Improve Your Understanding of College Level Statistics Problems

What is the shape of a chi-squared distribution?

Finding the Expected Value of $T=$$\sum <!--\; \sum \; -->X_i^2$

Which of the following statements is true about the t-distribution?
a) For large sample sizes, the t-distribution has the same properties as the normal curve.
b) For small sample sizes, the t-distribution has the same properties as the normal curve.
c) Like the Normal distribution, the t-distribution is symmetric for small n.
d) Since population standard deviation is usually unknown, the standard error uses the sample standard deviation to estimate population standard deviation.

How do you find the exact value for $\cos <!--\; \; -->240$?

When conducting a hypothesis test for the population mean, you will not know the population standard deviation, so you will have to use the sample standard deviation instead. How will this affect the process?

Let $\{Xi\}\sim <!--\; \sim \; -->N(i\theta ,1)$ for $i=1,....,n$ be an independent, but not identically distributed sample. Check that $T=\underset{i}{\sum <!--\; \sum \; -->}{X}_i$ it is a sufficient statistic for $\theta$.

Let $Xi$ be a random variable distributed as $N(i,i^2),i=1,2,3$. As-sume that the random variables $X_1,X_2$ and $X_3$ are independent. Using only the three random variables $X_1,X_2$ and $X_3$ give an example of a statistic that has a $t$ distribution with two degrees of freedom.

How do you find the exact value of ${\sin }^{-<!--\; -\; -->1}<!--\; \; -->(\sin <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}))$?

Bayesian Statistics - Basic question about prior
I try to get an understanding of bayesian statistics. My intuition tells me that in the expression for the posterior
$$p(\mathrm{\vartheta <!--\; \vartheta \; -->}|x)=\frac{p(x|\mathrm{\vartheta <!--\; \vartheta \; -->})p(\mathrm{\vartheta <!--\; \vartheta \; -->})}{{\int <!--\; \int \; -->}_{\mathrm{\Theta <!--\; \Theta \; -->}}p(x|\mathrm{\theta <!--\; \theta \; -->})p(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}}$$
the term $p(\mathrm{\vartheta <!--\; \vartheta \; -->})$ is the marginal distribution of the likelihood-function $p(\mathrm{\vartheta <!--\; \vartheta \; -->},x)$. It is obtained by
$$p(\mathrm{\vartheta <!--\; \vartheta \; -->})={\int <!--\; \int \; -->}_{X}p(\mathrm{\vartheta <!--\; \vartheta \; -->}|x)p_X(x)dx$$
where $p_X(x)$ should be the marginal distribution of the Observable data. Does that make sense?
To this point it makes sense with this example: Offering somebody a car insurance without knowing the person's style of driving (determined by $\mathrm{\vartheta <!--\; \vartheta \; -->}\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}$) to feed some statistical model, we still can make use of the nation's car-crash statistics as our prior, which is a pdf on $\mathrm{\Theta <!--\; \Theta \; -->}$. That would be the marginal distribution of the "driving styles" across the population.
Maybe I am just oversimplifying here, because my resources did not mention this.

Solve PDE using method of characteristics with non-local boundary conditions.
Given the population model by the following linear first order PDE in u(a,t) with constants b and $\mathrm{\mu <!--\; \mu \; -->}$:
$$u_a+u_t=-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tua,t>0$$
$$u(a,0)=u_0(a)a\ge 0$$
$$u(0,t)=F(t)=b{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$
We can split the integral in two with our non-local boundary data:
$$F(t)=b{\int <!--\; \int \; -->}_0^{t}u(a,t)\mathrm{d}a+b{\int <!--\; \int \; -->}_t^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$
Choosing the characteristic coordinates $(\mathrm{\xi <!--\; \xi \; -->},\mathrm{\tau <!--\; \tau \; -->})$ and re-arranging the expression to form the normal to the solution surface we have the following equation with initial conditions:
$$(u_a,u_t,-<!--\; -\; -->1)\bullet <!--\; \bullet \; -->(1,1,-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tu)=0$$
$$x(0)=\mathrm{\xi <!--\; \xi \; -->},t(0)=0,u(0)=u_0(\mathrm{\xi <!--\; \xi \; -->})$$
Characteristic equations:
$$\frac{\mathrm{d}a}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=1,\frac{\mathrm{d}t}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=1,\frac{\mathrm{d}u}{\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}}=-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}tu$$
Solving each of these ODE's in $\mathrm{\tau <!--\; \tau \; -->}$ gives the following:
$$(1)\int <!--\; \int \; -->\mathrm{d}a=\int <!--\; \int \; -->\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}(2)\int <!--\; \int \; -->\mathrm{d}t=\int <!--\; \int \; -->\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}(3)\int <!--\; \int \; -->\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}tu\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$a=\mathrm{\tau <!--\; \tau \; -->}+F(\mathrm{\xi <!--\; \xi \; -->})t=\mathrm{\tau <!--\; \tau \; -->}+F(\mathrm{\xi <!--\; \xi \; -->})$$
$$\therefore <!--\; \therefore \; -->a=\mathrm{\tau <!--\; \tau \; -->}+\mathrm{\xi <!--\; \xi \; -->}\therefore <!--\; \therefore \; -->t=\mathrm{\tau <!--\; \tau \; -->}$$
$$\int <!--\; \int \; -->\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}\mathrm{\tau <!--\; \tau \; -->}u\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$\int <!--\; \int \; -->\frac{1}{u}\mathrm{d}u=-<!--\; -\; -->\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}\mathrm{\tau <!--\; \tau \; -->}\mathrm{d}\mathrm{\tau <!--\; \tau \; -->}$$
$$\ln <!--\; \; -->u=-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2+F(\mathrm{\xi <!--\; \xi \; -->})$$
$$u=G(\mathrm{\xi <!--\; \xi \; -->})e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2}$$
$$\therefore <!--\; \therefore \; -->u=u_0(\mathrm{\xi <!--\; \xi \; -->})e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\tau <!--\; \tau \; -->}}^2}$$
Substituting back the original coordinates we can re-write this expression with a coordinate change:
$$\mathrm{\xi <!--\; \xi \; -->}=a-<!--\; -\; -->t\mathrm{\tau <!--\; \tau \; -->}=t$$
$$\therefore <!--\; \therefore \; -->u(a,t)=u_0(a-<!--\; -\; -->t)e^{-<!--\; -\; -->\frac{1}{2}{t}^2}$$
Now this is where I get stuck, how do I use the boundary data to come up with a well-posed solution?
$$u(0,t)=u_0(-<!--\; -\; -->t)e^{-<!--\; -\; -->\frac{1}{2}\mathrm{\mu <!--\; \mu \; -->}{t}^2}=b{\int <!--\; \int \; -->}_0^{t}u(a,t)\mathrm{d}a+b{\int <!--\; \int \; -->}_t^{\mathrm{\infty <!--\; \infty \; -->}}u(a,t)\mathrm{d}a$$

Using a "population" consisting of probabilities to predict accuracy of sample
Since I'm not sure if the title explains my question well enough I've come up with an example myself:
Let's say I live in a country where every citizen goes to work everyday and every citizen has the choice to go by bus or by train (every citizen makes this choice everyday again - there are almost no citizens who always go by train and never by bus, and vice-versa).
I've done a lot of sampling and I have data on one million citizens about their behaviour in the past 1000 days. So, I calculate the "probability" per citizen of going by train on a single day. I can also calculate the average of those calculated probabilities of all citizens, let's say the average probability of a citizen going by train is 0.27. I figured that most citizens have tendencies around this number (most citizens have an individual probability between 0.22 and 0.32 of going by train for example).
Now, I started sampling an unknown person (but known to be living in the same country) and after asking him 10 consecutive days whether he went by train or by bus, I know that this person went to his work by train 4 times, and by bus 6 times.
My final question: how can I use my (accurate) data on one million citizens to approximate this person's probability of going by train?
I know that if I do the calculation the other way around, so, calculate the probability of this event occurring given the fact that I know this person's REAL probability is 0.4 this results in: ${0.4}^4\cdot <!--\; \cdot \; -->{0.6}^6\cdot <!--\; \cdot \; -->10C4=\sim <!--\; \sim \; -->25\mathrm{\%<!--\; \%\; -->}$. I could calculate this probability for all possible probabilities between 0.00 and 1.00 (so, $0\mathrm{\%<!--\; \%\; -->}-<!--\; -\; -->100\mathrm{\%<!--\; \%\; -->}$ without any numbers in between) and sum them all, which sums to about 910%. I could set this to 100% (dividing by 9.1) and set all other percentages accordingly (dividing everything by 9.1 - so, our 25% becomes ~2.75%) and come up with a weighted sum: $2.75\mathrm{\%<!--\; \%\; -->}\cdot <!--\; \cdot \; -->0.4+X\mathrm{\%<!--\; \%\; -->}\cdot <!--\; \cdot \; -->0.41$ etc., but this must be wrong since I'm not taking my accurate samples of the population into account.

Derive an expression for the $p$-value using a test with test statistic $T=\sqrt{n}({\stackrel{¯<!--\; ¯\; -->}{X}}_n-<!--\; -\; -->{\mathrm{\theta <!--\; \theta \; -->}}_0)/\mathrm{\sigma <!--\; \sigma \; -->}$

Let $X1,...,Xn$ be independent and identically distributed with density $P_{\theta }(x)=$
$$\{\begin{array}{ll}2x/{\theta }^2& \text{for }0\le <!--\; \le \; -->x<\theta \\ 0& \text{else}\end{array}$$
Demonstrate that $m_n=max(x_1,...,x_n)$ is a sufficient statistic .

How do you use a power series to find the exact value of the sum of the series $1-<!--\; -\; -->\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^2}{2!}+\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^4}{4!}-<!--\; -\; -->\frac{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{)}^6}{6!}+$...?

Derive the Cramer von Mises test statistic
$$n{C}_n=\frac{1}{12n}+\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{(U_{(i)}-<!--\; -\; -->\frac{2i-<!--\; -\; -->1}{2n})}^2$$
where $U_{(i)}=F_0(X_{(i)})$ the order statistics

We're given a random sample of $X_1,X_2,....,X_n$ from a distribution with a pdf of
$f(x;\mathrm{\theta <!--\; \theta \; -->})={\mathrm{\theta <!--\; \theta \; -->}}^x(1-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}),x=0,1,2,....$ and $0<\mathrm{\theta <!--\; \theta \; -->}<1$
Show that that $Tn=\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i$ is a complete sufficient statistic for $\mathrm{\theta <!--\; \theta \; -->}$.

The occurrence of a disease is $\frac{1}{100}=P(D)$
The false negative probability is $\frac{6}{100}=P(-<!--\; -\; -->|D)$, and the false positive is $\frac{3}{100}=P(+|\mathrm{\neg <!--\; \neg \; -->}D)$
Compute $P(D|+)$
By bayes formula,
$P(+)=P(+|D)P(D)+P(+|\mathrm{\neg <!--\; \neg \; -->}D)P(\mathrm{\neg <!--\; \neg \; -->}D)=\frac{97}{10000}+\frac{297}{10000}=\frac{394}{10000}$
Similarly $P(D|+)=\frac{P(+|D)P(D)}{P(+)}=\frac{97}{394}=0.246$
Is this correct?

_____ is a sampling technique where the entire population is divided into groups and a random sample of these groups are selected and all observations of the selected group are included in the sample.
a. Random sampling
b. Systematic sampling
c. Cluster sampling
d. Stratified sampling

Separating populations and estimating line-fit parameters
Given a dataset containing two populations, each of which can be described by a linear relationship between two variables in each sample with high $R^2$, how does one separate the two populations (and incidentally compute the line-fit)?
This is fairly easy to do graphically - just create a scatterplot and the two lines are pretty apparent. But how does one do this algorithmically?
More generally, given a dataset containing an unknown number n of populations, each of which can be fit to a line with some lower bound on $R^2$ (e.g., .95), how does one separate the data into the minimum number of populations satisfying the $R^2$ criterion?

An investigator computes a $95\mathrm{\%<!--\; \%\; -->}$ confidence interval for a population mean on the basis of a sample of size 75 if she wishes to compute a $95\mathrm{\%<!--\; \%\; -->}$ confidence interval that is half as wide, how large a sample does she need?