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February 11[edit]

Distance between points on Earth[edit]

According to North Magnetic Pole, the distance between "Latitude 72 degrees N and 100 degrees Longitude West" and "73 degrees 15' and 99 degrees 45’ W" is less than 20 miles. It seems to me that it must be signficantly more than this. Could anyone check? I am assuming that "73 degrees 15'" must mean 73 degrees 15' North. 86.160.220.213 (talk) 01:42, 11 February 2014 (UTC)[reply]

No, not right, as more than a degree of difference in latitude is going to be more than 60 miles. If it was 72 degrees 15', then the asserted 20 miles would be approximately correct. --Mark viking (talk) 01:54, 11 February 2014 (UTC)[reply]

(edit conflict) I get 86 miles (139 kilometres) for those locations. If the second coordinate had been 72°15' N rather than 73°15', then the two locations would be 18 miles (29 kilometres) apart, which is less than the noted 20 miles. Perhaps one of the coordinates was misstated in the article? Dragons flight (talk) 01:58, 11 February 2014 (UTC)[reply]

It was addded in [1]. The only other edit by the user is [2] which indicates the whole thing may be original research. User:Cinderkimberstammy has email enabled so somebody (I'm not mailing the user) might be able to at least determine whether there is a typo. PrimeHunter (talk) 02:18, 11 February 2014 (UTC)[reply]

The issue is being discussed at Talk:North Magnetic Pole#Project Polaris confusion. -- ToE 13:38, 11 February 2014 (UTC) (Good job 86...213 on catching this. It amazes me how long simple and blatant mathematical errors remain in place. In this case, that error has been viewed more than a quarter million times in the year-and-a-half it has been in place!)[reply]

Got the reference (White's World in Peril) and found that the problem was due to a typo in our article. The latitudes in the quote are 73°N & 73°15′N. I've corrected the article. -- ToE 23:31, 21 February 2014 (UTC)[reply]

A cryptic message[edit]

I received an equation and was asked to help decrypt it for a friend. Apparently it is way beyond my expertise. I was wondering if anyone on here could help out. Seems like you'll need to know at least multivariate calc to figure it out.
The image is located here from my Google Drive. --Agester (talk) 03:06, 11 February 2014 (UTC)[reply]

There are two unrelated things there: the top two lines are an application of the axiom of extensionality to some (set-theoretic) construction of the real numbers. The integral is simply divergent, as is any integral of an even-degree polynomial over all space. (Even odd-degree polynomials only have a Cauchy principal value.) If it has a meaning as a message, it is not in its normal mathematical value. --Tardis (talk) 04:29, 11 February 2014 (UTC)[reply]

The top line seems to say "merrier", but I can't make anything of the rest. Looie496 (talk) 22:25, 11 February 2014 (UTC)[reply]

Haha merrier :) I don't think that's part of the message but thank you for your insight. Tardis, with my limited calculus knowledge that has helped a bit in interpreting the message. We are told the integral equation is supposed to tell us something... I guess it may translate to words / a picture. But it's way beyond my ability to understand without calculating it by hand... which would take me a long time time to do, if I dust off my integral solving skills... Nonetheless, thank you! --Agester (talk) 02:51, 12 February 2014 (UTC)[reply]

Displaying data spreads[edit]

At work I have weighed an item 10 times and got 10 very slightly different results. (I was expecting this, since I am trying to prove that our measuring devices are not accurate enough). I have then repeated this process for 10 different items. I want to show the "spread" of these results in a simple chart - what is the best way to display this? The software I have access to is Microsoft Excel and Minitab. Thanks in advance :) — sparklism ^hey! 20:58, 11 February 2014 (UTC)[reply]

You can give the average and standard deviation of your sampled data. I'm not sure how you'd do that with the software you mentioned but I'm sure if you Google "how to do a standard deviation / average with X software" I'm sure you'll get your instructions.--146.95.40.101 (talk) 21:21, 11 February 2014 (UTC)[reply]

The most common way of emphasizing data that I have seen is to limit the minimum and maximum values on a vertical weight y-axis to within 10% away from the top and bottom values the of values of the data. You only have one dimension (weight) , so one axis is enough. Vertical seems to be the easiest to understand. Make the values in red. Some use horizontal red bar chart lines, some use large red dots. Presenting the info in different ways at the same time can also help understanding. DanielDemaret (talk) 22:07, 11 February 2014 (UTC)[reply]

What you are trying to measure is Statistical dispersion, I would recommend that you weight the same item 100 times. In fact the more times you measure the same item, the better you can get dispersion data. You would calculate Coefficient of variation by calculating the mean and standard deviation. To make sure you can the correct Coefficient of variation, perform the same calculations for various items. The Coefficient of variation value should be the same. The other thing you may want to do is to find the shape of the distribution. The shape should be normal/gaussian but you may never know if the device have a tendency to cluster. This is why you may need to take as many measurements as possible for a single item. 202.177.218.59 (talk) 22:35, 11 February 2014 (UTC)[reply]

If you have a small number of different values (e.g., 9.8, 10.0, 10.2) that are repeated often, then use Excel to plot a histogram (bar graph with measurements along x-axis and repetition counts on y-axis). If you have a large number of different values (e.g., 9.799, 9.801, 9.803, 9.850, 9.991, 9.999, ...) then use Excel to sort them, then draw a simple chart showing sorted ordinal rank along the x-axis and measurements along the y-axis. And as others have suggested, dress it up with statistics. Poslfit (talk) 04:02, 12 February 2014 (UTC)[reply]

These are all great answers; thanks to everyone who has contributed. I've actually measured 30 items, 10 times each, but I don't want to produce 30 different charts. In my head I have a chart that shows 30 different data points (one for each item), and the spread of recorded data associated with each point - I think DanielDemaret is sort of closest to what I am getting at here, and the dispersion/variation stuff is also very useful. Also, although I am yet to carry out any proper analysis, I have a feeling that the "spread" of data is somehow less on the lighter products than it is on the heavier - is this (statistically) to be expected, or are our measuring devices less reliable than I thought? Thanks again! — sparklism ^hey! 10:22, 12 February 2014 (UTC)[reply]

the "spread" of data is somehow less on the lighter products than it is on the heavier - is this (statistically) to be expected, or are our measuring devices less reliable than I thought? Yes, it is to be expected! A weight w cannot be negative or zero. The spread, dw, is usually proportional to w, and so the relative spread dw/w is the interesting quantity to consider. The relative spread is the same thing as the spread of the natural logarithm. dw/w=d(log(w)). So the decent thing to do is to take the natural logarithm of all your weights before computing mean values and standard deviations. log(w)≈μ±σ. Then the weights are w≈e^μ⋅(e^σ)^±1. Something like this:

w ≈ 10.4 ⋅ 1.023 ^±1 kg (rather than w ≈ 10.4 ± 0.24 kg) . :This method will guarantee against estimating negative weights.

Programming in J:

   M=.+/%# NB. Mean value
   R=.M&.:*: NB. Root mean square
   D=.-M NB. Deviation from mean
   S=.R@D NB. Standard dev
   A=.M,.S NB. Arithmetic mean and standard deviation
   G=.A&.:^. NB. Geometric mean and standard deviation

Test:

   A 10.4 10.16 10.23 10.02 10.29 10.91 10.57 10.43 10.55 10.22
10.378 0.241446
   G 10.4 10.16 10.23 10.02 10.29 10.91 10.57 10.43 10.55 10.22
10.3752 1.02336

Note that (98,2) ≈ 50 ± 48 and also (98,2) ≈ 14 ⋅ 7^±1 .

Bo Jacoby (talk) 06:41, 13 February 2014 (UTC). Bo Jacoby (talk) 15:28, 14 February 2014 (UTC).[reply]

The absolute error might be more on heavier items, however, as a percentage of the total weight, I'd expect the error to be less on the heavier items. StuRat (talk) 18:45, 12 February 2014 (UTC)[reply]

The OP could consider using a box plot display for the weights, with that for each of the 30 items on the same sheet, if the nominal weights are not too widespread. With 10 repeats arranged in ascending order, Q1 is the third, Q2 is midway between the fifth and sixth, Q3 is the eighth - easy enough to determine by hand.→31.54.113.130 (talk) 16:39, 12 February 2014 (UTC)[reply]

Yeah, I use box plots for this kind of stuff fairly often. I think the problem that I have here is that the actual spread of weights per product is relatively small, and the absolute weights of each product are relatively diverse, so when I try to plot these on the same chart the y-axis scale is too wide ranging to show the upper/lower limits of each product effectively. I think I will resort to just looking at the standard deviation at product level. Thanks to everybody who has contributed to this thread :) — sparklism ^hey! 08:28, 13 February 2014 (UTC)[reply]

How about if instead of the actual weight and measured weights, you plot percentage error in each case ? So, you could just order the items horizontally from light to heavy (not to scale), but have 100% be the correct weight in each case for the vertical, with data points above or below that line to show percentage deviation. StuRat (talk) 17:35, 13 February 2014 (UTC)[reply]

Great idea! I'll have a go at that - thanks!! — sparklism ^hey! 20:37, 13 February 2014 (UTC)[reply]

You're quite welcome. StuRat (talk) 17:44, 14 February 2014 (UTC)[reply]

Be careful here, it really depends on what you're trying to show. The relative error described by Stu can be misleading, depending on your application. E.g. 1 +/- 0.5 will have a huge relative error, while 100 +/- 0.5 will have a tiny error. The question you have to answer is: are those errors the same or different for your application? SemanticMantis (talk) 15:05, 14 February 2014 (UTC)[reply]

But isn't that exactly what they want to show, that the relative error is unacceptably large for certain weight ranges, and hence purchasing another measuring device is justified (one more accurate in those ranges) ? StuRat (talk) 17:46, 14 February 2014 (UTC)[reply]

I don't know, maybe? I just wanted to point out that relative and absolute errors are pretty different, and you have to think carefully about which one is more meaningful in a certain case. E.g. "doing X increases your risk of Y by 1000 times!" is often seen in popular health headlines. But if the original risk of Y is 10e-18, then the relative change in risk is not very meaningful, because 10e-15 is also pretty insignificant. I'm sure you know all this, I was just cautioning the OP to be careful in the interpretation of relative error. SemanticMantis (talk) 17:43, 16 February 2014 (UTC)[reply]

Yes, I imagine the chances of dying in a plane crash are around 1000 times higher if you actually fly in planes (with the lesser risk from being killed on the ground by a plane crash). StuRat (talk) 19:18, 16 February 2014 (UTC)[reply]

SemanticMantis' example is

   A 1e_15 1e_18
5.005e_16 4.995e_16
   G 1e_15 1e_18
3.16228e_17 31.6228

A computes the absolute error: the probabilities are 5⋅10^-16±5⋅10^-16. G computes the relative error: the probabilities are 3.2⋅10^-17⋅32^±1. Bo Jacoby (talk) 07:40, 17 February 2014 (UTC).[reply]