures were just 5 scoops in sequence then you are looking at repeated measures which means your estimates for shift variance are much smaller than they really are.
Before you do anything else you need to understand the cause of the difference in scoop means between shifts. If the material is uniform, the front end loaders are the same, the 6 individuals all have the same level of skill in handling a front end loader and all other things of importance to the process are equal between the two shifts then you shouldn’t have a difference like the one you have in your data.
 This reply was modified 1 year, 4 months ago by Robert Butler.
0June 14, 2019 at 8:16 am #239794
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Thanks for the feedback Robert. I think I understand the difference between the shifts. Shift 1 measures were done on during the day and the process was monitored by day shift staff. Night shift decided to do the measurements on their own prior to mgmt. being available to monitor. I believe the data from night shift is actually more realistic because they left a mess on the way to and from the scale due to the large scoops while day shift I believe intentionally was more modest with the scoop size to not leave the same mess. We are using a single loader and took tare weights with each operator before taking the 5 measurements back to back. I plan on standardizing the scooping process to reduce variation but just wanted to use some mini tab tools to graphically show this difference while I am analyzing the data.
Which tools did you use? do you have a recommendation on how i should show this difference as well as the variation between operators and between each of their own scoops?
Again Thanks for your time.
Brandon
0June 14, 2019 at 8:21 am #239795
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Attached is the actual data for my question in a MPJ. I didn’t have my computer last night when i initially submit the question so i estimated what i remembered. If someone could show me how to analyze this in mini tab it would be greatly appreciated.
0June 14, 2019 at 10:50 am #239797
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Sorry, you can’t upload MPJ. files to this forum? Data in excel below.
Attachments:
 weightdata.xlsx
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0June 14, 2019 at 11:16 am #239799
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.I don’t know Minitab but I do know they have boxplots graphics which I would recommend you use. You would have sidebyside boxplots of the two shifts. What you will want to do is to make boxplots which also shows the raw data that way you will not only have the boxplot with its summary statistics but you will be able to see visually just how the data distributes itself with each shift. See Attached for example.
For your repetition you will want to take five measurements but not backtoback – rather randomize the pick through the time of the shift. This will mean extra work for you but it will guarantee that your measurements are representative of what is happening in the course of a shift.
0June 14, 2019 at 11:53 am #239801
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.I see I spoke too soon. What I said above still applies with respect to the graphical presentation and sampling plan you should use when repeating your test but since you provided the actual data there are a couple of other things to consider.
Again, if we pretend the sequential measures within a given operator are independent (they’re not) then for all of the data we find there is a significant difference between shift means 1817 and 2826 (P < .0001) and we also find there is a significant difference in shift variability with standard deviations of 193 and 98 respectively (P = .016).
A quick check of the boxplots suggests the single measure of 2280 in the first shift is the cause for the differences in standard deviations. If we rerun the analysis with that value removed what we get are means of 1784 and 2826 – still significantly different (P < .0001) and standard deviations of 150 and 98 (P = .12). So, overall you first shift has a significantly lower mean scoop measure and a numerically higher (but not significantly different) variability.
You could go ahead and try to parse the data down to the individual operator but with just five repeated measures per person I wouldn’t recommend doing this for two reasons 1) the measures within each operator are not independent and 2) even if the measures were independent it is highly unlikely that the variability of a sequential 5inarow sample is at all representative of the variability one would see within a given operator over the course of a shift.
I would recommend summarizing your findings in the form of the graph below along with a few simple summary statistics of the means and the variances and then, before trying to do anything else – address and solve the issues you mentioned in your second post and then rerun the sampling plan in the form I mentioned above.
0June 14, 2019 at 2:38 pm #239804
Michael CygerKeymaster@michaelcyger Include @michaelcyger in your post and this person will
be notified via email.(Side note: @Naquin81, we’re going to allow .mpj files in the near future. Sorry for the inconvenience.)
0June 14, 2019 at 3:12 pm #239805
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Thanks Robert, this is of great help!
0June 17, 2019 at 11:55 am #239835You have a 1st shift problem. There is a statistically significant difference between all three operators when comparing the two shifts. See attached .pdf’s Reply with email address if you want the MT file w/version you are using.
Attachments:
 GraphicalAnalysisofNetLbs.pdf
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 ANOVAComparisonofNetLbs.pdf
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 NetLbsWS.pdf
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0June 17, 2019 at 12:32 pm #239840
Naquin81Participant@Naquin81 Include @Naquin81 in your post and this person will
be notified via email.Thanks Rhb! I would love the MT file, my email is [email protected]
0June 18, 2019 at 10:35 am #239853
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.@michaelcyger you and your team is always an example of customer service and response!
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