Improve Your Understanding of College Level Statistics Problems

monte carlo simulation - confidence intervals construction
I am starting with Monte Carlo Simulation. I have run simulation to estimate the mean and the variance of the exponential distribution.
Simulation:
I have generated random sample from uniform distribution (sample size n) and then using the inverse function method generated the exponential distribution. I have run this simulation N1, N2, and N3 times where $N1<N2<N3$.
Results:
All worked well, when I plot the results I can see that the distribution of means and variances tends to the normal distribution as the number of simulation increases.
Questions:
1) Would you please help and clarify how the empirical and theoretical confidence intervals should be derived?
2) what is the manifestation of the Law of Large Numbers in this exercise?
3) and what is manifestation of Central Limit Theorem in this exercise?

How to calculate confidence level from confidence interval?
A simple random sample of size 10 from $N(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$ gives 98% confidence interval(20.49,23.51). The null hypothesis is $H_0:\mathrm{\mu <!--\; \mu \; -->}=20.5$ against $H_1:\mathrm{\mu <!--\; \mu \; -->}\ne <!--\; \ne \; -->20.5$.What will be the confidence level for accepting this hypothesis? I am trying to solve this problem but unable to understand how to extract confidence level from the interval, Can anyone help please?

Confidence interval for a bounded distribution
Given 1000 observations that come from a distribution that is bounded between 0 and 1. How do you calculate correct 95% Confidence intervals for the MEAN when dealing with a bounded distribution?
R code
set.seed(10)
data =runif(1000, min = 0, max = 1)
mean(data)
mean(data)+1.96*sd(data)/sqrt(length(data)) #usual CIs
mean(data)-1.96*sd(data)/sqrt(length(data)) #usual CIs
Are there any references on how to calculate 95% CIs for bounded distributions?

Exponential Confidence Interval
It is known that, for large n:
$$\sqrt{n}(\mathrm{\lambda <!--\; \lambda \; -->}\stackrel{¯<!--\; ¯\; -->}{x}-<!--\; -\; -->1)\sim <!--\; \sim \; -->\mathrm{N}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}(0,1)$$
With this approximation, show that the 95% confidence interval for $\mathrm{\lambda <!--\; \lambda \; -->}$ is:
$$\frac{\sqrt{n}-<!--\; -\; -->1.96}{\sqrt{n}\stackrel{¯<!--\; ¯\; -->}{x}},\frac{\sqrt{n}+1.96}{\sqrt{n}\stackrel{¯<!--\; ¯\; -->}{x}}$$
I think I need to manipulate the formula for confidence intervals for exponential distributions but I'm not sure where to start (simplified version since large n?)

Determining when two binary strings represent the same necklace or when one binary string is periodic
An equivalence relation on binary strings calls two strings equivalent if one can be obtained from the other by a cyclic permutation of the characters. Combinatorialists call the equivalence classes of such strings "necklaces".
Given two binary strings of length n, the brute-force method for determining if they are equivalent is just iteratively cycling one of them n times and comparing to see if the other string is ever obtained. Is there a faster algorithm? Perhaps there is a list of invariants that are fast to compute and uniquely determine the necklace? Or perhaps through a bijection with some other set, like the irreducible polynomials over ${\mathbb{F}}_2$ with degree dividing n, it is possible to more efficiently solve the problem and transfer back?
A related question: is there an algorithm better than iterative cycling and comparison to get a computer to recognize if a string is periodic? That is, is there a fast algorithm to compute the stabilizer subgroup of a string under the action of the cyclic group of order n?

Let X and Y be independent random variables each having the uniform density on {0,1,...,N}. Find P(X=Y).

Classify each of the following random variables as discrete or continuous. R = the temperature of the turkey meat in a randomly chosen location in a cooked turkey.

Comparing 2013! and ${1007}^{2013}$
I have to compare the following two numbers:
$$2013!\text{ and }{1007}^{2013}$$
where $n!=1\times <!--\; \times \; -->2\times <!--\; \times \; -->\cdots <!--\; \cdots \; -->\times <!--\; \times \; -->(n-<!--\; -\; -->1)\times <!--\; \times \; -->n$
I tried in different ways to group the $1\times <!--\; \times \; -->2\times <!--\; \times \; -->\cdots <!--\; \cdots \; -->\times <!--\; \times \; -->2012\times <!--\; \times \; -->2013$ to obtain some kind of association with the 1007 from ${1007}^{2013}$ but no luck.
Is there any standard approach for this kind of problem?

Why does the standard deviation change from confidence intervals to hypothesis tests?
When considering two-sample data that involves a difference of proportions, both a confidence interval and a hypothesis test can be done.
The standard deviation used for a difference of proportions in creating a confidence interval is $\sqrt{\frac{p_1(1-<!--\; -\; -->p_1)}{n_1}+\frac{p_2(1-<!--\; -\; -->p_2)}{n_2}}$
However, the standard deviation used for confidence intervals is $\sqrt{\frac{p(1-<!--\; -\; -->p)}{n_1}+\frac{p(1-<!--\; -\; -->p)}{n_2}}$, where $p=\frac{x_1+x_2}{n_1+n_2}$, $x_1=p_1{n}_1$, and $x_2=p_2{n}_2$
What I don't understand is why these are different. They're both the standard deviation of the same proportion, so why should they differ?

Suppose that X and Y are independent continuous random variables. Show that
$${\mathrm{\sigma <!--\; \sigma \; -->}}_{x}y=0$$
$$\to <!--\; \to \; -->$$
$${\mathrm{\sigma <!--\; \sigma \; -->}}_{xy}$$

Standard deviation used in confidence interval for mean
I am a novice to Confidence intervals. To figure out the confidence interval for mean, one could either use the Z distribution or t distribution depending on the sample size and population standard deviation. When the size is less than 30 and standard deviation is unknown, we go for t distribution. On the other hand, when the standard deviation is known, we go for Z distribution.
Confidence intervals for mean is used to quantify the uncertainty by providing a lower limit and upper limit that represent a range of values that will represent the true population mean with a specified level of confidence.
Now, in the case of Z distribution, how is the population standard deviation alone known prior to the estimation of population mean? Or in other words, what are the cases when population standard deviation is known before estimating the population mean?

First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: $\mathbb{Q}(\sqrt[3]{2},\mathrm{\gamma <!--\; \gamma \; -->})$, where ${\mathrm{\gamma <!--\; \gamma \; -->}}^3=1$ and ${\mathrm{\gamma <!--\; \gamma \; -->}}^2=1-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}$. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.
I know that my Galois group will consist of the following automorphisms:
${\mathrm{\sigma <!--\; \sigma \; -->}}_i$ = identity
${\mathrm{\sigma <!--\; \sigma \; -->}}_1:\sqrt[3]{2}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}\mathrm{\gamma <!--\; \gamma \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_2:\sqrt[3]{1}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_3:\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\sqrt[3]{2}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_4:\sqrt[3]{2}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}\mathrm{\gamma <!--\; \gamma \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$.
${\mathrm{\sigma <!--\; \sigma \; -->}}_5:\sqrt[3]{1}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$.
If I didn't miss something or make some error.
My specific question is, when constructing my automorphisms, why exactly is it that I can't send $\sqrt[3]{2}$ to something like $\sqrt[3]{4}$ or $\mathrm{\gamma <!--\; \gamma \; -->}$ or some other random thing? Why are these automorphisms so strictly defined?

Suppose that $$Y_1$$ and $$Y_2$$ are independent, standard normal random variables. Find the density

Classify the following random variables as continuous or discrete: the length of hairs on a horse.

How do you find a cubic function ${\displaystyle y=a{x}^3+b{x}^2+cx+d}$
whose graph has horizontal tangents at the points(-2,6) and (2,0)?

100 random samples are taken to estimates the mean $\mathrm{\mu <!--\; \mu \; -->}$. A 95% confidence interval on the mean is $0.49\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->0.82$. Consider the following statement:
There is a 95% chance that $\mathrm{\mu <!--\; \mu \; -->}$ is between 0.49 and 0.82.
Is the statement correct? Explain your answer.
I suppose the statement is wrong and the right one must be: There is a 95% chance that $\mathrm{\mu <!--\; \mu \; -->}$ is actually in some interval that are found.For example, by researching, I found 100 different intervals and 95 of them contains $\mathrm{\mu <!--\; \mu \; -->}$ in average.

Calculate a 90% confidence interval
There's a report saying that 67% of teachers surveyed think that computers are now essential tools in the classroom. Suppose that this information was based on a random sample of $n=200$ teachers.
Calculate a 90% confidence interval for $\mathrm{\pi <!--\; \pi \; -->}$, the true proportion of teachers who think the computer is essential in the classroom.
I know how to find the confidence interval for mean, variance and s.d., but I'm not sure of how to calculate the confidence interval for $\mathrm{\pi <!--\; \pi \; -->}$, can anyone give me a hand and show me how it works please?

Classify each of the following random variables as discrete or continuous. T = winning time in the men's 100-meter dash at a randomly selected international track meet.