A sample of 15 observations has a standard deviation of 4. The sum of the squared deviations from the sample mean is?
1.) 19
2.)56
3.)60
4.) 224
5.)240
I'm stuck on this problem is you can please help me I would much appreciate it, thanx!
A sample of 15 observations has a standard deviation of 4. The sum of the squared deviations from the sample mean is?
1.) 19
2.)56
3.)60
4.) 224
5.)240
I'm stuck on this problem is you can please help me I would much appreciate it, thanx!
$\displaystyle s_x = \sqrt{\frac{1}{n-1} \sum_i (x_i - x_{\text{mean}})^2} $
$\displaystyle 4 = \sqrt{\frac{1}{15-1} \sum_i (x_i - x_{\text{mean}})^2} $
$\displaystyle 16 = \frac{1}{14} \sum_i (x_i - x_{\text{mean}})^2 $
$\displaystyle 16\cdot 14 = 224 = \sum_i (x_i - x_{\text{mean}})^2 $
The sample SD uses a divisor of N not N-1, the latter is the usual estimator for the population SD.
So this changes your post to:
$\displaystyle s_x = \sqrt{\frac{1}{n} \sum_i (x_i - x_{\text{mean}})^2} $
$\displaystyle 4 = \sqrt{\frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2} $
$\displaystyle 16 = \frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2 $
$\displaystyle 16\times 15 = 240 = \sum_i (x_i - x_{\text{mean}})^2 $
For a sample you must use a divisor of N-1 to have an unbiased estimation of the real standard deviation.
For a population you must use a divisor of N to get the real standard deviation.
http://en.wikipedia.org/wiki/Standar...dard_deviation
Does the question ask for an estimate of a population variance based on the sample variance .....? That would be a completely different question and using N - 1 as the divisor would be appropriate.
When calculating a sample variance in the context of plain and simple univariate statistics, a divisor of N is used. The wikipedia link says this.
No. Eg. You might have data from a test done by a class of 20 students and you're interested in the mean and the standard deviation ......
A key quote from the Wiki link:
"Note that if the above data set represented only a sample from a greater population, a modified standard deviation would be calculated (explained below) to estimate the population standard deviation, which would give 6.93 for this example."