Statistics Marathon & Questions (3 Viewers)

Flop21 · Jun 11, 2017

http://i.imgur.com/tYDWeYf.png

Where am I going wrong here? The answers show my answer but with only a square root (not a 4-root), and no 'x' at the end.

InteGrand · Jun 11, 2017

Flop21 said:
http://i.imgur.com/tYDWeYf.png

Where am I going wrong here? The answers show my answer but with only a square root (not a 4-root), and no 'x' at the end.

It looks like your log term was not differentiated correctly.

He-Mann · Jun 11, 2017

Flop21 said:
http://i.imgur.com/tYDWeYf.png

Where am I going wrong here? The answers show my answer but with only a square root (not a 4-root), and no 'x' at the end.

Just a tip for log-likelihood stuff: after taking the log of the likelihood, continue to simplify the result using log laws such that fractions are eliminated. It'll make differentiation much easier.

Flop21 · Jun 12, 2017

Is there a TLDR or summary of stats that could help with a person SO SCREWED for the exam???

Notes for MATH2801, anywhere?

kawaiipotato · Jun 12, 2017

Flop21 said:
Is there a TLDR or summary of stats that could help with a person SO SCREWED for the exam???

Notes for MATH2801, anywhere?

Try asking on the Facebook MathSOC page or the Higher Year Mathematics group.

Flop21 · Jun 12, 2017

With hypothesis testing...

How do I evaluate things like P(Z>4.38). ??

InteGrand · Jun 12, 2017

Flop21 said:
With hypothesis testing...

How do I evaluate things like P(Z>4.38). ??

That's pretty much 0 (if Z is standard normal).

Flop21 · Jun 12, 2017

InteGrand said:
That's pretty much 0 (if Z is standard normal).

Yeah but what is it, it was like 0.0001 or something, like how do I evaluate these things? What is Z?

kawaiipotato · Jun 12, 2017

Flop21 said:
Yeah but what is it, it was like 0.0001 or something, like how do I evaluate these things? What is Z?

Z is a random variable following the standard normal distribution and we denote it by writing

$Z \sim N(0,1)$
which means it has probability distribution function

$f_{Z}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} .$

When you want something like P(Z>4.38), it's the area under the density (f_Z) curve for x>= 4.38.
You wouldn't be able to evaluate it by hand (since there isn't an elementary function for the integral of that function), so we'd use a standard normal table.
I'm not sure if MATH2801 still uses tables, but for MATH2901, we don't anymore so (in the test) we'd just be given the outputs (so the actual data and what, for example, P(Z>4.38) would equal to.), so if you guys don't get tables anymore, then it's fine.

InteGrand · Jun 12, 2017

Flop21 said:
Yeah but what is it, it was like 0.0001 or something, like how do I evaluate these things? What is Z?

It's very close to 0. Here's a numerical calculation via WolframAlpha: http://www.wolframalpha.com/input/?i=Pr(Z+>+4.38),+Z+~+N(0,+1) .

Flop21 · Jun 12, 2017

kawaiipotato said:
Z is a random variable following the standard normal distribution and we denote it by writing

$Z \sim N(0,1)$
which means it has probability distribution function

$f_{Z}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} .$

When you want something like P(Z>4.38), it's the area under the density (f_Z) curve for x>= 4.38.
You wouldn't be able to evaluate it by hand (since there isn't an elementary function for the integral of that function), so we'd use a standard normal table.
I'm not sure if MATH2801 still uses tables, but for MATH2901, we don't anymore so (in the test) we'd just be given the outputs (so the actual data and what, for example, P(Z>4.38) would equal to.), so if you guys don't get tables anymore, then it's fine.

Thank you!! I literally had no idea what it was.

Yeah we don't use tables anymore either, so will it literally tell us P(Z>4.38) = something or whatever?

kawaiipotato · Jun 12, 2017

Flop21 said:
Thank you!! I literally had no idea what it was.

Yeah we don't use tables anymore either, so will it literally tell us P(Z>4.38) = something or whatever?

If you guys use R I think it's likely that they'd use the functions
pnorm(a,mu,sigma) = some value, and this equals to P(X<a) for a r.v. with distribution N(mu,sigma^2)

So for that they would've given:
1 - pnorm(4.38,0,1) = some value

which of course equals P(Z>4.38) (since Z is usually defined to be standard normal)

Flop21 · Jun 13, 2017

What is this thing highlighted. I see it a lot, does it have a name?

(Sorry for the silly question, I just am going through so much content in a few days).

Flop21 · Jun 13, 2017

How do I do these? Someone point me in the right direction or go through one example for me? Thanks

kawaiipotato · Jun 13, 2017

Flop21 said:
How do I do these? Someone point me in the right direction or go through one example for me? Thanks

I'll do (v) which uses the result in (iv).
$$\noindent (iv). If $X_i$ follows $N(\mu, \sigma ^2) $ in distribution, then $\frac{X_i - \mu}{\sigma}$ follows $N(0,1)$ (this is the same process as your other question above your latest post).$

$\noindent (v). Define $Y:= \frac{X_i - \mu}{\sigma}$ (just for simpler writing) so $Y \sim N(0,1)$. You can prove that $Y^2 = \left(\frac{X_i - \mu}{\sigma}\right)^2$ follows a chi-squared distribution with parameter $1$ denoted by $\chi^2_{1}$ (easiest way is to show via moment-generating functions). Note that if you add two chi-squared independent random variables, $A$ and $B$, with parameters $a$ and $b$ respectively, then the result is another chi-squared random variable with parameter $a+b$. That is, if$

$\noindent $A \sim \chi^2_{a}$ and $B \sim \chi^2_{b}$ then $A+B \sim \chi^2_{a+b}$.$

$\noindent That means the random variable $\sum_{i=1}^{n} \left(\frac{X_i - \mu}{\sigma}\right)^2$ follows a chi-squared distribution with parameter $n$ (since you add up $n$ independent chi-squared random variables with a parameter of $1$ each.) denoted by $\chi^2_{n}$.$

kawaiipotato · Jun 13, 2017

Actually your part (ii) follows N(0,1) and in part (iii) it asks for the square of that. So your answer in part (iii) will also be a chi-squared random variable with parameter 1.

InteGrand · Jun 13, 2017

Flop21 said:
What is this thing highlighted. I see it a lot, does it have a name?

(Sorry for the silly question, I just am going through so much content in a few days).

$\noindent It's a \emph{standardised} version of $\overline{X}$. I can't see the context but I'm guessing $\overline{X}$ is the average of $n$ i.i.d. $\mathcal{N}\left(\mu, \sigma^{2}\right)$ random variables. This implies that $\overline{X}$ is a normal random variable with mean $\mu$ and variance $\frac{\sigma^{2}}{n}$, using properties of linear combinations of independent normals. Thus we can standardise $\overline{X}$ by writing $\frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$, and this will then follow a standard normal distribution. The purpose of doing this is that it is often simpler to work in terms of standard normals (e.g. because many standard values for it are typically known or calculated, such as 1.96 being the $97.5$-th percentile of it).$

$\noindent Note in general if $X$ is a normal random variable with mean $\mu$ and variance $\sigma^{2}$ ($\sigma > 0$), the way to standardise it is $\frac{X - \mu}{\sigma}$, which is then a standard normal random variable.$

He-Mann · Jun 13, 2017

Flop21 said:
How do I do these? Someone point me in the right direction or go through one example for me? Thanks

It's important that you do these in order because it acts as a guide. You'll use part (j) to do part (j+1).

Flop21 · Jun 14, 2017

If X is a binomial variable, and Y = 2X, the pdf of Y, would that just be the pdf of X, but every x = 2x????

There was also a question similar saying find the MGF of Y, and I found that to by solving E(e^(u2X)).

leehuan · Jun 14, 2017

Flop21 said:
If X is a binomial variable, and Y = 2X, the pdf of Y, would that just be the pdf of X, but every x = 2x????

There was also a question similar saying find the MGF of Y, and I found that to by solving E(e^(u2X)).

The binomial random variable is discrete, so you'd have a pmf as opposed to a pdf.

$\text{If }\mathbb{P}(X=x) = \binom{n}{x}p^x(1-p)^{n-x}\text{ where }x=0,1,\dots,n\\ \text{Then }\mathbb{P}(Y=y) = \mathbb{P}(2X=y) = \mathbb{P}\left(X=\frac{y}{2}\right) = \binom{n}{y/2}p^y (1-p)^{n-y/2}\\\text{ for }y=0,2,\dots,2n$

Also, yes, if you had to find the MGF of Y := 2X, you would consider E[e^(u(2X))]

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