Author Topic: Statistics Help (Read 1616 times) Tweet Share

chenchenpizza · « **on:** September 17, 2016, 02:23:17 pm »

Hi, so my teacher hasn't been very helpful in covering stats, and I've been struggling to understand almost everything from stats. I know it's a big ask, but can someone make a guide for statistics to help a struggling student? I'm so confused with the p's and obscure explanation of the normal approx. to the binomial distribution in the textbook. My teacher won't cover it in class because he said it was already covered in methods. But even then, I still can't get my head around the sample jargon and simple terminology of stats

RuiAce · « **Reply #1 on:** September 17, 2016, 03:33:37 pm »

Quote from: chenchenpizza on September 17, 2016, 02:23:17 pm

Hi, so my teacher hasn't been very helpful in covering stats, and I've been struggling to understand almost everything from stats. I know it's a big ask, but can someone make a guide for statistics to help a struggling student? I'm so confused with the p's and obscure explanation of the normal approx. to the binomial distribution in the textbook. My teacher won't cover it in class because he said it was already covered in methods. But even then, I still can't get my head around the sample jargon and simple terminology of stats

$\text{The normal approximation occurs due to the central limit theorem.}\\ \text{A bunch of random variables }X_1, X_2,\dots , X_n\text{ are summed together, and we assume }\\ \text{all }X_i\text{ are independent random variables.}$
$\text{Suppose all the random variables have }\boxed{\text{mean }\mu}\text{ and }\boxed{\text{variance }\sigma^2}$
$\text{If all the random variables are independent, then for the sum of ALL random variables:}\\ \text{The mean of the sum is the sum of the individual means}\\ \text{The variance of the sum is the sum of the variances}$
$\text{Refer }X_i\text{ being ANY of the above random variables}$
$\text{So for such a scenario, let }Y\text{ be the sum of }X_1, X_2, \dots, X_n\\ \text{i.e. }\boxed{Y=\sum_{i=1}^n X_i}$
$\text{We have, using the above text, the mean and the variance.}\\ \boxed{\begin{align*}\mathbb{E}[Y]&=n\mu\\ Var(Y)&=n\sigma^2\end{align*}}$
That was the background information.
$\text{Suppose now, that all the random variables }X_i\text{ follow a Bernoulli distribution.}\\ \text{That is, }\boxed{X_i\sim \mathcal{B}(1,p)}\\ \text{Then, the expectation on random variables }X_1,X_2,\dots,X_n\text{ is:}\\ \boxed{\begin{align*}\mathbb{E}[X_i]&=p\\ Var(X_i)&=p(1-p)\end{align*}}$
$\text{Using the above analogy, we then have }Y\sim \mathcal{B}(n,p)\\ \text{implying }Y\text{ is binomial}\\ \text{ and that }\\ \mathbb{E}(Y)=np\\ Var(Y)=np(1-p)$
Finally, with everything set up, we use the central limit theorem.
$\text{By the statement of the central limit theorem, if a very large number of random variables are summed,}\\ \text{no matter what happens, they start to follow a normal distribution.}\\ \text{What's being summed can be any distribution you like. It will always start to become a normal distribution.}$
$\text{When the CLT is applied, the sum of the mean and variance are }\textit{both preserved.}\\ \text{That is to say:}\\ \boxed{\sum_{i=1}^nX_i \sim \mathcal{N}(n\mu, n\sigma^2)\text{, approximately}}\\ \text{ in every circumstance.}$
$\text{So since we forced }X_i\text{ to be Bernoulli}\\ \text{and know what the mean and variance are from earlier:} \\ \boxed{\sum_{i=1}^nX_i = Y\sim \mathcal{N}(np, np(1-p) )}$
Honestly, this might be too much for a high school student to digest. But this is almost the full, proper approach.

MightyBeh · « **Reply #2 on:** September 17, 2016, 05:00:30 pm »

Rui did an excellent job explaining but I also thought I'd drop in and mention this thread: Statistics for VCE - Welcome to the New Study Design. Worth a skim

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Author Topic: Statistics Help (Read 1616 times) Tweet Share

chenchenpizza

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