Statistically compare an experimental data and a theory value - statistics

I appreciate if someone could answer my question. I have two values. The first one is experimental data deduced from only one measurement. The uncertainty for this values is determined. The second value is theory result. My question is how to statistically compare these two values? I tried to use t-test, but failed because the number of freedom is df = 1-1 =0 (only one experiment was conducted to measure the first value).

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Calculate risk using Cox model coefficients and mean values

I'm trying to understand the example presented in Appendix C here
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6481149/
Equation C1 is clear to me.
But in Equation C2 they use the mean values.
Such mean values are clear to me in the case of categorical variables for example 1.548 is the mean value of the Sex variable (as shown in the Table 3). Please correct me if I'm wrong.
But in numerical variables I don't understand which mean values are they using. For example for the Age variable they use 3.768, if I understand right, that value is the log of the mean age, should be log(44.15)=1.64. Instead the used value is 3.768.
Please could anybody clarify where does this value come from?
In statistics log often means the natural logarithm, sometimes denoted ln. The four values they take the logarithms of are:
Variable
Reported Mean
ln(Mean)
Reported
Age
44.15
3.788
3.768
BMI
25.61
3.243
3.230
BP Syst
138.6
4.932
4.913
Pulse Rate
75.61
4.326
4.311
The calculated values are not exactly equal to the reported values. But it looks close enough that this is probably the calculation they used. Without the data and/or code they used it's hard to say why the results are different. The study mentions excluding 130 participants because of ethics protections. So, perhaps one table was calculated using a slightly different group of participants than the other table?

stats, t-test, lists, and only one output

t-tests have been yielding just one output...but I want 10 t-tests.
Each t-test should compare each one of the 10 values in list to 0.
I have tried the below:
import scipy
from scipy import stats
list2=[0.10415380403918414, 0.09142102934943379, 0.08340408682911706, 0.07791383429638124, 0.0738177221067812, 0.07111840615962706, 0.0673345711222398, 0.06431875318226271, 0.06074216826770115, 0.052948996685723906]
print(scipy.stats.ttest_ind(list2,[0]*10))
Each t-test should compare each one of the 10 values in list to 0. that is, I should get 10 t-test comparison, so 10 t-tests should be outputted
All of this is to say: I am seeking 10 rows of output (each corresponding to a unique t-test, therefore I am seeking 10 t-tests), but the code I have now just provides me with one row output, i.e. just one test
listofzeros=[0,0,0,0,0,0,0,0,0,0]
for i in range(10):
print(scipy.stats.ttest_ind(list2,listofzeros))
Firstly, there is no need to use stats.ttest_ind and create a list of zeros with the same length as the sample. You just can use stats.ttest_1samp, as follows:
print(scipy.stats.ttest_1samp(list2,0,))
That will lead to the same result but without tweaking returning the r-static value and the p-value for the mean of the input sample not returning the results per value per sample.
To be more comprehensive, The t-test is used to determine whether the sample "mean" is statistically significantly different from the population "mean".
What you are trying to do is to perform a two-sample T-test which will work on the mean of the two lists, not on every two associated values of the two samples.

Comparing two contingency tables

I have two 6x4 contingency tables for frequency data. They are based on the same type of sampling criteria of a number of discreet variables but for two condition (before and after). I would like to compare these statistcally to see how much - or not - they differ.
A Chi square related test seems appropriate but normally this gives a result in comparison to the theoretical to calculate the statistic. So in other words I need to swap the theoretical for the second table. Of course it doesn't have to be a basic chi square test - any other appropriate test would be ok.
I have access to XLSTAT, Excel and SPSS. And would appreciate some help on this.

Compute statistical significance with Excel

I have 2 columns and multiple rows of data in excel. Each column represents an algorithm and the values in rows are the results of these algorithms with different parameters. I want to make statistical significance test of these two algorithms with excel. Can anyone suggest a function?
As a result, it will be nice to state something like "Algorithm A performs 8% better than Algorithm B with .9 probability (or 95% confidence interval)"
The wikipedia article explains accurately what I need:
http://en.wikipedia.org/wiki/Statistical_significance
It seems like a very easy task but I failed to find a scientific measurement function.
Any advice over a built-in function of excel or function snippets are appreciated.
Thanks..
Edit:
After tharkun's comments, I realized I should clarify some points:
The results are merely real numbers between 1-100 (they are percentage values). As each row represents a different parameter, values in a row represents an algorithm's result for this parameter. The results do not depend on each other.
When I take average of all values for Algorithm A and Algorithm B, I see that the mean of all results that Algorithm A produced are 10% higher than Algorithm B's. But I don't know if this is statistically significant or not. In other words, maybe for one parameter Algorithm A scored 100 percent higher than Algorithm B and for the rest Algorithm B has higher scores but just because of this one result, the difference in average is 10%.
And I want to do this calculation using just excel.
Thanks for the clarification. In that case you want to do an independent sample T-Test. Meaning you want to compare the means of two independent data sets.
Excel has a function TTEST, that's what you need.
For your example you should probably use two tails and type 2.
The formula will output a probability value known as probability of alpha error. This is the error which you would make if you assumed the two datasets are different but they aren't. The lower the alpha error probability the higher the chance your sets are different.
You should only accept the difference of the two datasets if the value is lower than 0.01 (1%) or for critical outcomes even 0.001 or lower. You should also know that in the t-test needs at least around 30 values per dataset to be reliable enough and that the type 2 test assumes equal variances of the two datasets. If equal variances are not given, you should use the type 3 test.
http://depts.alverno.edu/nsmt/stats.htm

Statistically removing erroneous values

We have a application where users enter prices all day. These prices are recorded in a table with a timestamp and then used for producing charts of how the price has moved... Every now and then the user enters a price wrongly (eg. puts in a zero to many or to few) which somewhat ruins the chart (you get big spikes). We've even put in an extra confirmation dialogue if the price moves by more than 20% but this doesn't stop them entering wrong values...
What statistical method can I use to analyse the values before I chart them to exclude any values that are way different from the rest?
EDIT: To add some meat to the bone. Say the prices are share prices (they are not but they behave in the same way). You could see prices moving significantly up or down during the day. On an average day we record about 150 prices and sometimes one or two are way wrong. Other times they are all good...
Calculate and track the standard deviation for a while. After you have a decent backlog, you can disregard the outliers by seeing how many standard deviations away they are from the mean. Even better, if you've got the time, you could use the info to do some naive Bayesian classification.
That's a great question but may lead to quite a bit of discussion as the answers could be very varied. It depends on
how much effort are you willing to put into this?
could some answers genuinely differ by +/-20% or whatever test you invent? so will there always be need for some human intervention?
and to invent a relevant test I'd need to know far more about the subject matter.
That being said the following are possible alternatives.
A simple test against the previous value (or mean/mode of previous 10 or 20 values) would be straight forward to implement
The next level of complexity would involve some statistical measurement of all values (or previous x values, or values of the last 3 months), a normal or Gaussian distribution would enable you to give each value a degree of certainty as to it being a mistake vs. accurate. This degree of certainty would typically be expressed as a percentage.
See http://en.wikipedia.org/wiki/Normal_distribution and http://en.wikipedia.org/wiki/Gaussian_function there are adequate links from these pages to help in programming these, also depending on the language you're using there are likely to be functions and/or plugins available to help with this
A more advanced method could be to have some sort of learning algorithm that could take other parameters into account (on top of the last x values) a learning algorithm could take the product type or manufacturer into account, for instance. Or even monitor the time of day or the user that has entered the figure. This options seems way over the top for what you need however, it would require a lot of work to code it and also to train the learning algorithm.
I think the second option is the correct one for you. Using standard deviation (a lot of languages contain a function for this) may be a simpler alternative, this is simply a measure of how far the value has deviated from the mean of x previous values, I'd put the standard deviation option somewhere between option 1 and 2
You could measure the standard deviation in your existing population and exclude those that are greater than 1 or 2 standard deviations from the mean?
It's going to depend on what your data looks like to give a more precise answer...
Or graph a moving average of prices instead of the actual prices.
Quoting from here:
Statisticians have devised several methods for detecting outliers. All the methods first quantify how far the outlier is from the other values. This can be the difference between the outlier and the mean of all points, the difference between the outlier and the mean of the remaining values, or the difference between the outlier and the next closest value. Next, standardize this value by dividing by some measure of scatter, such as the SD of all values, the SD of the remaining values, or the range of the data. Finally, compute a P value answering this question: If all the values were really sampled from a Gaussian population, what is the chance of randomly obtaining an outlier so far from the other values? If the P value is small, you conclude that the deviation of the outlier from the other values is statistically significant.
Google is your friend, you know. ;)
For your specific question of plotting, and your specific scenario of an average of 1-2 errors per day out of 150, the simplest thing might be to plot trimmed means, or the range of the middle 95% of values, or something like that. It really depends on what value you want out of the plot.
If you are really concerned with the true max and true of a day's prices, then you have to deal with the outliers as outliers, and properly exclude them, probably using one of the outlier tests previously proposed ( data point is x% more than next point, or the last n points, or more than 5 standard deviations away from the daily mean). Another approach is to view what happens after the outlier. If it is an outlier, then it will have a sharp upturn followed by a sharp downturn.
If however you care about overall trend, plotting daily trimmed mean, median, 5% and 95% percentiles will portray history well.
Choose your display methods and how much outlier detection you need to do based on the analysis question. If you care about medians or percentiles, they're probably irrelevant.

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