Can some one help me with Standard deviation
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 This topic has 14 replies, 4 voices, and was last updated 12 years, 9 months ago by Robert Butler.

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February 11, 2009 at 10:28 pm #25729
I made a histogram, calculated the mean at: 21.1935 (from the data below) used the formula for standard deviation and got: 6.9041
the problem I am having is getting the proper calculation for the lowest point of expected variation and the upper point of expected variation.
can someone help?
here is the data:Data for ‘Wait Times’ for last month
2427
18
11
22
27
17
23
17
5
17
28
24
17
8
21
26
23
17
31
18
27
22
27
17
40
22
18
17
18
28
0February 12, 2009 at 2:04 pm #62260Tony,
I assume that with “the lowest point of expected variation” you actually mean “the lower bound of the confidence interval of the population Variation (sigmasquared) based on the data”. If that is not true my answer doesnot answer your question.
If the data follows the normal distribution then from the data not only the S (st.dev of sample) can be calculated but also the Confidence Interval (CI) of the Sigma (st.dev of population). The lower value of that CI is the square root of the answer you are looking for. In Minitab: Use Stat > Basic Statistics > Graphical Summary.The pvalue of the AD test tells you if the sample follows a normal distribution. The lower bound and upper bound of the CI of St.Dev. are on the last line of output. Squaring those gives you your answer.
If the data is not normal distributed the calculated bounds are very probably wrong.
Hope this helps.0February 12, 2009 at 4:20 pm #62261
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.As has been noted the question is too vague. Usually for homework problems of this type the requirements are to compute the mean and then find the 95% C.I. around the mean. In this case, for the 95% C.I. for a mean with N = 31 you would have 21.1935 +2.04*6.90/sqrt(31) which would give 18.66 and 23.72.
Now, if they are asking for the C.I. for a future mean value comprised of fewer points then you would put that value in for N. So, if in the future you were only going to take 4 samples and compute a mean then the estimates of the C.I. for those small sample means – using your historical grand mean as “truth” would be 21.1935 +2.04*6.90/sqrt(4).
There are other options besides these, however, addtional discussion really should wait until you provide a better desciption of what you mean by “lowest point of expected variation and the upper point of expected variation.”.0February 12, 2009 at 4:36 pm #62262Yes a class project assignment.
The mean is 21.1935 . The standard deviation is 6.9013 . I am having trouble figuring out how to calculate the correct expected variation + or – 3 standard deviations from the mean. So far every answer I have given is wrong. The last hint from the instructor was: my mean is correct, my standard deviation is correct but my lower and higher points are not. Then he hinted: How wide do I beleive this distrubution is? In essence how far to the right and how far to the left do I have..(I think is what he is looking for)0February 12, 2009 at 4:59 pm #62263
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.The use of the phrase “from the mean” would suggest he is looking for confidence limits on individual measurements. This is buttressed with the hint “how wide do I believe this distribution is”.
In this case N = 1 and if we treat the computed sample mean and the computed sample standard deviation as truth then the answer would be 21.1935 + 3*6.9013/sqrt(1) or, if you wanted to be a purist about it substitute 3.09 for 3 – the t value for 99.9 given an infinte number of degrees of freedom.
If he had used the phrase “of the mean” this would most likely have meant he was looking for the + 3 standard deviation limits associated with a mean comprised of 31 future measurements – in this instance you would put 31 in place of 1 inside the square root.0February 12, 2009 at 5:15 pm #62264i think it is of the mean. I am coming up with .0879 and 7.5250
if it is from the the mean I am coming up with 0.4896 and 41.89740February 12, 2009 at 5:53 pm #62265
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.You are making a mistake in your calcuiations. The first set of limits – .0879 and 7.525 don’t include the mean value therefore they are wrong. The second group is, as noted, for individuals from the mean and if you take a look at it you can see it inlcudes both your smallest (5) and your largest (40) values.
0February 12, 2009 at 6:05 pm #62266god I am lost…..My last answer was 5 and 40 and it was wrong. I am confused on the below where “the .0879 and 7.525 don’t include the mean value”
by the way thank you very much for your help. I so need it0February 12, 2009 at 7:13 pm #62268
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Ok, let’s take this from the top. You are confusing a number of terms and concepts and you really need to do some reading. However, before we go there let’s look at what you have.
Given –
1. 31 numbers which are said to represent wait times.
2. The minimum number in the sample is 5
3. The maximum number in the sample is 40
4. A computation of the average (mean) of the 31 numbers give 21.19.
5. A computation of the standard deviation of the mean of 31 numbers gives 6.9.
The sample is just that – it is not the entire universe of possible wait times that would occur with that process given that you recorded every wait time for that process from the time it began until it crumbled into dust (or until everyone associated with the wait time retired – whichever comes first :) )
Since it is a sample it is very unlikely that the minimum and the maximum of the sample actually happen to be THE absolute minimum and maximum that one could get for a wait time from that process.
Therefore, if you want to have some idea of what the REAL minimum and maximum of that distribution of wait times is you are going to have to estimate them.
The way you estimate the limits of the distribution of the individuals is you compute the mean of the sample (that is of the 31 measurements you have recorded), you compute the standard deviation of that sample mean, and you use the equation for the confidence interval about the mean to compute these values.
The basic equation is:
Average + Q*Std. Dev/sqrt(N)
where Average = sample average, Std. Dev = standard deviation of that sample average, N = the number of data points used to compute the average and Q = either the Z or t value for whatever limits you are trying to compute….and it is here that the vagueness of your instructions can cause problems.
For a normal population 3 standard deviations either side of the mean will encompass 99.68% of the data. Therefore you will want to know the location of the values that are located at the .16% and the 99.84% points
For a quick estimate of these values it is common to substitute 3 for the value of Q and run the calculations. The exact values for a Z score would be 2.96 and for t (the best I can do is a table with 99.8%) is 3.09.
Depending on your book/instructor/whatever there are those who will use a sample of 30 as a cut point and declare that for more than 30 points you use the Z value instead of the t value and there are those who say otherwise and still others who don’t care – so you will have to find out what your instructor wants in this regard.
In order to find the extremes of the distribution corresponding to the above percentage points you will need to let N = 1 since you are interested in the spread associated with the individual numbers in the distribution and not the spread of a collection of sample averages from the distribution where the sample averages are each computed from repeated collections of 31 independent samples from the wait times of your process.
Therefore – if we are looking for an estimate of the minimum and maximum number of the distribution of wait times where the minimum and the maximum are defined as being 3 standard deviations either side of the mean and if we use the Z statistic the numbers will be
21.19 + 2.96*(6.9)/sqrt(1) or 41.6 (rounded would be 42) and .77 (rounded would be 1).
Since you sound like you are in class you might want to look up the book – The Cartoon Guide to Statistics by Gonick and Smith and check pages 128136.0February 12, 2009 at 8:41 pm #62275thank you so much that helped. Very much appreciated.
Tony0February 15, 2009 at 7:59 pm #62279Tony,
I am working on the same problem. Was Robert’s answer correct?
I came up with .4896 for the lower limit and 41.8974 for the upper limit.
0February 15, 2009 at 8:51 pm #62280your calculation is correct.
0February 15, 2009 at 11:42 pm #62281thanks, we must be in the same Black Belt Class at Villanova
0February 16, 2009 at 12:06 am #62282no i am actually an instructor monitoring this board to see who is cheating… JUST KIDDING! No we are not in the same class or school.
0February 16, 2009 at 1:24 pm #62283
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.To answer your question – yes the answer I provided was correct…and so is yours….and so would be at least one other. The “correct” answer depends on the value of Q you put in the equation (see the longer post on this subject). I mentioned three possibilities 3.09, 2.96, or 3 and I mentioned the usual practice for quick estimates was to use 3.
I used the Z value for my calculations because I thought the instructor might be playing games with the choice of 31 numbers in the sample. Based on your answer and the fact that Tony said it was “correct” it would appear that instead of playing games your instructor is using the very straight forward approach of just using 3. The take away for you now and in the future is to remember that for a 3 standard deviation estimate any of the three numbers (2.96, 3.09, and 3) are acceptable and thus so are any of three different sets of answers.0 
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