# Thread: Statistics Marathon & Questions

1. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
$As the population is normally distributed, \bar{x} is normally distributed. Therefore, as the population standard deviation \sigma is also known, the standardised test statistic Z has a standard normal distribution.$

$Test statistic :$

$Z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}$

$Z=\frac{103-100}{10/ \sqrt{50}}=2.12$

$P(Z > 2.12) =0.017$
dont you have to use the FCF ?

2. ## Re: University Statistics Discussion Marathon

Originally Posted by Rhinoz8142
dont you have to use the FCF ?
Think me and Trebla got the same answer, you have to use the n value which in this case is the n=50. (not the finite value)

3. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
Think me and Trebla got the same answer, you have to use the n value which in this case is the n=50. (not the finite value)
but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.

4. ## Re: University Statistics Discussion Marathon

Originally Posted by Rhinoz8142
but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.
if it isn't a 1st year level of statistics then you may be right.

$Finite Population Correction
Some formulas used to compute standard errors are based on the idea that (1) samples are selected from an infinite population or (2) samples are selected with replacement. In most actual surveys, neither of these ideas are correct.$

$This does not present much of a problem when the sample size (n) is small relative to the population size (N); that is, when the sample is less than 5% of the population. However, when the sample is larger, it is best to apply a correction to the formulas used to compute standard error (SE). This correction is called the finite population correction or fpc.$

$Here is the formula for the finite population correction, when the random variable is a mean score or proportion.$

$f_{pc} = \sqrt{\frac{N - n}{N - 1}}$

$Here is how the finite population correction is used to compute the standard error of a mean score.$

$SE = [ \frac{standard \ deviation}{\sqrt(n)} ] * f_{pc}$

$And here is how the it is used to compute the standard error of a proportion (p).$

$SE =\sqrt{\frac{p(1 - p)}{n}} * f_{pc}$

$Note that when n is very small relative to N, the finite population correction is almost equal to one; and it has only a small effect on the standard error. However, when n is large relative to N, the finite population correction can have a big effect on the standard error.$

5. ## Re: University Statistics Discussion Marathon

Originally Posted by Rhinoz8142
but don't you use the finite correction factor since n/N = 50/250 = > 0.05. Thus using the equation sqrt(N-n/N-1).

I could also be wrong.
This is needed if there is a sampling without replacement which I guess now looking at the question again is somewhat implied but not immediately obvious. If it is a sampling with replacement then what I had earlier is correct.

6. ## Re: University Statistics Discussion Marathon

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

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

$B) F-Statistic=0$

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

$D) SS(Total)=0$

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

7. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
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.

8. ## 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.

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

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

$p =0.62 (true population)$

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

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

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

$P(Z<-0.42)= 0.3372$

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

10. ## Re: University Statistics Discussion Marathon

$Simple Aggregate Price Index$

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

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

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

11. ## Re: University Statistics Discussion Marathon

We can disregard quantity in this question.

12. ## Re: University Statistics Discussion Marathon

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

13. ## Re: University Statistics Discussion Marathon

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

14. ## Re: University Statistics Discussion Marathon

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

15. ## Re: University Statistics Discussion Marathon

Another one you can do similarly:

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

16. ## Re: University Statistics Discussion Marathon

Originally Posted by InteGrand
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.

17. ## 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?

18. ## Re: University Statistics Discussion Marathon

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

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

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

19. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
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.

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

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

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

21. ## Re: Statistics

Originally Posted by leehuan
It's a geometric distribution but suppose we didn't know that

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

$\text{Proven: }\mathbb{E}[X]=\frac32$
$\noindent One way is to essentially first derive the formula for \sum_{k\geq 0} k^{2}z^{k} (|z| < 1) by applying the operator z\frac{\mathrm{d}}{\mathrm{d}z} a couple of times to the geometric series \sum _{k\geq 0} z^{k} = \frac{1}{1-z} (|z| < 1).$

22. ## Re: Statistics

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

24. ## Re: Statistics

Originally Posted by Confound
Explain what? Every single detail? InteGrand's sum derivation?

25. ## 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

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