Master Confidence Intervals: Examples, Equations, and Questions

The fill of a cereal box machine is required to 18 ounces, with the variance already established at 0.24. The past 150 boxes revealed an average of 17.96 ounces. A testing error of ${\displaystyle \alpha =2\mathrm{\%}}$ is considered acceptable. If the box overfills, profits are lost; if the box underfills, the consumer is cheated. (a) Can we conclude that machine is set at 18 ounces? (b) Find a 98% confidence interval for ${\displaystyle \mu }$. (c) Would the result change if the sample variance were 0.24 from the data rather than knowing ${\displaystyle {\sigma }^2=0.24}$?
My Answer:
(a) I calculated the positive square root of the variance (0.24) to get the standard deviation of the population - ${\displaystyle \sigma =0.49}$. I found the acceptable region to be between -2.33 and 2.33, so, for a value of ${\displaystyle z=-0.9998}$, I concluded that the machine is set at 18 ounces.
(b) ${\displaystyle 17.87<\mu <18.05}$
(c) I understood the question as: "will there be a change in the result of part (b) if 0.24 changed from being a population variance to a sample variance". My answer is that when only the standard deviation of a sample is known and ${\displaystyle n>30}$, the same formula will be used to calculate the 98% confidence interval, so there will be no change in the result of part (b).

The bad debt ratio for a financial institution is defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. A random sample of 6 Ontario banks is selected and that the bad debt ratios (in percentages) for these banks are:
${\displaystyle 7,4,6,7,5,8}$
Compute and interpret a ${\displaystyle 95\mathrm{\%}}$ confidence interval for the mean bad debt ratio. What needs to be true in order for this interval and interpretation to be valid?

Statistics - Confidence Intervals -
${\displaystyle P\left(X>1\right)}$
${\displaystyle X_1,\cdots c,X_n\sim U(0,2p)}$

Sampling Distribution of the Sample Mean
Concerning estimations/intervals:
True or False: if the three conditions Random, Normal, and Independent for using a confidence interval for a population mean are not met, then the sampling distribution of ${\displaystyle \bar{x}}$ is unknown.
I'm unsure about the criteria - would you not know the distribution of the sample mean even if these criteria are unfilled? Are there specific conditions for doing so?

Sample mean to population mean
How to use sample mean to know the population mean? Now I have sample mean, sd and population mean. They are 37.28, 25 and 34. Sample size is 25 Here is my thought ${\displaystyle N<25}$. Sample mean may not equal to population mean. population mean is ${\displaystyle 37.28+\frac{25}{\sqrt{25}}=42}$ Am i correct?

I assume that this problem is trivial but I am not sure what to do: In a survey of 900 people in the US a journalist says that ${\displaystyle 60\mathrm{\%}}$ of people support a new law. If the margin of error is ${\displaystyle 2.7\mathrm{\%}}$ for the percentage, what is the level of confidence?

Confidence Intervals for ${\displaystyle \beta 1}$ and ${\displaystyle \beta 2}$
'm having a bit of trouble with the question below.
Using this Data -> Data
Question: Establish 95% confidence intervals for ${\displaystyle \beta 2}$ and ${\displaystyle \beta 3}$.
I know I use the formula: ${\displaystyle \beta 2\pm }$ (t-value)(standard error)
How do I find the standard error using the data given in the pic?

Confidence interval for a density function parameter
I am trying to solve this problem: Given ${\displaystyle X_1,\dots ,X_n}$ a random sample of a population of random variables with p.d.f.
${\displaystyle f(x,\theta )=e^{-(x-\theta )}I{\left(x\right)}_{x\ge \theta }}$, find a confidence interval of level ${\displaystyle 1-\alpha \text{for}\theta }$.SNKFollowing a hint, I first looked into the distribution of ${\displaystyle Y=\text{min}\left(X_i\right)-\theta }$, which I found to be Exponential(n).
Taking into account that the indicator function in the pdf limits to values of x greater than ${\displaystyle \theta }$, and using the exponential's cdf. and some bounds, I got that the confidence interval should be ${\displaystyle I_{1-\alpha }(X_1,\dots ,X_n)=[0,\frac{\ln \left(\alpha \right)}{n}]}$.
Would that be correct? I am not sure I am using the hint correctly. If it is wrong, what else could I try?

Constructing confidence Interval (standard normal)
I am asked to construct 95 percent confidence interval of the variance,
sigma squared using this: ${\displaystyle \frac{\sum _{j=1}^n\left(\frac{\left|Y_j\right|}{\sigma }|-\mathbb{E}|Z\mid \right)}{\sqrt{nVar\left(\left|Z\right|\right)}}}$?
I know the expression given above is standard normal with mean, 0 and variance 1. Note I have I have calculated E|z| (the expectation of a standard normal) to be ${\displaystyle \sqrt{\frac{2}{\pi }}}$ and Var|z| to be ${\displaystyle 1-\frac{2}{\pi }}$. Note Y1...Yn is a sample from N(0,sigma squared).
Note: ${\displaystyle \sum _{j=1}^n\left|\frac{T_j}{\sigma }\right|}$ are pivots, meaning is a pivot for variance, distribution does not depend on variance.
PROBLEM: The problem I'm having in constructing the confidence interval is isolating sigma squared (variance) from the expression given at the top to get an interval for variance. Please provide help, suggestions. Thank you

Confidence of a failure
A joint of steel pipe (casing) has a 1% failure rate. 400 joints of casing are in a typical well. How may wells can I drill or joints can I run, before I am 90% confident of having at least one casing failure? You can leave out the well to simplify the problem: Casing has a 1% failure rate. How many joints of casing can I have before I am 90% confident in having a failure? Thanks for re teaching me. I think it is called gambler's ruin.

Confidence Intervals using Normal Distribution with variance known
From a sample of 32 observations from a population with ${\displaystyle {\sigma }^2=24.602}$, we obtained ${\displaystyle x=79.47}$. Construct a 99% confidence interval for ${\displaystyle \mu }$.
${\displaystyle \left(a\right)79.47\pm 1.4424}$
${\displaystyle \left(b\right)79.47\pm 1.7186}$
${\displaystyle \left(c\right)79.47\pm 2.0430}$
${\displaystyle \left(d\right)79.47\pm 2.2622}$

Confidence Interval for the Difference between Two Binomial Proportions
I have this stats problem where I am to create a 98% confidence interval for the difference between proportions. ${\displaystyle X_1=21,n_1=90\text{and}{X}_2=39,n_2=70}$.
I did a 2propzinterval test and my confidence interval came out as ${\displaystyle (-0.497,-0.151)}$.
The question is: A marketing manager claims that the proportion of customers who purchased pretzels did not change after the app campaign. Does the confidence interval contradict this claim?
How do I know if this confidence interval contradicts the claim? Thank you for your help!

Confidence interval when both ${\displaystyle \mu }$ and ${\displaystyle {\sigma }^2}$ are unknown
I have the following problem in my problem book:
Let ${\displaystyle X\sim N(\mu ,{\sigma }^2)}$ where both ${\displaystyle \mu }$ and ${\displaystyle {\sigma }^2}$ are unknown. I have to find a confidence interval for the mean.
What I have so far:
I know that when ${\displaystyle {\sigma }^2}$ is unknown I can use t-distribution for finding a confidence interval, i.e.:
${\displaystyle \bar{x}\pm t_{n-1}^{\cdot }\frac{s}{\sqrt{n}},}$
where ${\displaystyle t_{n-1}^{\cdot }}$ is a t-distribution with ${\displaystyle n-1}$ degrees of freedom and ${\displaystyle \bar{x}}$ is the sample mean from a sample of size n.
The whole problem seems kind of vague to me. Is this enough for describing the confidence interval? I don't know the sample mean ${\displaystyle \bar{x}}$ so is it okay to describe the solution in such a way?

Confidence interval for population proportion given two means
I am trying to estimate a confidence interval containing the true number of people who would vote for Trump. So far, I have calculated the mean number of men who would vote for Trump and the mean number of women who would vote for Trump. I have also rejected the null hypothesis that the means are equal.
How would I go about constructing a 95%-confidence interval for the true number of people (men and women) who would vote for him GIVEN my estimates and rejection of equality between the means?

Confidence interval and standard deviation
I'm currently talking an intermediate course in finance where we want to calculate Value-at-Risk for portfolios and bonds. To use this VaR formula I need to know the standard deviation for different confidence intervals. Now my teacher have put up the following standard deviation for different confidence intervals:
${\displaystyle C.I90=\frac{+}{-}1,64S.d}$
${\displaystyle C.I95=\frac{+}{-}1,96S.d}$
${\displaystyle C.I98=\frac{+}{-}2,33S.d}$
${\displaystyle C.I99,9=\frac{+}{-}3,09S.d}$
When I watched an old exam for calculating VaR, the C.I was 99% and the student wrote that the S.d was equal to 2,33. How is this possible? (P:s the student got an A on this exam).

How can I find the Margin of Error E
The professor hasn't gone over this section but I want to understand it so that I won't confused in class. I want to learn so that I can do the other problems.
So this is the problem, If anyone could teach me so I could take note I'd deeply appreciate it. Thank you for your time.
Question: Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. ${\displaystyle n=550}$, x equals 330, 90 % confidence
I know how to obtain the alpha/2 to find the confidence.
I just went over the Binomial Distribution and Limit Theorem, which understand now this is next. If anyone would care to spent their time explaining I'd be forever grateful, thank you again for your time.

Constructing a confidence interval for population variance
Given the following data set, how does one construct a 99% confidence interval for the population variance? I should mention now I don't have a very good understanding of when to divide by n and when to divide by ${\displaystyle (n-1)}$.
$$\begin{array}{|ccccc|}\hline 22.2& 24.7& 20.9& 26.0& 27.0\\ 24.8& 26.5& 23.8& 25.6& 23.9\\ \hline\end{array}$$
I've already tried a couple approaches, none seem to give the correct answers. The first was to find the mean of the samples(sum divided by number of samples), giving
${\displaystyle \bar{X}=\frac{\sum X_n}{n}=24.54}$
Then obtaining the variance of the samples as
${\displaystyle s^2=\frac{\sum {(X_n-\bar{X})}^2}{n}=3.2924}$
Given that there are 9 degrees of freedom, and I'm looking for 99% confidence for a two-sided interval, I used a ${\displaystyle {\chi }^2}$ table to find values of ${\displaystyle {\chi }^2}$ of 1.73 and 23.59. Then the following values using the following formula.
${\displaystyle \frac{(n-1)s^2}{{\chi }_1^2}\le {\sigma }^2\le \frac{(n-1)s^2}{{\chi }_2^2}}$
${\displaystyle 1.256\dots \le {\sigma }^2\le 17.128\dots }$
Upon seeing that this was wrong, I tried to obtain mean and variance by dividing by (n-1) instead of n. This gave.
${\displaystyle \bar{X}=11.919\dots }$
${\displaystyle s^2=3.292\dots }$
Then, using the same ${\displaystyle {\chi }^2}$ values as above, I got the following for a confidence interval
${\displaystyle 4.547\dots \le {\sigma }^2\le 62.002\dots }$
This was also wrong. I cannot tell why these would both be wrong, whether I should have divided by n or ${\displaystyle (n-1)}$, and how I would obtain the correct answer. Does anyone know where I went wrong here?