I implemented my mathematical model using Ilog Cplex ver 2.7. the decimal part of the objective function is very small and cplex returns 0 so the cplex abandons part of the objective function (so the objective function is not really optimized). Is there a way to increase the accuracy so that the cplex takes the max of decimal into account?
I have created a file ops to change the decimal prision from 4 to 10 but cplex always does not take into account the figures after the decimal point for you well understands to see the image below.?
In that part you cannot change the display precision. But as said in this post you may see more in the statistics tab.
Or you may use scripting to get any precision you need.
Related
I'm trying to do a simple comparison of two samples to determine if their means are different. Regardless of whether their standard deviations are equal/unequal, the formulas for a t-test or z-test are similar.
(i can't post images on a new account)
t-value w/ unequal variances:
https://www.biologyforlife.com/uploads/2/2/3/9/22392738/949234_orig.jpg
t-value w/ equal/pooled variances:
https://vitalflux.com/wp-content/uploads/2022/01/pooled-t-statistics-300x126.jpg
The issue here is the inverse and sqrt of sample size in the denominator that causes large samples to seem to have massive t-values.
For instance, I have 2 samples w/
size: N1=168,000 and N2=705,000
avgs: X1=89 and X2=49
stddev: S1=96 and S2=66 .
At first glance, these standard deviations are larger than the mean and suggest a nonhomogeneous sample with a lot of internal variation. When comparing the two samples, however, the denominator of the t-test becomes approx 0.25, suggesting that a 1 unit difference in means is equivalent to 4 standard deviations. Thus my t-value here comes out to around 160(!!)
All this to say, I'm just plugging in numbers since I didn't do many of these problems in advanced stats and haven't seen this formula since Stats110.
It makes some sense that two massive populations need their variance biased downward before comparing, but this seems like not the best test out there for the magnitude of what I'm doing.
What other tests are out there that I could try? What is the logic behind this seemingly over-biased variance?
I am stuck in a statistics assignment, and would really appreciate some qualified help.
We have been given a data set and are then asked to find the 10% with the lowest rate of profit, in order to decide what Profit rate is the maximum in order to be considered for a program.
the data has:
Mean = 3,61
St. dev. = 8,38
I am thinking that i need to find the 10th percentile, and if i run the percentile function in excel it returns -4,71.
However I tried to run the numbers by hand using the z-score.
where z = -1,28
z=(x-μ)/σ
Solving for x
x= μ + z σ
x=3,61+(-1,28*8,38)=-7,116
My question is which of the two methods is the right one? if any at all.
I am thoroughly confused at this point, hope someone has the time to help.
Thank you
This is the assignment btw:
"The Danish government introduces a program for economic growth and will
help the 10 percent of the rms with the lowest rate of pro t. What rate
of pro t is the maximum in order to be considered for the program given
the mean and standard deviation found above and assuming that the data
is normally distributed?"
The excel formula is giving the actual, empirical 10th percentile value of your sample
If the data you have includes all possible instances of whatever you’re trying to measure, then go ahead and use that.
If you’re sampling from a population and your sample size is small, use a t distribution or increase your sample size. If your sample size is healthy and your data are normally distributed, use z scores.
Short story is the different outcomes suggest the data you’ve supplied are not normally distributed.
so here is my situation. I have to solve a math problem on server end and could expect tens of thousands of requests a second so I'm trying to find the most efficient path to solving the problem.
Client will submit some number, let's call it A, and I need to determine base of the exponent in a geometric series (see below) where the result will be as close to A as possible without exceeding it.
The problem is that in the real-world, each value of the geometric series is rounded, so standard math can't apply.
round(x^1)+round(x^2)+round(x^3).
I can use the partial sum of geometric series equation to find some rough upper and lower limits using:
((x)^(n+1)-1)/((x)-1)
So say x=2 is a lower limit and x=2.03 is an upper limit... and the value i'm solving for is x=2.02392372838123.
So far the only solution i found was to use a recursive function to go through decimals individually testing until I find the number, but the load on the server is too high at the volume of requests I expect. (I am using node.js).
Does anyone have any thoughts or suggestions on a more efficient way to solve this? Again the only reason I can't solve this with math alone (to the best of my skill) is because of the real-world rounding of numbers in the sum.
Thanks.
In Maple, there is some feature that allows you to calculate the pdf of a function of a random variable. For example, if X is exponentially distributed, and you want to know the distribution of X^2, then there is a function that will do that for you.
My question is , is there a functionality in matlab that allows you to do so? I have looked through the matlab's guide, but I didn't see it.
The Statistics toolbox includes many probability distributions for you to choose from, both parametric and non-parametric distributions. For each it provides functions for PDF, CDF, fitting, random number generation, etc..
I suggest you start with the "Distribution Fitting app": dfittool.
EDIT:
In addition, MuPAD has support for a number of distributions, which you can manipulate symbolically. Example:
The function intlib::changevar might be of interest here, though it seems intended for integrals...
Also, if you're interested in getting the values of the PMF, or discrete PDF, then, given x some RV with some distribution,
my_pmf = hist(x)/sum(x);
So try,
doc hist
I'm trying to find confidence intervals for the means of various variables in a database using SPSS, and I've run into a spot of trouble.
The data is weighted, because each of the people who was surveyed represents a different portion of the overall population. For example, one young man in our sample might represent 28000 young men in the general population. The problem is that SPSS seems to think that the young man's database entries each represent 28000 measurements when they actually just represent one, and this makes SPSS think we have much more data than we actually do. As a result SPSS is giving very very low standard error estimates and very very narrow confidence intervals.
I've tried fixing this by dividing every weight value by the mean weight. This gives plausible figures and an average weight of 1, but I'm not sure the resulting numbers are actually correct.
Is my approach sound? If not, what should I try?
I've been using the Explore command to find mean and standard error (among other things), in case it matters.
You do need to scale weights to the actual sample size, but only the procedures in the Complex Samples option are designed to account for sampling weights properly. The regular weight variable in Statistics is treated as a frequency weight.