Question 1197336: We are given a Data Set with n=10 entries in it. Which one of the following two ways will produce a larger result for the Standard Deviation?
1) If we compute the Standard Devation of this Data Set using the formula for the Population.
2) If we compute the Standard Devation of this Data Set using the formula for Samples.
Can you explain why this way produces a larger result?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: (2) If we compute the Standard Deviation of this Data Set using the formula for Samples.
Reason:
The process of calculating the standard deviation follows this outline- Compute the sample mean (xbar). Recall this sub-process is where we add up all the values and divide by the sample size n.
- Subtract xbar from each data item
- Square each result from the previous step
- Add up the squares to get the Sum of the Squared Error (SSE)
- Divide the SSE by n if you want the population variance OR divide SSE by (n-1) if you want the sample variance.
- Apply the square root to the variance to get the standard deviation. Standard deviation = sqrt(variance)
Steps 1 through 4 handle computing the SSE.
Then there's a bit of branching depending if you use the population or sample.
population variance = SSE/n
sample variance = SSE/(n-1)
The sample variance involves a smaller denominator, so the overall fraction will be larger compared to SSE/n
It's the same idea as to how 1/9 > 1/10 for example.
Therefore, the sample standard deviation is the larger of the two items.
A numeric example:
Consider this set of n = 10 values
{1,2,3,4,5,6,7,8,9,10}
Use the outline mentioned above, or a calculator, to get the following- population variance = 8.25
- sample variance = 9.167 (approximate)
We see in this example, the sample variance is larger.
By extension, the sample standard deviation will be larger compared to the population standard deviation
You can use your calculator to find that- population standard deviation = sqrt(8.25) = 2.8723
- sample standard deviation = sqrt(9.167) = 3.0277
each result is approximate.
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