Master Confidence Intervals: Examples, Equations, and Questions

Confidence interval multiplication
The question looks pretty simple but I can't get my hands on it:Say I have a probability which is the product of two other independent probabilities $p=p_1{p}_2$.
I have estimated each probability $p_1$ and $p_2$ and found some 95% confidence interval for each. How do I obtain a 95% confidence interval for p?
Taking the product of the bounds of the interval won't work as I would be taking 95% of a 95% confidence interval resulting in an approximately 90% which is not what I want.So conversely I would be taking 97.5% confidence interval for each and by multiplying the bounds I will obtain a 95% confidence interval, is that right?
I feel like something is going wrong. In my situation I deal with probabilities but it could be anything so this question can be generalised to any type of confidence intervals.
If my reasoning is correct, could someone convince me that's the correct way of doing so?

Confidence Interval. Bernoulli Distribution
I am reviewing the construction of confidence intervals for a random sample with Bernoulli distribution. The book uses the statistics of the central limit theorem that distributes N(0,1) to estimate the interval :
$$Z_n=\frac{X_1+X_2+\cdots <!--\; \cdots \; -->+X_n-<!--\; -\; -->n\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}\sqrt{n}}$$
Why are the intervals constructed from these statistics symmetrical around the origin?
The book says: "Since it is desirable that the length of the interval be as small as possible and since the standard normal distribution is symmetrical around the origin, it turns out that the minimum length interval must also be symmetric around the origin", but I don't understand this.

If a 95% confidence interval for the mean was computed as (25,50), then if several more samples were taken with the same sample size, then 95% of them would have a sample mean between (25,50)
And I know this statement is false, but I want to know exactly why.

Confidence interval interpretation difficulty
When we are dealing with 95% confidence interval we mean that if we repeat process of collecting samples of same size and calculate 95% intervals for those samples then 95% of those intervals will contain the true population parameter.
Let the infinite number of intervals be represented by 100 for simplicity. Then 95 of these intervals will contain true population parameter.
Suppose we got an interval at the starting of the above process (L,U). Then if I ask what is the probability that this interval (L,U) contains the true population parameter then shouldn't it be $95/100=0.95$? (Because this interval (L,U) can be anyone of 100 and it would contain true population parameter of its one of those 95).
But this interpretation of confidence interval is considered incorrect. Can someone explain me why is this so?

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?)

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?

Confidence Intervals for an Exponential Distribution
$y_1$ is distributed $f_Y(y\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})=\mathrm{\theta <!--\; \theta \; -->}{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}y}{I}_{(0,\mathrm{\infty <!--\; \infty \; -->})}(y)$, where $\mathrm{\theta <!--\; \theta \; -->}>0$. Analyze the confidence interval for $\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}$ given by $[L(Y),U(Y)]=[Y,2Y]$
1: Determine the confidence coefficient of this interval.
Thoughts: I realized that my single data point is distributed
$\mathit{\text{exponential}}(\mathrm{\beta <!--\; \beta \; -->}=\frac{1}{\mathrm{\theta <!--\; \theta \; -->}})=\mathit{\text{gamma}}(\mathrm{\alpha <!--\; \alpha \; -->}=1,\mathrm{\beta <!--\; \beta \; -->}=\frac{1}{\mathrm{\theta <!--\; \theta \; -->}})$. So I did the following steps to get my confidence coefficient:
$$\begin{gathered} P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\in <!--\; \in \; -->[Y,2Y]) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->2Y) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}) \\ =\underset{\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}}{\overset{\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}}{\int <!--\; \int \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}y}\mathrm{d}y \\ =e^{-<!--\; -\; -->.5}-<!--\; -\; -->e^{-<!--\; -\; -->1} \\ =.23865 \end{gathered}$$
So I got .23685 as my confidence coefficient, and I'm fairly confident that that is the correct answer.
2: Determine the expected length of this interval.
Thoughts: So I'm not too sure what to do here, but this is what I have written down.
$$\begin{gathered} P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\in <!--\; \in \; -->[Y,2Y]) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->2Y) \\ =P_{\mathrm{\theta <!--\; \theta \; -->}}(\frac{1}{2\mathrm{\theta <!--\; \theta \; -->}}\le <!--\; \le \; -->Y\le <!--\; \le \; -->\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}) \\ =P(e^{-<!--\; -\; -->1}\le <!--\; \le \; -->Y\le <!--\; \le \; -->e^{-<!--\; -\; -->.5})=.23865 \end{gathered}$$
Since my 'a' and 'b' values were $e^{-<!--\; -\; -->1}$ and $e^{-<!--\; -\; -->.5}$, respectively. Now I don't know where to go from there to find the expected length. Do I use the gamma probability table? Any help would be greatly appreciated.
3: Find a different confidence interval for $\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}$ with the exact same confidence coefficient, but the expected length is smaller than part 2's expected length. Find an interval where the characteristics, same confidence coefficient and expected length is smaller than part 2's, are satisfied.
Thoughts: I let $W=Y^{\frac{1}{\mathrm{\theta <!--\; \theta \; -->}}}$, and I performed the transformation to get the distribution of W, and I ended with:
$$f_W(w\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})={\mathrm{\theta <!--\; \theta \; -->}}^2{e}^{-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{w}^{\mathrm{\theta <!--\; \theta \; -->}}}{w}^{\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->1}$$
, which I think looks sorta like the gamma distribution, but then I don't know what to do from there. Any help would be appreciated.

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?

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.

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?

Statistics and Confidence Intervals
Given the following set of values:
10,11,14,95,73,30,29,9,97,94,70
How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10
Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:
$x-<!--\; -\; -->z_{\frac{a}{2}}\frac{{\mathrm{\sigma <!--\; \sigma \; -->}}^2}{sqrt(n)}$ and $x+z_{\frac{a}{2}}\frac{{\mathrm{\sigma <!--\; \sigma \; -->}}^2}{sqrt(n)}$ where x denotes the mean. Here the significance level (a) is 0.01.

Sample size in Confidence Intervals
In repeating confidence interval experiments, are we allowed to take samples of different size every time? Because a confidence interval of 95% means that if the sampling process is repeated infinite times, 95% of all the intervals obtained will have the parameter of interest.
So wrt this process of repeating infinite times - do we have to repeat with the same sample size or can we vary it?

I am trying to write a C++ program for parameter estimation(with Confidence Interval information) of an Exponentially distributed data set. I understand that $\mathrm{\lambda <!--\; \lambda \; -->}\stackrel{¯<!--\; ¯\; -->}{X}\sim <!--\; \sim \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(n,n)$. To come up with numeric values of lower/higher confidence intervals for a specified $\mathrm{\alpha <!--\; \alpha \; -->}$, I need to be able to compute Inverse Gamma. Could you please point me to some algorithms?

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?

$(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ Confidence Intervals
Let $Y_1,Y_2,...,Y_n$ be i.d.d. random variables from a gamma distribution with shape parameter ${\mathrm{\alpha <!--\; \alpha \; -->}}_0>0$ and unknown scale parameter $\mathrm{\beta <!--\; \beta \; -->}$. Find a $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ confidence interval for the parameter $\mathrm{\beta <!--\; \beta \; -->}$. Note that the minimum variance unbiased estimator for $\mathrm{\beta <!--\; \beta \; -->}$ is $\stackrel{^<!--\; ^\; -->}{\mathrm{\beta <!--\; \beta \; -->}}=\frac{\stackrel{¯<!--\; ¯\; -->}{Y}}{{\mathrm{\alpha <!--\; \alpha \; -->}}_0}$ where $\stackrel{¯<!--\; ¯\; -->}{Y}=\frac{1}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{Y}_i$.
I have no clue where to start here. Any hints?

This is a question from a sample mid-term for a first course in Statistics
Let p denote the proportion of elderly people with artificial hips. In an earlier study, the estimated proportion of the number of elderly people with artificial hips was found to be 11%.
Q.1. Using earlier study's estimate, find the number of samples we would need in order to create a 95 confidence interval for the proportion of elderly people with artificial which is no more than 0.05 away from the true proportion.
Ans. I have no clue how to do this. If could get the formula on how to approach this I would appreciate it.
Q.2. What is the maximum sample size needed to create a 93 confidence interval for the proportion of elderly people with artificial hips which is no more than 0.04 away from the true proportion ? Ans I would appreciate a formula for this as well.

Confidence intervals on maximum likelihoods of observed data
I observed 400 episodes of nursing care in a hospital. I tracked the movement of the nurses between 5 rooms $A-<!--\; -\; -->E$. The maximum likelihood of them moving from room $i\to <!--\; \to \; -->j$ is given by:
$$P_{ij}={\displaystyle {\displaystyle \frac{\text{\# of times from room }i\to <!--\; \to \; -->j}{{\displaystyle \text{Total \# of transitions to any room}}}}}$$
- Is there a way of defining a confidence interval on this maximum likelihood estimate $P_{ij}$?
- And for all maximum likelihood estimates of all possible room combinations?

Confidence intervals review
It is known that monthly spending by individuals in a community is normally distributed. To estimate the average monthly spending, a statistician asked 4 people randomly about their monthly spending. The following are the number of dollars these for spent in April: 480, 510, 501, 505. With 98% confidence, over what interval doesthe mean of people's monthly spending lie?