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ggxoxo · « **on:** June 02, 2013, 01:12:17 pm »

Does E(X2)-[E(X)]^2 also work for sample variance or does it just work for population variance?

alondouek · « **Reply #1 on:** June 02, 2013, 01:19:12 pm »

I think sample variance might be better given by $(\frac{\sigma}{\sqrt{n}})^2$ .

ggxoxo · « **Reply #2 on:** June 03, 2013, 11:34:05 pm »

thank you- but why is the denominator merely n and not (n-1)

alondouek · « **Reply #3 on:** June 03, 2013, 11:44:45 pm »

This is only what I've learned from STA1010, but for a sample (rather than the population), standard deviation is given by $\frac{\sigma}_{\sqrt{n}}$ - sample variance is simply the square of that.

If your question is more conceptual than that, I don't think I can help

TrueTears · « **Reply #4 on:** June 04, 2013, 12:11:05 am »

Quote from: ggxoxo on June 02, 2013, 01:12:17 pm

Does E(X2)-[E(X)]^2 also work for sample variance or does it just work for population variance?

What's your question exactly?

If you assume the random variable X follows a probability density function of p(x) and this density function is an accurate characterization of the population frequency distribution, then $E(X^2) - [E(X)]^2 = \sigma^2$ where $\sigma^2$ is the population variance. If p(x) is not an accurate characterization then $E(X^2) - [E(X)]^2 \neq \sigma^2$ .

Now assume we take a random sample $X_1, X_2, \cdots, X_n$ from the population, because this is a random sample, they are iid (independent and identically distributed), that is $E(X_i) = \mu$ and $V(X_i) = \sigma^2$ . Now the question is how do we estimate $\sigma^2$ ? In other words, we need to find a point estimator. With that, we also need a criteria to assess the "goodness" of our point estimator, there are many favourable properties one wants to satisfy when obtaining a good "estimator", unbiasedness, consistency, sufficiency, efficiency etc. One can take a guess and estimate $\sigma^2$ with the following expression:

$S'^2 = \frac{1}{n}\sum_{i=1}^n (X_i- \bar{X})^2$

However $S'^2$ is itself a random variable, and it can be shown that $E(S'^2) \neq \sigma^2$ , ie, it is is a biased estimator. It can also be easily shown that the unbiased estimator of $\sigma^2$ is in fact given by:

$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i- \bar{X})^2$

that is $E(S^2) = \sigma^2$ .

We define $S^2$ to be the sample variance. In fact it can be easily shown that $S^2$ is not only unbiased, but also consistent, sufficient and efficient.

With regards to the above posts $\left(\frac{\sigma}{\sqrt{n}}\right)^2$ is not the sample variance. That expression is the variance of the sample mean (which is itself a function of random variables, hence why we can compute its variance).

ggxoxo · « **Reply #5 on:** June 04, 2013, 12:27:26 am »

Thank you sooo much TT; I am soo sorry though... but I still don't understand...

So with E(X^2)-[E(X)]^2 can that equal both population variance AND sample variance??? (so take for example, they only gave me a sample;if I have the mean of that sample and the E(X^2) can I use this formula?)

Also, with alondouek's formula, I just really want to clarify what you are saying... if we have a sample distribution, then the variance of all of those sample means will be the formula alondouek gave me right? (so it's not an alternative to the "normal" variance: sum(x-mean)/(n-1) right?)

TrueTears · « **Reply #6 on:** June 04, 2013, 12:34:16 am »

$E(X^2) - [E(X)]^2$ will only equal the population variance iff the underlying probability distribution of X is an accurate characterization of the population frequency distribution.

If not, then we will never know the population variance, we can only hope to estimate it.

The aforementioned expression has nothing to do with the sample variance. The sample variance (commonly denoted as $S^2)$ is defined to be the unbiased estimator of the population variance, that is:

$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i- \bar{X})^2$

The expression, $\left(\frac{\sigma}{\sqrt{n}}\right)^2$ is the variance of the random variable $\overline{X}$ .

ggxoxo · « **Reply #7 on:** June 04, 2013, 12:44:47 am »

Thanks to both of you! I think I got it now!

alondouek · « **Reply #8 on:** June 04, 2013, 12:56:20 am »

My bad, thanks for clearing that up TT!

ggxoxo · « **Reply #9 on:** June 04, 2013, 05:30:59 pm »

Suppose we played the coin tossing game at Casino A 8,000 times and found we had 5,000 tails. What would be the test statistic associated with this information?

Note 1: under the null of a fair coin, the average number of heads is µ = np, with variance σ^2 = np(1-p), where n is the number of coin tosses and p the probability of a head.

Note 2: assume a Normal approximation, eg X~N(µ, σ^2).
Select one:
a. Z-stat = -22.36
b. Z-stat = -0.5
c. Z-stat = 22.36
d. Z-stat = 0.5
e. No answers are correct
Feedback
The correct answer is: Z-stat = -22.36

Hi; can someone please help me with this problem?

TChen · « **Reply #10 on:** June 05, 2013, 12:17:53 am »

Use Z=(result-u)/SD
We use the given formulas to work out u and SD.
note: n=8000 and p=0.5 (given in the question). Also, question is relating to the number of heads and this is 3000 (since there were 5000 tails).

b^3 · « **Reply #11 on:** June 05, 2013, 12:34:54 am »

We will apply the normal approximation to the binomial here. That is we have a sample proportion, $\hat{p}$ . Now if we have $X$ following a binomial distribution above, then we will have $X~\text{Bi}\left(8000,\frac{1}{2}\right)$ , since $n$ is large, we can approximate this by applying the central limit theorem, so we can approximate it by a normal distribution.
$\begin{alignedat}{1}X & \sim N\left(np,\sqrt{np\left(1-p\right)}\right) \\ X & \sim N\left(4000,20\sqrt{5}\right) \end{alignedat}$

Now the Z-stat we want here is for the sample proportion. $\begin{alignedat}{1}\hat{p} & =\frac{X}{n}\end{alignedat}$ which means that the distribution for the sample proportion will be
$\begin{alignedat}{1}\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} & \sim N\left(0,1\right)\text{\:\text{\ensuremath{\left(approximately\right)}}}\end{alignedat}$
That is our Z-stat is
$\begin{alignedat}{1}\hat{p} & =\frac{3000}{8000} \\ & =\frac{3}{8} \\ \sqrt{\frac{p\left(1-p\right)}{n}} & =\sqrt{\frac{\frac{1}{2}\left(\frac{1}{2}\right)}{8000}} \\ & =\frac{\sqrt{5}}{400} \\ \frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} & =\frac{\frac{3}{8}-\frac{1}{2}}{\frac{\sqrt{5}}{400}} \\ & =-10\sqrt{5} \\ & =-22.36 \end{alignedat} $

EDIT: Note, your unit looks like they use $N(\mu,\sigma^{2})$ while my stats unit uses $N(\mu,\sigma)$ , which is why in my working I had just $\sigma$ , so just take note that you will need to square that value for your working.

ggxoxo · « **Reply #12 on:** June 05, 2013, 03:45:58 am »

Quote from: b^3 on June 05, 2013, 12:34:54 am

We will apply the normal approximation to the binomial here. That is we have a sample proportion, $\hat{p}$ . Now if we have $X$ following a binomial distribution above, then we will have $X~\text{Bi}\left(8000,\frac{1}{2}\right)$ , since $n$ is large, we can approximate this by applying the central limit theorem, so we can approximate it by a normal distribution.
$\begin{alignedat}{1}X & \sim N\left(np,\sqrt{np\left(1-p\right)}\right) \\ X & \sim N\left(4000,20\sqrt{5}\right) \end{alignedat}$

Now the Z-stat we want here is for the sample proportion. $\begin{alignedat}{1}\hat{p} & =\frac{X}{n}\end{alignedat}$ which means that the distribution for the sample proportion will be
$\begin{alignedat}{1}\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} & \sim N\left(0,1\right)\text{\:\text{\ensuremath{\left(approximately\right)}}}\end{alignedat}$
That is our Z-stat is
$\begin{alignedat}{1}\hat{p} & =\frac{3000}{8000} \\ & =\frac{3}{8} \\ \sqrt{\frac{p\left(1-p\right)}{n}} & =\sqrt{\frac{\frac{1}{2}\left(\frac{1}{2}\right)}{8000}} \\ & =\frac{\sqrt{5}}{400} \\ \frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} & =\frac{\frac{3}{8}-\frac{1}{2}}{\frac{\sqrt{5}}{400}} \\ & =-10\sqrt{5} \\ & =-22.36 \end{alignedat} $

EDIT: Note, your unit looks like they use $N(\mu,\sigma^{2})$ while my stats unit uses $N(\mu,\sigma)$ , which is why in my working I had just $\sigma$ , so just take note that you will need to square that value for your working.

Thank you so much!

I just have some questions:

(1) how did you know the Z-stat was for the sample distribution? I was trying to find another "value"

(2) how did you know that 1/2 was the "mean" in the numerator?

Thanks!

TrueTears · « **Reply #13 on:** June 05, 2013, 03:53:10 am »

^ I agree, the question isn't actually worded very well, but the question is clearly testing whether you know the central limit theorem, so in that sense, the z statistic refers to the sample mean.

(2) from mu = np

ggxoxo · « **Reply #14 on:** June 10, 2013, 10:15:22 pm »

Hi can someone please help me with (c) (ii) and (c) (iii)

With c(ii) I don't see how we can solve for the height without being given income (we can't just omit income from the model right?)

For (c) (iii) I just want to double check... would you just do a hypothesis test? With H0: age coefficients are the same and H1: age coefficients not the same??? But then I don't know any critical values or which t-stat to use, if that makes sense...

Are there any answers to ETC1000 practice exams???

Thank you!

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