What is the correct interpretation of the standard error of the sample mean?

[Addez123]

Homework Statement

X is the amount of calls a phone will recieve each day.
$$X \in Po(u)$$
For n days the following amount of calls were recorded
115, 82, 108, 106, 118, 87, 99, 92

1. Calculate μ.
2. Estimate the mean error of μ.

Relevant Equations

$$μ* = (x1 + x2 + x3..+xn)/n$$
$$D(μ*) = \sigma/\sqrt n$$
$$\sigma ~= s$$
$$s^2 = \frac 1 {n-1} * \sum_{j=1}^n {x_j - μ*}$$

1. μ* is easily calculated to be 100.88, textbook confirms its 100.9.
Its just the mean of all those values.

2. To estimate mean error I first calculate s^2, which is:
$$s^2 = 1/7 * ((115 - 100.9)^2 + (82 - 100.9)^2 ... + (92-100.9)^2) = 1200.88$$
$$s = \sqrt s = 34.65$$
Then I use this formula except instead of sigma I use s
$$D(μ*) = \sigma/\sqrt n ~= s/\sqrt n = 34.65/\sqrt 8 = 12.25$$
The book says the answer is 3.55, which first of all doesn't feel resonable either.

Shouldnt most numbers (66%) lies within the scope of μ +- s? (88.65 - 113.15, if s = 12.25)?
As that is the definition of a standard deviation.

 

[hutchphd]

Redo (2), it cannot be correct.

 

[Mark44]

$$s^2 = \frac 1 {n-1} * \sum_{j=1}^n {x_j - μ*}$$

Your formula for the sample variance is wrong, but I suspect this is a typo. Each term in the summation should be squared, like so:
$$s^2 = \frac 1 {n-1} * \sum_{j=1}^n (x_j - μ*)^2$$

$$s^2 = 1/7 * ((115 - 100.9)^2 + (82 - 100.9)^2 ... + (92-100.9)^2) = 1200.88$$

You have a typo here, also. The number you calculated is just the sum of the squared terms, without being multiplied by 1/7.

After the multiplication by 1/7, I get $s^2 \approx 171.6$, from which the variance s is a bit larger than 13.

$$s = \sqrt s = 34.65$$

The line above is missing an exponent. It should be $s = \sqrt{s^2} \dots$

With all that being said, you aren't using the fact that this is a Poisson distribution. The mean is calculated as you did it, but what are the variance and standard deviation in a Poisson distribution?

If I calculate the Poisson variance, and divide by sqrt(8), I get the book's answer.

 

[uart]

As already pointed out, you've made a simple arithmetic error somewhere in your calculation of the standard deviation.

Further, it seems that the question might be asking for something other than just the standard deviation of the distribution of the sample mean. While we might sometimes loosely refer to the standard deviation as an estimate of the error in a measurement, it is of course actually the square root of the mean squared error.

Technically the mean (signed) error in the sample mean is always zero, but it seems like this question may be asking for the mean absolute error (of the sample mean).

My approach would be to assume that the mean of the 10 Poisson random variable was close enough to normal to take the absolute error as approx a half normal distribution with \sigma = s/\sqrt{n}, and find it's mean.

 

[uart]

If I calculate the Poisson variance, and divide by sqrt(8), I get the book's answer.

Good catch Mark, that's probably what they are looking for then. My approach gets an answer of 3.69, which is approx 4% higher than the book answer.

 

[hutchphd]

My approach would be to assume that the mean of the 10 Poisson random variable was close enough to normal to take the absolute error...

The issue here is the omniscient mean. If for some reason God tells you the actual mean of your distribution, you divide the sum of the calculated variances by n. This is seldom true (at least for agnostic me). If instead you estimate the mean from the data then you have a potentially biased mean and you must divide by n-1. This represents that you have reduced the number of independent data by one because of the mean.
So here you would use n-1 for any distribution. (Usually I just use n anyway because if it really makes a difference your data set is pretty danged small...)

 

[uart]

The issue here is the omniscient mean. If for some reason God tells you the actual mean of your distribution, you divide the sum of the calculated variances by n. This is seldom true (at least for agnostic me). If instead you estimate the mean from the data then you have a potentially biased mean and you must divide by n-1

Yes, you need to use n-1 to calculate the sample variance and standard deviation (estimate of \sigma), but the standard deviation of the sample mean is \sqrt{n} times less than the standard deviation of the individual initial data, and here we really do need to use n rather than n-1.

The main difference between Mark's approach and mine is that the Poisson distribution allows you to imply the standard deviation directly from the mean, rather than actually calculate the sample standard deviation.

 

[hutchphd]

Yes, you need to use n-1 to calculate the sample variance and standard deviation (estimate of σ), but the standard deviation of the sample mean is n times less than the standard deviation of the individual data samples, and this term really is n rather than n-1.

I have no idea what you are talking about.

There are 8 samples here and the OP is estimating the $\sigma$. There is no estimate of the standard deviation of the "individual" data. Only if they were themselves averaged would your statement make sense.

So he divides by n-1

 

[uart]

I have no idea what you are talking about.

There are 8 samples here and the OP is estimating the $\sigma$. There is no estimate of the standard deviation of the "individual" data. Only if they were themselves averaged would your statement make sense.

So he divides by n-1

Yes, to find the standard deviation (hence estimate \sigma) of the sampled data (x), that's correct.

However the sample mean itself is also a random variable, and that random variable has a \sigma of 1 / \sqrt{n} times that of x.

\sigma_{\bar{x}} = \sigma_{x} / \sqrt{n}

BTW. Sorry, I meant to say "initial data" rather than "individual data points", will edit to make that clearer.

 

[hutchphd]

But the OP only made one estimate of the sample mean so $\sigma_{sample mean}=\sigma_x$ because N=1. Not to be unkind, but the comment is correct but irrelevant and slightly confusing.
Had he repeated the experiment, it would be both true and relevant.

 

[uart]

But the OP only made one estimate of the sample mean so $\sigma_{sample mean}=\sigma_x$ because N=1.

That is incorrect. The relevant value of n to use when calculating the standard deviation of the sample mean is the number of data points in the sample.

Just to be clear, what I'm saying is that the relevant value of n to use in the following equation is n=8
\sigma_{\bar{x}} = \sigma_{x} / \sqrt{n}

 

[hutchphd]

That is incorrect. The relevant value of n to use when calculating the standard deviation of the sample mean is the number of data points in the sample.

Just to be clear, what I'm saying is that the relevant value of n to use in the following equation is n=8
\sigma_{\bar{x}} = \sigma_{x} / \sqrt{n}

What is $\sigma_x$ in your equation and how does the OP calculate it from his data ?

 

[hutchphd]

Oh I see what you mean...yes we agree. Sorry. They do seem to want the error in the mean.

 

[Addez123]

First of all, sorry for all the typos in the description. I was really tired when I wrote that.

Secondly,

With all that being said, you aren't using the fact that this is a Poisson distribution. The mean is calculated as you did it, but what are the variance and standard deviation in a Poisson distribution?

Ok so I calculated variance by the poison defintion of variance: V(X) = μ
That makes
$$D(μ*) = \frac {\sqrt μ}{\sqrt 8} = 3.55$$
which is the correct answer.

Thanks everyone for your help!

 

[uart]

The book says the answer is 3.55, which first of all doesn't feel resonable either.

Shouldnt most numbers (66%) lies within the scope of μ +- s?

Good that you now have the right answer, but I feel that you may still be unclear on what it means.

The quoted interpretation is not correct. The value you are calculating is the standard error of the sample mean, not that of the data points. That is, there is about a 68% probability that the true mean (\mu) will be within the range 100.88 +/- 3.55

As for your first attempt, if not for the arithmetic errors then the result that you would have got using the sample standard deviation was 4.63. This is actually also a valid estimate (of the above), however it as not as reliable as the value found by deducing the standard deviation from the distribution and mean (3.55).