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I'm having difficulty in determining what exactly the difference is between the 2, especially when given an exercise and I have to choose which of the 2 to use. These is how my text book describes them:

Sum standard deviation

Given is a population with a normally distributed random variable $X$. When you have a sample $n$ from this population the population is:

$X_{sum} = X_1 + X_2 ... + X_n$ with

$\mu_{Xsum} = n \times \mu_x$ and $\sigma_{Xsum} = \sqrt{n} \times \sigma_x$.

Standard error

When you have a normally distributed random variable $X$ with mean $\mu_X$ and standard deviation $\sigma_X$ and sample length $n$, the sample mean $\bar{X}$ is normally distributed with $\mu_{\bar{x}} = \mu_X$ and $\sigma_{\bar{x}} = \dfrac{\sigma_X}{\sqrt{n}}$

These 2 are awefully similair to me to the point I can't at all decide which to use where. Here are the problems where I discovered I couldn't:

Problem 1

A filling machine fills bottles of lemonade. The amount is normally distributed with $\mu = 102 \space cl$.

$\sigma$ = $1.93\space cl$.

  • Calculate the chance that out of 12 bottles the average volume is $100 \space cl$.

The problem itself is easy, however the troublesome part is what to choose for the standard deviation of the sample. Here they use $\dfrac{1.93}{\sqrt{12}} $ which I can live with, until I encountered the second problem.

Problem 2

A tea company puts 20 teabags in one package. The weight of a teabag is normally distributed with $\mu = 5.3 \space g$ and $\sigma = 0.5 \space g.$

  • Calculate the chance that a package weighs less than 100 grams.

Here I thought they'd also use $\dfrac{0.5}{\sqrt{20}}$, but instead they use $\sqrt{20} \times 0.5$.

Can someone clear up the confusion?

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You should tag this as "homework" as well, since it seems to be a homework question. –  Placidia Jan 20 '13 at 17:33
3  
@Placidia Are you kidding me?! This isn't homework, this is about understanding and differentiating 2 general concepts in statistics, which then could be implemented in homework questions.. like every other mathematical concept.. –  JohnPhteven Jan 20 '13 at 17:36

2 Answers 2

up vote 1 down vote accepted

The sum standard deviation is, as the name suggests, the standard deviation of the sum of $n$ random variables. The standard error you're talking about is just another name for the standard deviation of the mean of $n$ random variables. As you noted, the two formulas are closely related; since the sum of $n$ random variables is $n$ times the mean of $n$ random variables, the standard deviation of the sum is also $n$ times the standard deviation of the mean:

$\sigma_{X_{sum}} = \sqrt n\sigma_X = n \times \frac{\sigma_X}{\sqrt n} = n\times \sigma_\bar{X}$.

In the first problem you are dealing with a mean, the average of twelve bottles, so you use the standard deviation of the mean, which is called standard error. In the second problem you are dealing with a sum, the total weight of 20 packages, so you use the standard deviation of the sum.

Summary: use standard error when dealing with the mean (averages); use sum standard deviation when dealing with the sum (totals).

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But what I think is that theyask one about the sum of 12 bottles, and the mean of that sum? In other words, they're too similair to me.. –  JohnPhteven Jan 20 '13 at 18:37
    
There's no sum in question one. Each bottle is filled with an amount given by a normal distribution with mean 102, the question asks about the mean of twelve bottles. Where do you see a sum? –  Jonathan Christensen Jan 20 '13 at 18:45
    
Oh wait nevermind, I was being a little bit blind! In the first one they ask about the MEAN (i.e. average) out of a sample, in the second they transform a sample of 6 into '1' object (>namely, the box of teabags), with its own SD and M! –  JohnPhteven Jan 20 '13 at 18:45
    
I think I was writing my response the same time you were doing yours. Nice answer. –  Placidia Jan 20 '13 at 18:46
1  
I meant 20, my brain is random, no idea how I got to 6 –  JohnPhteven Jan 20 '13 at 18:48

The first standard deviation formula you gave is the SD for a sum. The standard error is the SD of the sample mean. Remember that: $\text{Var}(aX)=a^2 \text{Var}(X)$ and the variance of the sum is the sum of the variances (First formula). So

$\text{Var}(\bar{X})=\frac{n\sigma^2}{n^2}=\sigma^2/n$. Taking the square root gives the result.

Recall:

$\text{Var}(\sum X_i)=\sum (\text{Var}(X_i)=n \sigma^2.$ The Variance of the sums.

Problem 1 is looking for a statement about the sample mean; Problem 2 is about the sum, since the weight of the package is the sum of the weights of individual tea bags.

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Nice answer, +1, but I gave the other one a best answer since I read it first and it answered my question first. –  JohnPhteven Jan 20 '13 at 18:48

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