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95% confidence interval

 
 
Reply Tue 2 Mar, 2010 08:47 am
I am currently producing an audit project in a hospital. However, I have reached a small amount of a brick wall. I am working out the 95% CI of the standards I have measured against, but I am unsure of what to do in order to work out the 95% CI of something that was measured as 100% since I am sure that slight differences will exist in the process due to differences in where you are measuring in the distribution.
 
View best answer, chosen by jonnypdavies
High Seas
 
  1  
Reply Tue 2 Mar, 2010 09:39 am
@jonnypdavies,
Are you modeling a process distributed according to a Weibull distribution? If so, check your data fit according to this link:
Quote:
Reliability analysis (or survival analysis) is a branch of statistics that deals with death in biological organisms and failure in mechanical or electronics systems, which may be within a localized test time. This topic is called reliability theory or reliability analysis in engineering, and duration analysis or duration modeling in economics or sociology.

http://demonstrations.wolfram.com/FittingLifetimeDataToAWeibullModel/
If it's a different distribution, try the links on the same page. If you're unsure of the distribution, start testing confidence intervals here:
http://demonstrations.wolfram.com/ConfidenceIntervalExploration/
0 Replies
 
Brandon9000
  Selected Answer
 
  3  
Reply Tue 2 Mar, 2010 09:54 am
@jonnypdavies,
jonnypdavies wrote:

I am currently producing an audit project in a hospital. However, I have reached a small amount of a brick wall. I am working out the 95% CI of the standards I have measured against, but I am unsure of what to do in order to work out the 95% CI of something that was measured as 100% since I am sure that slight differences will exist in the process due to differences in where you are measuring in the distribution.

To calculate a confidence interval for a mean, you need to have two things - the data points, and a knowledge of the distribution of the underlying random variable. In general, phenomena which are the result of many factors, none of which dominates, can be assumed to be Normally distributed, in which case you use Student's T-distribution for your calculation.
High Seas
 
  1  
Reply Tue 2 Mar, 2010 12:52 pm
@Brandon9000,
Whoever picked Student's T distribution must somehow be certain that the original poster has no clue at to the variance of his data >
Quote:
Student's t-distribution is defined as the distribution of the random variable t which is (very loosely) the "best" that we can do not knowing sigma

http://mathworld.wolfram.com/Studentst-Distribution.html
> for which no basis is given in the original query. Who decided that, and on what basis, exactly? Naturally I'm assuming the moderator is an expert in statistical applications for medical data generated by hospitals ...
engineer
 
  2  
Reply Tue 2 Mar, 2010 02:07 pm
@jonnypdavies,
Welcome to A2K! If you can give us some more insight into what you are measuring, we might be able to provide more help. What type data are you collecting? Why are you trying to calculate the confidence interval?
0 Replies
 
Brandon9000
 
  1  
Reply Tue 2 Mar, 2010 02:38 pm
@High Seas,
High Seas wrote:

Whoever picked Student's T distribution must somehow be certain that the original poster has no clue at to the variance of his data >
Quote:
Student's t-distribution is defined as the distribution of the random variable t which is (very loosely) the "best" that we can do not knowing sigma

http://mathworld.wolfram.com/Studentst-Distribution.html
> for which no basis is given in the original query. Who decided that, and on what basis, exactly? Naturally I'm assuming the moderator is an expert in statistical applications for medical data generated by hospitals ...

Student's T-distribution is the only correct distribution to use to find a confidence interval for the mean of a Normally distributed random variable. Fisher's F-distribution, if I recall correctly, is the distribution to use to find a confidence interval for the variance of a Normally distributed random variable.

Whether his variable actually is Normally distributed is another question, but I mentioned this, without knowing his situation, because the Central Limit Theorem states that every random variable with a value made up from many factors, none of which dominates, approaches a normal distribution as the number of factors increases.
High Seas
 
  1  
Reply Tue 2 Mar, 2010 03:20 pm
@Brandon9000,
Yes - am in agreement with both you and with engineer: the original poster didn't provide enough data, as my original post here makes clear in providing 2 different options. The first option is habitually used for risk modeling of medical data, the second is the same as yours and engineer's idea. My subsequent query was simply addressed to the nameless individual in this forum affixing ribbons to statistical distribution suggestions - I wanted to know the basis for his or her selection. Here's another link:
http://demonstrations.wolfram.com/ComparingNormalAndStudentsTDistributions/
0 Replies
 
 

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