Master Confidence Intervals: Examples, Equations, and Questions

Construct confidence interval from proportions
Suppose you have a population of count data, i.e., 1,2,3,…,k, you have a sample of the population of size n, and you have a confidence interval for the proportion of 1's , 2's,\dotsn's in the population. Is it possible to construct a confidence interval for the sum of the counts in your sample.
For example,
Suppose you have a sample of size 1000 from a population of count data that can take any value from 1 to 5 and the
95% CI for 1's is [.75,.76]
95% CI for 2's is [.12,.13]
95% CI for 3's [0.04,.045]
95% CI for 4's is [.048,.058]
95% CI for 5's is [.03,.035]
Can I construct a 95 percent confidence interval for $n\mathrm{\mu <!--\; \mu \; -->}$, where $n=1000$.

90% Confidence Interval
A random sample of 504 out of 1600 students in a school play soccer. How large should the sample of students be if a 90% confidence interval is desired with a margin of error of 2.2 percentage points for the proportion of students that play soccer in the school?
My main problem here is the interpretation of margin of error. The confidence interval I came up with was $\frac{504}{16000}+-<!--\; -\; -->1.645\sqrt{\frac{\left(\frac{8631}{40000}\right)}{n}}$. The + and - back to back represent a "plus or minus". 1.645 is the respective z value for a 90% confidence interval and $\left(\frac{8631}{40000}\right)$ $=\frac{504}{1600}(1-<!--\; -\; -->\frac{504}{1600})$ is the variance I calculated.
I thought then that I needed to find a value for $1.645\sqrt{\frac{\left(\frac{8631}{40000}\right)}{n}}$ that was 2.2% of $\frac{504}{1600}$. This means that I got a final n value of approximately 12158, but I am not sure if this is what is required in this situation.

Confidence intervals and sample size.
Are confidence intervals wider if sample size is larger? Similarly are confidence intervals narrower if sample size is smaller?

Confidence intervals for means
[126.4, 132.2] is a 95% confidence interval for the mean $\mathrm{\mu <!--\; \mu \; -->}$ of a normally distributed random variable with known variance. Find a 98% confidence interval for $\mathrm{\mu <!--\; \mu \; -->}$, based on the same sample.
I got gamma $=0.98gamma/2=0.49soZ\ast <!--\; \ast \; -->=2.33$ but then I am lost on what to do next.

Confidence Intervals - Doubts on Interpretations
Suppose we use a sample mean $\stackrel{¯<!--\; ¯\; -->}{X}$ to construct a 95% confidence interval [a,b].
I was told that it is incorrect to say that there is a 95% probability that the population mean lies between a and b. Because the population mean is a constant and not a random variable. The probability that a constant falls within any given range is either 0 or 1.
However, from the textbook it is said that we expect 95% of the confidence intervals to include the population mean.
If 95% of the confidence intervals are expected to include the population mean, then each confidence interval has 95% probability to include the population mean. Therefore I think it is correct to say there is a 95% probability that the population mean lies between a and b.
Where did I make mistakes?

"... This means that if we used the same sampling method to select different samples and computed an interval estimate for each sample, we would expect the true population parameter to fall within the interval estimates 95% of the time. "
However, to my knowledge, isn't this one of the common misconceptions of a confidence interval? And that the actual interpretation is that if we gather n amount of these confidence intervals, there is a 95% probability that the n collected intervals all contain the true parameter?
My notes give this definition:Let $L:=L(X_1,\dots <!--\; \dots \; -->,X_n)$ and $U:=U(X_1,\dots <!--\; \dots \; -->,X_n)$ be such that for all $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}$,
$$\mathbb{P}(L<\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->U)\ge <!--\; \ge \; -->1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}$$
and then it says, that this is the probability $1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}$ that the true parameter lies within this interval . Which one is correct?

98% confidence interval
For this question from a large amount of data I have calculated that the mean is 44.22, the sample size is 100 and the standard deviation is 22.0773.
From this I am asked to , make the 98% confidence intervals for the (1) true mean µ of the module mark (2) true variance of the module mark
And for each: (a) Determine what quantity to look at, and which distribution table to use, justifying your choice. (b) Determine the number of degrees of freedom, justifying your answer. (c) Calculate the actual intervals.
So far for 1, I have used the z table to look for 99% as I need 1% to the right of 2.33 and 1% to the left of −2.33, so 98% is between $\pm <!--\; \pm \; -->2.33$. Giving me
$$\stackrel{¯<!--\; ¯\; -->}{x}\pm <!--\; \pm \; -->2.33\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}$$
Which provides me with a 39.08 to 49.36 confidence interval, is this correct? And how would I determine degrees of freedom and go about answering part 2?

Confidence interval for difference confidence level
Lets say I measure quantity $\mathrm{\mu <!--\; \mu \; -->}$ and compute confidence interval for it $[\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_1,\mathrm{\mu <!--\; \mu \; -->}+{\mathrm{\delta <!--\; \delta \; -->}}_2]$ with a 68% confidence limit.
What if I want a confidence interval for 95% CL, can I simply scale errors like this? Let $x=\frac{0.95}{0.68}$
$[\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->x{\mathrm{\delta <!--\; \delta \; -->}}_1,\mathrm{\mu <!--\; \mu \; -->}+x{\mathrm{\delta <!--\; \delta \; -->}}_2]$

How we interpret confidence intervals.
For the formula $\stackrel{¯<!--\; ¯\; -->}{X}\pm <!--\; \pm \; -->z_{\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}}\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}$, we substitute ${\stackrel{¯<!--\; ¯\; -->}{x}}_{obs}$ for $\stackrel{¯<!--\; ¯\; -->}{X}$, but isn't that observed sample mean ${\stackrel{¯<!--\; ¯\; -->}{x}}_{obs}$ only for one particular sample?
Also why is it incorrect to interpret a confidence interval as the probability that the actual $\mathrm{\mu <!--\; \mu \; -->}$ lies in a interval (a,b) is $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})100\mathrm{\%<!--\; \%\; -->}$

T-Intervals and % Confidence Interval
A random sample of six 2009 sports cars is taken and their "in the city" miles per gallon is recorded. The results are as follows: 23, 19, 24, 17, 16, 22. Assuming the population distribution is normal, calculate the 99% confidence interval for μ, the population mean "in city" mpg for 2009 sports cars.

Confidence Intervals (A level)
Firstly let me apologise for asking this question on here - I see this as getting a sledgehammer to crack a nut, but I have nobody else that I can ask for advice on this topic.
I'm an A level Mathematics student so please forgive me for lack of/poor notation that you may normally come to expect/be familiar with.
I am looking at the specification for my exam that is coming up and these are three sections:
A) Construct symmetric confidence intervals for the mean of a normal distribution with known variance
B) Construct symmetric confidence intervals from large samples, for the mean of a normal distribution with unknown variance.
C) Construct symmetric confidence intervals from small samples, for the mean of a normal distribution with unknown variance using the t -distribution.

Confidence interval for Uniform($\mathrm{\theta <!--\; \theta \; -->}$, $\mathrm{\theta <!--\; \theta \; -->}+a$)
Consider a random sample $X_1,\dots <!--\; \dots \; -->,X_n$ from a Uniform($\mathrm{\theta <!--\; \theta \; -->}$, $\mathrm{\theta <!--\; \theta \; -->}+a$) distribution, where $\mathrm{\theta <!--\; \theta \; -->}$ is unknown and a is known. We wish to determine a confidence interval for $\mathrm{\theta <!--\; \theta \; -->}$.
The reader may verify the following details: The statistics $Y={\text{min}}_i{X}_i$ and $Z={\text{max}}_i{X}_i$ are jointly sufficient for $\mathrm{\theta <!--\; \theta \; -->}$. For $\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->c_1\le <!--\; \le \; -->c_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a$, $P\{c_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->c_2\}=[(c_2-<!--\; -\; -->c_1)/a{]}^n$. For $0<\mathrm{\gamma <!--\; \gamma \; -->}<1$, set $d_1=(1-<!--\; -\; -->\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}})/2$ and $d_2=(1+\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}})/2$. Then $\mathrm{\gamma <!--\; \gamma \; -->}=P\{\mathrm{\theta <!--\; \theta \; -->}+a{d}_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a{d}_2\}=P\{Z-<!--\; -\; -->a{d}_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->Y-<!--\; -\; -->a{d}_1\}$. Thus, $\mathrm{\gamma <!--\; \gamma \; -->}=P\{\mathrm{\theta <!--\; \theta \; -->}+a{d}_1\le <!--\; \le \; -->Y\le <!--\; \le \; -->Z\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}+a{d}_2\}=P\{Z-<!--\; -\; -->a{d}_2\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->Y-<!--\; -\; -->a{d}_1\}$ is a γ confidence interval for $\mathrm{\theta <!--\; \theta \; -->}$.
Now here's the difficulty: If we observe $Z-<!--\; -\; -->Y>a\sqrt[n]{\mathrm{\gamma <!--\; \gamma \; -->}}$, then $Z-<!--\; -\; -->a{d}_2>Y-<!--\; -\; -->a{d}_1$, so our formula yields a nonsensical answer. Have I made an error in my calculations? Or is this one of those problems with confidence intervals?

Why can we use empirical standard deviation when computing mean confidence intervals
I'm reviewing some basic statistics, and I'm asking myself questions on things I used to take for granted when I first saw them years ago. I'm going to state things as I understand them, so there might be a mistake in the following.
Consider a random variable X following a distribution of mean $\mathrm{\mu <!--\; \mu \; -->}$ and standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}$. We measure n samples of X, and observe an empirical mean $\stackrel{¯<!--\; ¯\; -->}{x}$ and empirical std s.
Because we're observing samples, $\stackrel{¯<!--\; ¯\; -->}{x}$ and s are random variables themselves. We should therefore not use $\stackrel{¯<!--\; ¯\; -->}{x}$ and $\mathrm{\mu <!--\; \mu \; -->}$ interchangeably, and the same goes for s and $\mathrm{\sigma <!--\; \sigma \; -->}$. Instead, people compute confidence interval on μ based on $\stackrel{¯<!--\; ¯\; -->}{x}$. By the central limit theorem, if n is large enough, we can say that $\stackrel{¯<!--\; ¯\; -->}{x}\sim <!--\; \sim \; -->N(\mathrm{\mu <!--\; \mu \; -->},\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}})$.
When computing confidence intervals, we usually use ${\mathrm{\sigma <!--\; \sigma \; -->}}_{\stackrel{¯<!--\; ¯\; -->}{x}}=\frac{s}{\sqrt{n}}$. If all of this is correct, my question is the following: since both s and $\stackrel{¯<!--\; ¯\; -->}{x}$ are random variables, why does it seems to be ok to consider $s=\mathrm{\sigma <!--\; \sigma \; -->}$ when computing confidence intervals for $\mathrm{\mu <!--\; \mu \; -->}$?

Confidence Interval problem
Which of the following will result in a wider confidence interval?
Larger sample size
Higher level of confidence
Larger population standard deviation
Larger population mean
Larger sample mean
So after running some numbers, I believe 2,3, and 4 are correct because they appeared to work however apparently that's incorrect. Could anyone please help me?

Calculate confidence level from a given confidence interval
I'm fairly comfortable calculating the confidence interval. But now I'm seeing a problem where I'm giving a confidence interval CI(27.6621,30.3379) and I'm requested to calculate the confidence level. I'm also given the sample size $n=85$ and a standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}=7.5$. I can't find a formula for that, I feel like it's something simple I'm not seeing. Any help would be appreciated.

Calculating the confidence interval for population variance given the confidence interval for population mean
I am given a sample size of 9 from a normal population, as well as the 95% confidence interval for population mean. I want to calculate the 95% population variance.
I see that the confidence interval for an unknown population mean is $\stackrel{¯<!--\; ¯\; -->}{Y}\pm <!--\; \pm \; -->t_{\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}}\frac{S}{\sqrt{n}}$. Then since the midpoint of the confidence interval for population mean is the sample mean, I can plug that into $\stackrel{¯<!--\; ¯\; -->}{Y}$, 9 into n, and 0.025 into $\mathrm{\alpha <!--\; \alpha \; -->}$. From there, I can solve for S. How can I find the confidence interval for population variance from there?

the situation in the question :so we have a number of scatter plots with each showing an estimated regression line (based on a valid model) and associated individual 95% confidence intervals (CI) for the regression function at each x-value, as well as the observed data. A professor asks 'I don't understand how 95% of the observations fall outside the 95% CI as depicted in the figures'. Briefly explain how is is entirely possible that 95% of the observations fall outside the 95% CI as depicted in the figures.(We weren't given actual figures)
Anyway I thought that it may have been due to the fact that a lot of outliers affected the regression line calculated, and so a confidence interval formed from a bad regression line would be bad - resulting in 95% of observations falling outside the 95% CI.

Interpretaton of confidence interval
I have read two books that explicitly state that the $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$% confidence interval should be interpreted as:
If you construct 100 such confidence intervals, $\mathrm{\alpha <!--\; \alpha \; -->}$ of them are expected to not contain the true population statistic and $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ of them are expected to contain the true population statistic.
and not as
There is a $(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ probability that the true population statistic is contained in the confidence interval.
In my view, they both equivalent: If you make the first statement, you implicitly make the second statement. You are looking at any one arbitrary confidence interval, which in itself is a random variable, the generic confidence interval should, therefore, be subject to the second statement. Do things change when this random variable is actually realized?

Estimators and Confidence intervals
I was curious as to what the relationship between probabilistic values, estimators and confidence intervals are. I was wondering, if I have an estimator of some parameter $\mathrm{\lambda <!--\; \lambda \; -->}$, and a probability value that depended on $\mathrm{\lambda <!--\; \lambda \; -->}$ and the estimator, what would be the relationship between those two and a confidence interval for an arbitrary distribution?

95% Confidence Interval for $\mathrm{\lambda <!--\; \lambda \; -->}$
Consider a random sample $X_1,X_2,..,X_n$ from a variable with density function
$$f_X(x)=2\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\pi <!--\; \pi \; -->}x{e}^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\pi <!--\; \pi \; -->}{x}^2}x>0$$
A useful estimator of $\mathrm{\lambda <!--\; \lambda \; -->}$ is
$$\stackrel{^<!--\; ^\; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{n}{\mathrm{\pi <!--\; \pi \; -->}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_1^2}$$
Derive a 95% confidence interval for $\mathrm{\lambda <!--\; \lambda \; -->}$ (a hint is provided suggesting to consider the distribution of $\frac{\mathrm{\lambda <!--\; \lambda \; -->}}{\stackrel{^<!--\; ^\; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}}).$).