Search results
Results from the WOW.Com Content Network
and the updated standard deviation would be just the square root. If you want the population variance or standard deviation replace N-1 with N and N-2 with N-1. Share
This is the formula for the 'pooled standard deviation' in a pooled 2-sample t test. If we may have two samples from populations with different means, this is a reasonable estimate of the (assumed) common population standard deviation $\sigma$ of the two samples.
The formula you cite is for the standard deviation of the ... to find the sample standard deviation. The ...
In order to "get the sample standard deviation," you need to specify a sample (a subset of the population). If you do not specify a sample, then you cannot get the sample standard deviation. If you do specify the sample, then you can get the sample standard deviation. In either case, knowledge of the population standard deviation is irrelevant.
Then from there to find the standard deviation i would use: srqroot(Sxx/n-1) ... Standard deviation from ...
At 4:30 of this video the author decided to estimate the standard deviation of the population with sample standard deviation (sample size was $100$). In the next video, the author mentioned that it was reasonable because the sample size greater than $30$.
I have an HP 50g graphing calculator and I am using it to calculate the standard deviation of some data. In the statistics calculation there is a type which can have two values: Sample Population I
The formula for standard deviation is sqrt([sample size][probability of success](1-[probability of success])). To find the sample size from the mean and success rate, you divide the mean by the success rate. If mean=10 and success=0.2, you do 10/0.2 to get your sample size, or 50 in this case. Then you do sqrt(50*0.2*(1-0.2) to get about 2.83.
Assume that the population standard deviation is $5$ cl. A sample of 100 bottles yields to an average of $48$ cl. Calculate a $90\%$ and $95\%$ confidence interval for the average content. Suppose the sample size is unknown. What sample size is required to estimate the average contents to be within $0.5$ cl at the $95\%$ confidence level?
I need the standard deviation. I generate about a hundred points and take the standard deviation of the points. This gives a hopefully good approximation of the true standard deviation, but it won't, of course, be exact. How do I estimate the uncertainty in the standard deviation?