Any one good with standard deviations please help

**jax760** · January 16th 2009, 05:21 PM

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!

**vincisonfire** · January 16th 2009, 05:42 PM

$s_x = \sqrt{\frac{1}{n-1} \sum_i (x_i - x_{\text{mean}})^2}$
$4 = \sqrt{\frac{1}{15-1} \sum_i (x_i - x_{\text{mean}})^2}$
$16 = \frac{1}{14} \sum_i (x_i - x_{\text{mean}})^2$
$16\cdot 14 = 224 = \sum_i (x_i - x_{\text{mean}})^2$

**Constatine11** · January 17th 2009, 01:32 AM

Originally Posted by vincisonfire

$s_x = \sqrt{\frac{1}{n-1} \sum_i (x_i - x_{\text{mean}})^2}$
$4 = \sqrt{\frac{1}{15-1} \sum_i (x_i - x_{\text{mean}})^2}$
$16 = \frac{1}{14} \sum_i (x_i - x_{\text{mean}})^2$
$16\cdot 14 = 224 = \sum_i (x_i - x_{\text{mean}})^2$

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:

$s_x = \sqrt{\frac{1}{n} \sum_i (x_i - x_{\text{mean}})^2}$

$4 = \sqrt{\frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2}$

$16 = \frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2$

$16\times 15 = 240 = \sum_i (x_i - x_{\text{mean}})^2$

**vincisonfire** · January 17th 2009, 04:34 AM

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

**mr fantastic** · January 17th 2009, 04:53 AM

Originally Posted by vincisonfire

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.
Standard deviation - Wikipedia, the free encyclopedia

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.

**vincisonfire** · January 17th 2009, 05:10 AM

Isn't it always the case that we are trying to figure out what the standard deviation of the population is?
I may be wrong but I've never seen a sample variance written with a divisor of N.

**mr fantastic** · January 17th 2009, 05:17 AM

Originally Posted by vincisonfire

Isn't it always the case that we are trying to figure out what the standard deviation of the population is?
[snip]

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."

**vincisonfire** · January 17th 2009, 05:19 AM

I understand what you mean.
But as I learned it, in your example, the 20 students would be a population.

**mr fantastic** · January 17th 2009, 01:45 PM

Originally Posted by vincisonfire

I understand what you mean.
But as I learned it, in your example, the 20 students would be a population.

So then you're not estimating the variance of a population from a sample, you're actually getting the 'exact' varience of the population.

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