Standard deviation
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 This topic has 28 replies, 11 voices, and was last updated 15 years, 1 month ago by NICOLAE.

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April 3, 2006 at 1:34 pm #42944
I have a question about combining two populations. I have 2 same populations and one different one, if i combine them, is their overall standard deviation pooled standard deviation?
Thanks0April 3, 2006 at 2:52 pm #135826That is an interesting question. I wondered that mayself since I am new to this methodology. Anyone out there that can answer this? Thank you.
0April 3, 2006 at 3:37 pm #135827
Iain HastingsParticipant@IainHastings Include @IainHastings in your post and this person will
be notified via email.In a word: No.
To illustrate this consider two populations each with n=500, both with a standard deviation of 1.0. One population has a mean of zero, the other has a mean of 10.
From the definition, pooled standard deviation is equal to 1, however the standard deviation of the combined populations will be about 5. (Try this at home by generating 2 columns of 500 data points in Minitab then combining them).
The reason for this is that each of the 2 populations is normal – one centred at zero, the other at 10. The combined populations form a bimodal distribution with a mean of 5. Each point in the combined distribution is now much further away from the population mean (on average a value of 5).0April 3, 2006 at 4:17 pm #135830So how would I then calculate overall standard deviation of these three populations that are to be combined into one? Thanks
0April 3, 2006 at 4:49 pm #135833
Iain HastingsParticipant@IainHastings Include @IainHastings in your post and this person will
be notified via email.J
It is not clear to me what you are trying to do here. Are you trying to perform some type of analysis or are you simply looking for the standard deviation of a single population (made up from 3 others)?
If it is the former, what is the specific analysis intended to do? If it is the latter, how much information do you have on the individual populations? BTW are they really populations or are they samples?0April 3, 2006 at 5:16 pm #135835Iain,
I have a compound that is made out of 2 parts of one ‘cake 1 ‘ and 1 part of ‘cake 2’. When the final compound ‘cake’ is done, I now have a finished product. I have a median and standard deviation information for cake 1 and cake 2. How do I calculated the standard deviation for the final cake? I know that median of my final cake is going to be calculated as follows: (Median Cake 1 + Median Cake 1 + Median Cake 2 ) / 3. How about standard deviation calculation? Can I just average the three or not? Thanks for the help…I hope I make sense. I am just starting this Six Sigma thing and am trying my hardest to understand it. Thanks again.0April 3, 2006 at 5:36 pm #135836I might be completely off here, but I’ll give it a shot…
If I’m not mistaken, “pooled” and “overall” standard deviation are pretty much two different things. “Pooled” is similar to “average” in that it is taking multiple estimates of standard deviation (based on several samples) and trying to come up with an estimate of the population standard deviation by taking the “average” of the sample standard deviations. It assumes that the various samples came from the same population.
“Overall” is simply the standard deviation of all of the data points that you have collected, regardless of how you define the various “populations” that exist.
Think of an Xbar chart. You have several subgroups, each of which has a different standard deviation. However, we assume that there is roughly the same amount of localized variation at any given moment, so the “average” standard deviation of all of the subgroups would give us an idea of how much short term variation there actually is. However, because we have “shift and drift” in the mean, those distributions don’t all line up nice an pretty, and therefore combining all of the data points together into one big heap makes a wider distribution. The standard deviation of this distribution would be the “overall” standard deviation. (Incidently, if the “shifts and drifts” are subtle enough, then the overall distribution can appear be normally distributed.)
This is what the “Within” versus “Overall” thing in capability study is all about, and closely parallels the example that Iain gave. (I think.) Also, this is what ANOVA is all about…if the means of the sampled populations are the same, then the pooled standard deviation and the overall standard deviation should be pretty close to equal (F statistic = 1)
In summary: “Pooled” std dev. = “average” of sample std devs regardless of mean value. “Overall” std dev = std dev of all samples together relative to the overall mean.
By the way, I don’t really get the cake thing…0April 3, 2006 at 6:03 pm #135839Thank you TwoCents.
The cake thing is really funny to explain. I have samples for cake 1 (think of it as an experiment) and samples for cake 2. For my finished product I have to combine two cake 1 measures and one cake 2 measures to get a finished product. But before I do that I have to ‘predict’ the final cake properties like standard deviation and mean. Does this make more sense?0April 3, 2006 at 6:11 pm #135843What are you measuring?
Is it a cumulative property, like weight or something? (The weight of the finished cake = 2*Cake1 + Cake2).
Or is it like a mixture effect, like reactivity or something?0April 3, 2006 at 6:18 pm #135844It is a cumulative property.
0April 3, 2006 at 6:29 pm #135845If it’s a cumulative, additive property, then you’re answers aren’t so simple…
For example, with weight:
Suppose that Cake 1 has a mean of 30 and an SD of 3. At the same time, Cake 2 has a mean of 40 and a SD of 3. Creating a “final cake” from these distributions will probably yield an average weight of 100 (2*(30) + 40) = 100), but your standard deviation will be something smaller than 3, because when you are adding these values, the Central Limit Theorem kicks in and your variation decreases. Try simulating this in Minitab, and you’ll see what I’m talking about.0April 3, 2006 at 6:32 pm #135846By the way, this happens because your odds of getting an “extreme value” from all three of the distributions at the same time are very, very small, so your distribution narrows up.
0April 3, 2006 at 6:40 pm #135847J:
J:The big
picture is that it depends on what you are trying to measure.You may
have three different standard deviations; overall, within group, and pooled. An
example should help show the difference.Customers
have been complaining because your inconsistent deliveries are creating
problems with their supply chain management. You are in a meeting to see how
big the problem is. About 2/3 of your business is domestic, so you have a
sample of delivery times for 20 domestic and 10 international shipments.Domestic
International
5
10
3
19
3
13
1
12
4
6
2
11
2
19
2
6
3
13
1
11
3
2
3
2
2
1
5
1
1
3
The stakeholders,
really fluent in the Six Sigma lingo, ask you to describe your data. You say
that they have a mean of 5.6 with a standard deviation of 5.4. The CIO,
assuming a normal distribution and doing a quick mental calculation, states
that the problem is that the spread of deliveries is too wide (2*5.4*1.96 = 21
days).The COO,
also pretty good, disagrees with the CIO and says the problem is that the
international deliveries take a long time and they skew the distribution. He
wants a measure of the spread broken out by domestic/international. You find
these results tell a different story.Group
Mean
Standard Deviation
All
5.63
5.30
Domestic
2.45
1.23
International
12.00
4.45
The
standard deviation within each group is much smaller than before (1.23 and 4.45).
The large standard deviation of 5.3 was caused by the large difference in the
average delivery times for the two groups (2sample ttest).The CEO wants
a single number to best describe the overall variation in delivery time for all
deliveries. You perform an ANOVA using the Domestic/International
classification as a group and find the pooled standard deviation is 2.72. You
report this number, but you inform the CEO that the test for equal variances
showed that there is a significant difference in the amount of variation for
the two groups.Cheers,
BTDT0April 3, 2006 at 6:50 pm #135850If it’s cumulative and additive, why wouldn’t the overall Variance be the sum of the individual Variances. (Root Summ of Squares)
0April 3, 2006 at 7:27 pm #135853You know what? I think I’m a liar…althought I’m trying to figure out why…
I was in the process of writing what is below, when I really started thinking about this. As I think about it, it’s becoming clear to me that my logic is broken somehow, but I’m not sure how. I’m screwing up the CLT somehow…maybe someone smarter than me can tell us.
—————————————————–
Because, in my little example, the three individual data points are no longer individual data points…they are each a component in one larger data point – the overall weight of the “finished cake”.
For example, I make my first cake…I pick the first Cake1, and it’s “average”. I pick the second Cake1, and it’s also “average”. I pick my Cake2, and it’s abnormally super heavy (it’s way out on the tail of my Cake2 distribution). The resulting “finished cake” will be heavier than average, but not that heavy, because the “average” Cake1’s soften the effect of the heavy Cake2.
In order for me to get a megaheavy Cake (I’ll add here that I feel a little silly using this example, but work with what I was given :) ), I would have to pick an annormally heavy Cake1, another abnormally heavy Cake1, and then an abnormally heavy Cake2. The odds of these three things happening simultaneously are much smaller than the odds of picking just any one abnormally large component. Therefore, the distribution of the combined components becomes much narrower than the distributions of the individual components.0April 3, 2006 at 7:52 pm #135856
Iain HastingsParticipant@IainHastings Include @IainHastings in your post and this person will
be notified via email.Yes, under those conditions it would be. The important point for the original poster is that variances can be added but standard deviations cannot. (Hence RSS).
0April 3, 2006 at 8:13 pm #135860Alright, Iain, help me out…why am I confused? (I know…that’s a loaded question…)
Why does the CLT work for means, but not for sums? Shouldn’t the principle work the same way? In other words…why does the logic of my last post not work?0April 3, 2006 at 10:42 pm #135870
Iain HastingsParticipant@IainHastings Include @IainHastings in your post and this person will
be notified via email.Now you’ve got me confused – I need to think about that.
But, the CLT is based on taking averages of samples from a distribution. The resulting mean of these samples forms a normal distribution with the same mean as the parent population. The variance of the sample is equal to the variance of the distribution divided by the sample size.
In the problem at hand I don’t think that we are following the rules for the CLT. We have a distribution which is really the sum of the other 2 (actually the sum of 3 distributions, 2 of them identical) in which case the mean of the distribution is the sum of the means of the individual distributions and the variances add also.
This is very similar to a statistical tolerancing exercise.
I’ll think about ut some more and try to give you a better answer.0April 3, 2006 at 11:18 pm #135872A coworker and I discussed this issue for quite some time this afternoon…and I think we made some headway in figuring out why I was so confused.
I’m not going to go into much detail with regard to our discussion, partially because it was pretty drawn out and more than a little confusing, but mostly because I don’t want to risk sounding any more like a dolt. :)
The problem is with squaring, adding, and then taking the square root (big shocker there, huh?)
The conclusion that we came to is that the really extreme values in the “final assembly” distribution aren’t necessarily the result of extreme values in the “component” distributions.
Ok, I got this far…what the heck…
Suppose that we’re making a wedding cake with three tiers and want to evaluate variation in the overall weight of the cake. The tiers are roughly the same size, and are therefore being pulled from the same population. We collect some data and find that the “tiers” population has a mean of 10oz., with a standard deviation of 2oz. So, that indicates that only 2.5% of the tiers should weigh less than 6oz. (2 StdDev.). This also means that the variance of the “tiers” population is equal to 4 (2 squared).
So now we look at the “finished cake” distribution: The three “tiers” are added together…to get a mean of 30oz., with a variance of 12oz**2 (4+4+4), and a resulting standard deviation of 3.46oz. So, that means that only 2.5% of the “finished cakes” should weigh less than 23oz (again, 2 StdDev).
Why is that important? Well, if you’ll notice, you don’t have to have the “smallest of the small” (outside of 2 StdDev) tiers to get the “smallest of the small” (outside of 2 StdDev) cakes! If I take three of those “unusually small” tiers and put all three into a cake, I get a cake that weighs only 18 oz, far below my 2 StdDev limit for my cakes (indicating that the odds of this happening by dumb luck are really, really small…just like I reasoned.)
The moral of the story…I think…is that the logic was right, I was just overestimating the impact on that this “smoothing by dumb luck” factor would have on the variation in the finished cakes. Not only that, but for whatever reason, you see the effect in the standard deviation, but not in the variance. (Still not quite sure about that one.)
I hope you don’t feel like you’re getting dumber by the second for having read this drivel…0April 4, 2006 at 10:19 am #135885Yeah, we can take sum of variances and then square root for the final standard deviation. only thing is that all the samples should follow normal distribution. Otherwise, get the probability function and get 2nd order momonet for standard deviation.
Whereas for a continuous distribution, it is given byThe variance is therefore equal to the second central moment .
You can refer Binormal or Multi Normal distribution to get the variance and standard deviation
0April 4, 2006 at 12:48 pm #135889You had to go and bring one of the those big, stretchedout “S” thingies into the discussion, didn’t you?
You couldn’t just let me bumble through life by adding, subtracting, multiplying, and dividing?
I see what you’re talking about, but what is “d” in the equation? :)
Ahhh, calculus…helping people “avoid” engineering since 1666!
It worked for me! :)
(Do you detect some mathematical resentment?)0April 4, 2006 at 1:13 pm #135890You guys are awesome. I can really learn a lot from you. It will probably take me decades to understand statistics as well as you do. What a humbling experience! Thanks for all your comments. J
0April 4, 2006 at 1:40 pm #135892BTDT,
I disagree with the way you have used the term within group and overall. You have mentioned within group as the one within the Domestic or International. But the term within group infact refers to the variation within in a subgroup and overall variation is based on the variaiton within a subgroup and between the subgroups.
Siva0April 4, 2006 at 2:02 pm #135893If there is one thing that studying statistics and Six Sigma will teach you it’s…humility.
I know that as soon as I start to feel like I’ve got a handle on this stuff, someone or something comes along and demonstrates for me just how much I don’t know.
But that’s a good thing, I think. I mean, if you could learn all about statistics in a three week training class, what would be the fun in that? (I’m going to call my shrink now…I just used fun and statistics in the same sentence.)
I tell my clients all the time: I hate math. I hate formulas. I hate calculations. But using some pretty basic math to describe and explain what’s going on around you? Now that’s pretty cool. :)0April 4, 2006 at 2:11 pm #135894
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Fun and statistics??? Of Course!
“Only the truly educated can be moved to tears by statistics.” – G.B. Shaw
Or, as I prefer to think of it – A single number by itself is nothing more than a dull, frivolous, plaything for the idle rich. However, join that single number with just one more number and the fun begins because now we have statistics!0April 4, 2006 at 4:14 pm #135898Siva:
Siva:You are right. When I used the phrase ‘within group,’ it can
be confused with the clearly defined statistical term, ‘within group variation,’
as measured by the Sum of Squares (SS) for each effect. I see your point about using
terms like overall rather than total. To make it more unequivocal, it would
have been better if I had used a term like, ‘variation within each group’ to
refer to the standard deviation for each of domestic and international orders.You say, But the term within group infact refers to the
variation within in a subgroup and overall variation is based on the variaiton
within a subgroup and between the subgroups.The accepted formula for total variation states that,SS(total) = SS(within) + SS(between)I assume you are referring to the relationship between the
components of the variation as calculated by the formula and summarized in an
ANOVA table. When I present the ANOVA table, I point out that the SS(within) is
like the ‘left over’ variation after accounting for SS(total) and SS(between).
I rearrange the equation like this,SS(within) = SS(total) – SS(between)I have people calculate the two right hand terms by hand and
verify that the left hand term is correctly calculated by the software.In the big picture, I see that J may be faced with the
problem of common and vague interpretations of overall standard deviation
rather than the nuances of the calculation.Cheers, BTDT0April 10, 2006 at 12:46 pm #136165It´s important to distinguish between sum and mixture of populations. An example of sum is if you are ensambling two parts of different and random height (caracterized by some distribution), then the height of the ensambled part will be the sum of both parts. The resulting mean will be the weighted average of both means, and the resulting variance will be the weighted average of both variances. If both variables are normal, then the resulting one will be normal.
If what you have is mixture of populations, for example, you have in one box some parts made by different machines with different means and variance in some characteristic, then the resulting distribution IS NOT the sum, as in the previous case. It´s the mixture.
If variables are normal, the resulting one IS NOT normal.
The mean of the population is calculated in the same previous way, and the variance is, for two populations:
If the proportions of populations 1 and 2 are P(1) and P(2) and the mean of the combined populations is MEAN(1,2) (average of all values):
V(1,2)=V(1)*P(1) + V(2)*P(2) + (MEAN(1)MEAN(1,2))^2 * P(1) + (MEAN(2)MEAN(1,2))^2 * P(2)
Hence, StdDev=SQRT(V)
It´s important to clarify that if the original populations are normal, the resulting distribution of the mixture IS NOT normal. Thus, the resulting MEAN and StdDev parameters ARE NOT parameters of a normal distribution.
Hope this clarifies, and not confuse because of the formulas.
It´s important to understand that in statistics, when you want to make more detailed and complex things, the complexity of the maths involved increases exponentialy, and misinterpretation and incorrect usage of this may cause huge mistakes!!! so be very careful when you try to use things beyond your knowledge!!! the only solution is to keep on studying….
Greatings,0April 10, 2006 at 12:54 pm #136166I commited 2 mistakes in the previous message. At the begining:
It´s important to distinguish between sum and mixture of populations. An example of sum is if you are ensambling two parts of different and random height (caracterized by some distribution), then the height of the ensambled part will be the sum of both parts. 1)The resulting mean will be the weighted average of both means, and the resulting variance will be the weighted average of both variances. If both variables are normal, then the resulting one will be normal.
If what you have is mixture of populations, for example, you have in one box some parts made by different machines with different means and variance in some characteristic, then the resulting distribution IS NOT the sum, as in the previous case. It´s the mixture.
If variables are normal, the resulting one IS NOT normal.
2)The mean of the population is calculated in the same previous way, and the variance is, for two populations:
Corrections are:
1) The resulting mean is the sum of both means
2)The mean of the population is calculated as the weighted average of both means, and the variance is, for two populations:
Sorry!!!0April 15, 2007 at 10:54 pm #154819i have this problem looked kinda similar. this is high school AP course question. manufacturer produces them.
~a nugget : mean weight 3oz stand.dev0.2oz
~fudge: mean weight 4oz st.dev0.3 oz
~box : mean weight 3oz st.dev0.1
so a box of candy contains 4 nuggets, 5 fudges. im lookin for mean and the standard deviation. i found the mean to be 35 oz.
but the standard deviation i cant find it. either i learnt it but forgot or we never studied it. can anybody help me? plz0 
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