The confidence interval is given by this formula
As you can see, if we want the CI to be divided by 3 with the sample size as a parameter, we thus need to multiply N by 9, it being under the square root.
I used a similar formula but i am a bit worried since we didn't see this in the slides- what other formulas are we expected to know that weren't explicitly reviewed this semester?
In the first exercise sheet, it is mentioned that having a basic understanding of probabilities and statistics are needed for the course, along with links to the material cited.
We've seen this in the slides of Generalisation. The probability of the error being grater than a a certain fraction (with sqrt 1/n) is smaller than the confidence interval. Without proof it says in below that delta grows with 1/|S_test|
I feel unconfortable with that problem. Because the CI is proportional to the std (here number of element by sample is fixed). When the number of sample tends to infinite the std will converge to the std of the distribution. So the CI will tend to a value different from zero. So we cannot divide the CI by 3 after a N enough big... Do you see my point?
Question 1 confidence intervals
Hi,
Can anyone explain as to why we need 9N samples for the CI to shrink to sigma/3?
I can't get this.
Thanks in advance.
The confidence interval is given by this formula
As you can see, if we want the CI to be divided by 3 with the sample size as a parameter, we thus need to multiply N by 9, it being under the square root.
1
I used a similar formula but i am a bit worried since we didn't see this in the slides- what other formulas are we expected to know that weren't explicitly reviewed this semester?
1
In the first exercise sheet, it is mentioned that having a basic understanding of probabilities and statistics are needed for the course, along with links to the material cited.
Link here : https://github.com/epfml/ML_course/blob/master/labs/ex01/exercise01.pdf
EDIT: note that I am not a TA, and don't actually know what will be on the exam and what will not :)
We've seen this in the slides of Generalisation. The probability of the error being grater than a a certain fraction (with sqrt 1/n) is smaller than the confidence interval. Without proof it says in below that delta grows with 1/|S_test|
2
I feel unconfortable with that problem. Because the CI is proportional to the std (here number of element by sample is fixed). When the number of sample tends to infinite the std will converge to the std of the distribution. So the CI will tend to a value different from zero. So we cannot divide the CI by 3 after a N enough big... Do you see my point?
1
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