Master Confidence Intervals: Examples, Equations, and Questions

How long will it take to drive 450 km if you are driving at a speed of 50 km per hour?
1) 9 Hours
2) 3.5 Hours
3) 6 Hours
4) 12.5 Hours

The manufacturer of a certain type of product claims that the machine that fills packages of these products is set up in such a way that the average net weight of the packages is 32 grams with a variance of 0.015 square grams. A consumer agency took a random sample of 25 packages from the production line, obtained a sample variance of 0.029 square grams and constructed a 95% confidence interval for the population variance. It is known that the machine is stopped and adjusted if either both or one of the two limits of the confidence interval is not in the interval 0.008 to 0.030. Based on the agent’s sample information, does the machine need an adjustment? Assume that the product net weights in all packages are normally distributed.
I know the confidence interval is defined by:
$[\stackrel{¯<!--\; ¯\; -->}{x}-<!--\; -\; -->\frac{cs}{\sqrt{n}},\stackrel{¯<!--\; ¯\; -->}{x}+\frac{cs}{\sqrt{n}}]$
I'm just not sure which mean or standard deviation to use.

About intervals of confidence
Suppose that $T_1$ is a $\mathrm{\alpha <!--\; \alpha \; -->}\ast <!--\; \ast \; -->100\mathrm{\%<!--\; \%\; -->}$ lower confidence limit for $\mathrm{\theta <!--\; \theta \; -->}$ and $T_2$ is a $\mathrm{\alpha <!--\; \alpha \; -->}\ast <!--\; \ast \; -->100\mathrm{\%<!--\; \%\; -->}$ upper confidence limit for $\mathrm{\theta <!--\; \theta \; -->}$. Further assume that $P(T_1<T_2)=1$. Find a $(2\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->1)\ast <!--\; \ast \; -->100\mathrm{\%<!--\; \%\; -->}$ confidence interval for $\mathrm{\theta <!--\; \theta \; -->}$
I thought that $(T_1,T_2)$ was a ${\mathrm{\alpha <!--\; \alpha \; -->}}^2\ast <!--\; \ast \; -->100\mathrm{\%<!--\; \%\; -->}$ confidence interval, but it is only if $T_1,T_2$ are independents (correct?)

Why is the "wrong" interpretation of confidence intervals still seemingly correct?
According to online sources, if you are operating at 95% confidence, it means if you repeated a sampling process many times and then looked at the 95% confidence intervals over all the results, 95% of the time the brackets would contain the true population mean.
But then they say that it is NOT the same as saying "you can be 95% confident that the intervals you computed contain the population mean."
Isn't it, though? For instance I do my first experiment and get a 95% confidence interval. Then a second. Third, fourth, ..., 100th. 95 of those intervals should contain the population mean. Is this correct so far?
If so, then why isn't it the same as me saying, from the moment I did the very first test, "this particular interval has a 95% chance at being one of the intervals that contain the true population mean"?

Statistics and confidence - intervals
An account on server A is more expensive than an account on server B. However, server A is faster. To see whether it's optimal to go with the faster but more expensive server, a manager needs to know how much faster it is . A certain computer algorithm is executed 20 times on server A and 30 times on server B with the following results,
$\begin{array}{}& ServerA& ServerB\\ \text{Sample means}& 6.7min& 7.5min\\ \text{Sample std. dev.}& 0.6min& 1.2min\end{array}$
A 95% confidence interval for the difference ${\mathrm{\mu <!--\; \mu \; -->}}_1-<!--\; -\; -->{\mathrm{\mu <!--\; \mu \; -->}}_2$ between the mean execution times on server A and server B is [-1.4,-0.2] . Is there a significant difference between the two servers?
(a) Use the confidence interval above to conduct a two-sided test at the 1% level of significance.
(b) Compute a p-value of the two-sided test in (a).
(c) Is server A really faster? How strong is the evidence? Formulate the suitable hypothesis and alternative and compute the corresponding p-value.

Computing 95% confidence interval
The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber 75 million years ago give these percents of nitrogen: 63.4, 65.0, 64.4, 63.3, 54.8, 64.5, 60.8, 49.1, 51.0. Construct a 95% confidence interval for the true average % nitrogen in the atmosphere at this time.

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?

A random sample of 10 motorists buying petrol are found to spend an average of £58.30 with estimated standard error £5.25. Calculate a 95% confidence interval for the expected spending of motorists at this petrol station.
I got 10 of these questions so if someone can help me do this one i think i can do the rest.

Confidence Intervals Calculation and Interpretation
Could someone verify if my reasoning is correct?
1. Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance) $\pm <!--\; \pm \; -->$ (critical value)(standard deviation of sample variance)] to estimate the population variance?I understand that the form statistic $\pm <!--\; \pm \; -->$ (critical value) ∗ (standard dev. of statistic) is applicable to mean and proportion, but is it applicable to any statistic? I believe there is a more complex formula for variance, as it follows a chi-squared distribution and is not always normally distributed, even when the population is.
2. A 90% confidence interval for the height, in meters, of adults in Switzerland is 1.78 $\pm <!--\; \pm \; -->$ 0.2. Are we 90% confident that the sample mean of the next sample of adults taken in Switzerland will be between 1.58 and 1.98 meters tall?
Would this also be false? I understand that a particular confidence interval of 90% calculated from an experiment does not mean that there is a 90% probability of a sample mean from a repeat of the experiment falling within this interval.

Confidence Interval Probabilty
1) We have discovered a new method to produce a liquid containing Zinc. We measure the concentration $\mathrm{\mu <!--\; \mu \; -->}$ four times independently (from the same solution) and find the values
0.3, 0.33, 0.3, 0.27
a) Assuming that the standard deviation of the measurement error is 0.02 (when we make one measurement) estimate μ and give a 90%-confidence interval.
Answer:
We assume the measurement errors to be normal with zero expectation. The estimate for $\mathrm{\mu <!--\; \mu \; -->}$ is
${\mu }_X=(X_1+X_2+X_3+X_4)/4=(0.3+0.33+0.3+0.27)/4=\mathrm{0.3.}$
The same procedure as usual leads to a confidence interval
$[0.3-<!--\; -\; -->b_{\text{standard deviation}}/\sqrt{n},0.3+b_{\text{standard deviation}}/\sqrt{n}]$
where the standard deviation equals 0.02 and $\sqrt{n}=2$, whilst b is defined by the equation
$P(-b\le <!--\; \le \; -->N(0,1)\le <!--\; \le \; -->b)=90\mathrm{\%<!--\; \%\; -->}.$
The last equation above is equivalent to $\mathrm{\lambda <!--\; \lambda \; -->}(b)=0.95$ which we can solve with the help of a table and find $b=1.645$. Hence the 90% confidence interval is equal to $[0.3-0.0164,0.3+0.0164]=[0.2836,0.3164]$. The true concentration is thus between 0.2836 and 0.3164 with at least 95% probability.
My question is how were they able to get 95% probability and b?

Conclusion for confidence interval
If I got, let's say, a 95 % confidence interval for the mean and a 95 % confidence interval for the variance.
Would it then be wrong to conclude:
The 95 % confidence interval for the mean contains with at least 95 % probability the true mean?
and
The 95 % confidence interval for the variance contains with at least 95 % probability the true variance?
What would be a more correct/precise way to express what the confidence intervals stand for?

Test of confidence intervals?
In one of my assignments I have to "test" if the confidence intervals for a set of parameters in a mixed effect model is accurate. I'm asked to simulate from fittet parameters and there after refit them using the same model many times, and lastly take 2.5% and 97.5% quantiles of them and compare with the original CIs. My question is, how does this procedure in anyway measure how accurate my original confidence intervals are?

Statistics, confidence interval
I have a sample of $x_i$ where $x_i=\mathrm{\xi <!--\; \xi \; -->}+\mathrm{\eta <!--\; \eta \; -->}$, $\mathrm{\xi <!--\; \xi \; -->}\sim <!--\; \sim \; -->N(0,{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$ and $\mathrm{\eta <!--\; \eta \; -->}\sim <!--\; \sim \; -->N(a,1)$ i.i.d. So I need to construct confidence interval with confidence level $\mathrm{\gamma <!--\; \gamma \; -->}$ for ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$ with unknown a. My attempt is, I used the following statistic: $S^2=\frac{\sum <!--\; \sum \; -->(x_i-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{x}{)}^2}{n-<!--\; -\; -->1}$ but I get an interval which could be with negative endpoints.

Defining confidence intervals
I have a set of n observations O. I want to calculate the mean of these observations and their confidence interval. I can do this when the observations are iid, by looking up the t-table for the required confidence value and then calculate the interval as --
$2t\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}$
However, if the $i^{th}$ observation of this sequence is related to $(i-<!--\; -\; -->1{)}^{th}$ observation, how do I now define the confidence interval. I believe the iid assumptions are no longer valid.

The confidence interval is [2.663;2.937]?
The concentration of calcium in the blood for a given population follows a normal law of mean $\mathrm{\mu <!--\; \mu \; -->}=2.8$ mmol / L and standard deviation $\mathrm{\sigma <!--\; \sigma \; -->}=0.7$ mmol / L. If we take a sample of 100 people, what is the confidence interval of the calcium concentration at the 95% confidence level.
I got the confidence interval [2.663; 2.937] but I think this is wrong.

What is the difference between confidence interval and the width of confidence interval?
Let $X_1,...,X_n$ be n random samples of a normal distribution $X\sim <!--\; \sim \; -->N(\mathrm{\mu <!--\; \mu \; -->},4)$ where $\mathrm{\mu <!--\; \mu \; -->}$ is unknown. Write down the 95% confidence interval of $\mathrm{\mu <!--\; \mu \; -->}$ in terms of n. What sample size n is required such that the width of this confidence interval is barely larger that 0.5?

Find confidence levels from given intervals
A set of 40 data items, produces a confidence interval for the population mean of $94.93<\mathrm{\mu <!--\; \mu \; -->}<105.07$. If $\sum <!--\; \sum \; -->x^2=424375$, find the confidence level.
So the idea of confidence intervals is still rather new to me and one that isn't fully clear in my mind, would someone be able to give me some hints about solving this and explain what they are doing to solve it.
I get that the confidence interval is given by
$$\stackrel{¯<!--\; ¯\; -->}{x}-<!--\; -\; -->z\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}<\mathrm{\mu <!--\; \mu \; -->}<\stackrel{¯<!--\; ¯\; -->}{x}+z\frac{\mathrm{\sigma <!--\; \sigma \; -->}}{\sqrt{n}}$$
and that $z={\mathrm{\Phi <!--\; \Phi \; -->}}^{-<!--\; -\; -->1}(c)$ where c is the confidence level, however this is as far as my knowledge goes.