1. ## stats

For a sample size of $n = 10$ I simulated the roll of a dice with $k = 200$ replications.

My sample mean and variance of the mean were $\bar{x} = 1.131, \ \ s_{\bar{x}}^{2} = 1.28$

And $\mu = 3.5, \ \ \sigma^{2} = 2.917, \ \ \ \frac{\sigma^2}{10} = 0.2917$.

$s_{\bar{x}}^{2} > 0.2917$. Is this because my sample size $n = 10$ was too small, and thats why they differ by a lot?

2. Originally Posted by heathrowjohnny
For a sample size of $n = 10$ I simulated the roll of a dice with $k = 200$ replications.

My sample mean and variance of the mean were $\bar{x} = 1.131, \ \ s_{\bar{x}}^{2} = 1.28$

And $\mu = 3.5, \ \ \sigma^{2} = 2.917, \ \ \ \frac{\sigma^2}{10} = 0.2917$.

$s_{\bar{x}}^{2} > 0.2917$. Is this because my sample size $n = 10$ was too small, and thats why they differ by a lot?
It is not clear what your experiment was from what you write.

What is the outcome of a replication? what are your $\bar{x}$ and $s_{\bar{x}}^{2}$ mean and variance of?

RonL

3. Basically I simulated the rolling of a dice 30 times, and repeated this 200 times. $\bar{x}$ is the average of the averages from each replication. $s_{\bar{x}}^{2}$ is the variance of those averages

4. Originally Posted by heathrowjohnny
Basically I simulated the rolling of a dice 30 times, and repeated this 200 times. $\bar{x}$ is the average of the averages from each replication. $s_{\bar{x}}^{2}$ is the variance of those averages

Suppose you have a fair die, and you roll it 10 times and compute the mean of the rolls.

On a single roll the mean is 3.5 and the sd is ~1.708. So for the mean of 10 rolls the mean is 3.5 and the se of that mean is ~1.708/sqrt(10) ~=0.540

Now you conduct 200 replications, and the grand mean will be 3.5 and the se of such a grand mean will be ~=0.540/sqrt(200) ~=0.0382.

Therefore if the die is fair the mean of your sample of 200 means should be in the interval [3.42, 3.58] ~95% or the time (200 being large enough for us to use a normal approximation). Your grand mean is not in this interval so we may conclude that it is likley that the die is not fair.

You can also compare the experimental variance 1.28 with the fair die prediction of (0.540)^2~=0.292 using the F test.

RonL
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