Statistics Marathon & Questions (1 Viewer)

davidgoes4wce · Jun 6, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
$If all the data points fall on the least-squares regression line in simple linear regression which of the following is true?$

$A) SS(Regression Model)=SS(Trial)$

$B) F-Statistic=0$

$C) MS(Regression Model)=100 \%$

$D) SS(Total)=0$

$E) None of the above statements are true.$

Just read up last night that if SSR=SST then its a perfect linear relationship.

Anyway I am taking a break from this forum for the next 1 month, got alot of work to do, so I'll look forward to bumping into you guys again in July. Just decided that I needed a break as I have been sitting in front of a computer all the time and have no social life as well.

davidgoes4wce · Jun 7, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
Just read up last night that if SSR=SST then its a perfect linear relationship.

Anyway I am taking a break from this forum for the next 1 month, got alot of work to do, so I'll look forward to bumping into you guys again in July. Just decided that I needed a break as I have been sitting in front of a computer all the time and have no social life as well.

I broke my promise I miss this forum too much, I'm back.

davidgoes4wce · Jun 10, 2016

Re: University Statistics Discussion Marathon

I am stuck on part (iv) on how to calculate the Power of the Test. I understand what a power test is but struggle to see it in this question.

Here is an excerpt from the Business Statistics textbook from Sharpe, De Veaux and Velleman book.

$$ Having done the earlier parts of this question, I believe the clearance rate (i) is $ \hat{p}=\frac{351}{574}=0.6115$

$p_0= 0.60 $ (which in this case was the steady state) $$

$p =0.62 $ (true population) $$

$$ I also think that vertical line cuts through both curves when $ \hat{p}=0.6115$

$Calculating the probability of $ \beta \ $ (Type II error)$

$z= \frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}=\frac{0.6115-0.62}{\sqrt{\frac{0.62 \times 0.38}{574}}}=-0.42$

$P(Z<-0.42)= 0.3372$

$Would love it if anybody could give me some feedback on this one.$

davidgoes4wce · Aug 4, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce · Aug 4, 2016

Re: University Statistics Discussion Marathon

$Simple Aggregate Price Index$

$A simple aggregate price index is defined as the ratio of the sum of the prices of the n commodities in the current period 1 over the sum the base period O, multiplied by 100. That is,$

$I_{1,0}=[ \frac{\sum_{i=1} ^n p_{i1}}{ \sum_{i=1}^n \ p_{i0}}] \times 100$

$=\frac{6.50+4.40+28.8}{5.95+4.15+22.95} \times 100 = 120.12$

davidgoes4wce · Aug 5, 2016

Re: University Statistics Discussion Marathon

We can disregard quantity in this question.

leehuan · Nov 7, 2016

Re: University Statistics Discussion Marathon

$\text{Prove that the sum of two binomials with the same probability parameter is still a binomial.}$

seanieg89 · Nov 7, 2016

Re: University Statistics Discussion Marathon

$X\sim \textrm{Bin}(n,p), Y\sim \textrm{Bin}(m,p)\\ \\ f_{X+Y}=\sum_{j=0}^k f_X(j)f_Y(k-j)\\ \\ = p^k(1-p)^{m+n-k}\sum_{j=0}^{k}\binom{n}{j}\binom{m}{k-j}\\ \\=\binom{m+n}{k}p^k(1-p)^{m+n-k}\\ \\ \Rightarrow X+Y\sim \textrm{Bin}(m+n,p).$

InteGrand · Nov 7, 2016

Re: University Statistics Discussion Marathon

leehuan said:
$\text{Prove that the sum of two binomials with the same probability parameter is still a binomial.}$

The probabilistic interpretation: X ~ Bin(n, p) counts the no. of sucesses in n independent Bernoulli trials (with success parameter p) and Y ~ Bin(m, p) counts the no. of successes in m independent Bernoulli trials. So if X and Y are independent, X + Y is counting the no. of successes in m+n independent Bernoulli trials (each with success prob. p), whence X+Y ~ Bin(m+n, p).

InteGrand · Nov 7, 2016

Re: University Statistics Discussion Marathon

Another one you can do similarly:

$\noindent Show that if $X$ and $Y$ are independent geometric random variables, then $Z := \min(X, Y)$ is also geometric, and find the parameter of $Z$ in terms of those of $X$ and $Y$.$

He-Mann · Nov 7, 2016

Re: University Statistics Discussion Marathon

InteGrand said:
Another one you can do similarly:

$\noindent Show that if $X$ and $Y$ are independent geometric random variables, then $Z := \min(X, Y)$ is also geometric, and find the parameter of $Z$ in terms of those of $X$ and $Y$.$

$$Suppose that $X \sim \mathrm{Geo}(p)$ and $Y \sim \mathrm{Geo}(q). $ Then considering the CDF of $Z,$ we get$\\ \begin{align*} F_Z(z) &= \mathbb{P}(Z\leq z) \\ &= 1-\mathbb{P}(Z > z) \dots (*)\\&= 1-\mathbb{P}(\mathrm{min}\{ X,Y\} > z) \\ &= 1-\mathbb{P}(X > z, Y > z) \\ &= 1-\mathbb{P}(X > z) \mathbb{P}(Y > z) \quad [\text{By independence}] \\ &=1 - (1-p)^z(1-q)^z \\ &= 1 - [1 - (p + q - pq)]^z\end{align*} \\ \begin{align*}Thus, Z \sim \mathrm{Geo}(p+q - pq) &= \mathrm{Geo}(p(1-q) - (1-q) + 1)\\ &= \mathrm{Geo}(1 + (p-1)(1-q))\\ &=\mathrm{Geo}(1 - (1-p)(1-q)) \end{align*}.$

Factorised the above since I saw a pattern, lel.

(*) If I didn't do this, the result will be (1 - (1-p)^z)(1 - (1-q)^z) which is too hard to put into the form (1 - r^z).

Not sure about a probabilistic interpretation though.

He-Mann · Nov 7, 2016

Re: University Statistics Discussion Marathon

Received this question from a friend so wanted to share it here.

You roll a conventional die repeatedly. If it shows 1, you must stop, but you may choose to stop at any prior time. Your score is the number shown by the die on the final roll. Consider the strategy: stop the first time that the die shows r or greater.

a) What stopping strategy (i.e. what value of r) yields the greatest expected score?

b) What strategy would you use if your score were the square of the final roll?

InteGrand · Nov 7, 2016

Re: University Statistics Discussion Marathon

$\noindent (It is clear we should have $r\geq 2$ for optimality.) One way to do it would be as follows. Let $D_{k}$ be the result of the $k$-th die roll ($k=1,2,3,\ldots$). Let $T_{r}$ be the first $n$ such that $D_{n}$ is either $1$ or $\geq r$ ($r \in \left\{2,3,4,5,6\right\}$). We are after the $r$ that maximises $D_{T_r}$'s expected value.$

$\noindent The numbers in the range of $D_{T_r}$ are $1$ and the whole numbers from $r$ to $6$ (there are $8-r$ such numbers for each $r$). Observe that by symmetry, $\mathbb{P}\left(D_{T_6} = 1\right) = \mathbb{P}\left(D_{T_6}= 6\right) = \frac{1}{2}$ (i.e. if we wait till the first time we get either $1$ or $\geq 6$ (i.e. $6$), the prob. we get the $1$ first is just half by symmetry). Similarly, $\mathbb{P}\left(D_{T_5} = j\right) = \frac{1}{3}$ for each $j$ in the range of $D_{T_5}$, and in general $\mathbb{P}\left(D_{T_r} = j\right) = \frac{1}{8-r}$ for each $j$ in the range of $D_{T_r}$, for each $r \in \left\{2,3,\ldots,6\right\}$.$

$\noindent So for a deterministic function $\phi$, we have $\mathbb{E}\left[\phi\left(D_{T_r}\right)\right] = \frac{1}{8-r}\cdot \phi\left(1\right) + \sum_{j=r}^{6} \frac{1}{8-r}\cdot \phi\left(j\right)$ (in part b), $\phi(t) = t^{2}$. In part a), $\phi(t)$ is just $t$.). These sums are finite and easy to compute for each $r$, and there are only five values of $r$ to check for each part, so we can find which $r$ is optimal.$

He-Mann · Nov 7, 2016

Re: University Statistics Discussion Marathon

davidgoes4wce said:
Just read up last night that if SSR=SST then its a perfect linear relationship.

Anyway I am taking a break from this forum for the next 1 month, got alot of work to do, so I'll look forward to bumping into you guys again in July. Just decided that I needed a break as I have been sitting in front of a computer all the time and have no social life as well.

What is SS(Trial)?

I understand that a perfect-fit occurs when sum of square residual is zero which means sum of square regression equals sum of square total.

leehuan · Feb 23, 2017

Statistics (thread merged)

It's a geometric distribution but suppose we didn't know that

$\\\text{Find, via first principles, }Var(X)\\ \text{if }Pr(X=k)=\frac23 \left(\frac13\right)^{k-1}\, k =1,2,\dots$

$\text{Proven: }\mathbb{E}[X]=\frac32$

InteGrand · Feb 23, 2017

Re: Statistics

leehuan said:
It's a geometric distribution but suppose we didn't know that

$\\\text{Find, via first principles, }Var(X)\\ \text{if }Pr(X=k)=\frac23 \left(\frac13\right)^{k-1}\, k =1,2,\dots$

$\text{Proven: }\mathbb{E}[X]=\frac32$

$\noindent One way is to essentially first derive the formula for $\sum_{k\geq 0} k^{2}z^{k}$ ($|z| < 1$) by applying the operator $z\frac{\mathrm{d}}{\mathrm{d}z}$ a couple of times to the geometric series $\sum _{k\geq 0} z^{k} = \frac{1}{1-z}$ ($|z| < 1$).$

InteGrand · Feb 23, 2017

Re: Statistics

$\noindent The final answer will be $\frac{q}{p^{2}}$, where $p$ is the ``success'' probability parameter, and $q = 1-p$. In your case, $p = \frac{2}{3}$.$

Confound · Feb 24, 2017

Re: Statistics

ELI5 Please

He-Mann · Feb 24, 2017

Re: Statistics

Confound said:
ELI5 Please

Explain what? Every single detail? InteGrand's sum derivation?

davidgoes4wce · Mar 3, 2017

Re: University Statistics Discussion Marathon

Which of the following R expressions are logical? (i.e. evaluation of them results in logical objects). Select all that apply.

- exp(x)
- y>=32
- B==’a”
- 1:50

Statistics Marathon & Questions (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Vexed?

Vexed?

Well-Known Member

Vexed?

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Vexed?

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)