Master Statistics and Probability Problems with Expert Help

What is the mathematical formula for the pooled variance of two populations?

In order to have population increasing, a country accepts A people a year from other counntries, and the number x(t) of people in a country changes by equation
$x^{\prime }(t)=-0.03x(t)+A$
where t is time in years. while $x(0)=16m$ and $x(-<!--\; -\; -->10)=15m$ find A.
This is what I did:
from the equation I know that each year the population of the country decrease in 0.03
I tried to use the data, since t is in years
$x(t_0)\ast <!--\; \ast \; -->1.03=x(t_0+1)$
Since we wish in 10 years the total popluation increase in million people therefore :
$x(t_0)\ast <!--\; \ast \; -->1.03=x(t_0+1)+100,000$
I get the feeling my solution is wrong, But I don't really find other ways to approach this.
Any ideas? I'll be glad if someone could tell me why my solution fails.

Let $T$ be a complete sufficient real-valued statistic for the parameter $\mathrm{\theta <!--\; \theta \; -->}$ and $S$ be another real-valued statistic whose distribution function $F$ does not depend on the parameter $\mathrm{\theta <!--\; \theta \; -->}$. Show that $P(S\le <!--\; \le \; -->s\mid <!--\; \mid \; -->T)-F(s)$ equals zero almost surely under each $P_{\mathrm{\theta <!--\; \theta \; -->}}$ and for each real $s$. Conclude from this that $S$ and $T$ are independent.

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.

What is the exact value of $\sec <!--\; \; -->180$?

Is a t distribution for a certain degree of freedom equivalent to the sample mean distribution for the corresponding sample size?
This may seem like a weird question, but hear me out. I'm essentially struggling to see the connection between a t-value from a t-table and a t-value that is calculated.
The following formula is used to calculate the value of a t-score:
$t=\frac{\stackrel{¯<!--\; ¯\; -->}{X}-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\frac{S}{\sqrt{n}}}$
It requires a sample mean, a hypothesized population mean, and the standard deviation of the distribution of sample means (standard error).
According to the Central Limit Theorem, the distribution of sample means of a population is approximately normal and the sample distribution mean is equivalent to the population mean.
So the t-score formula is essentially calculating the magnitude of difference between the sample mean in question and the hypothesized population mean, relative to the variation in the sample data. Or in other words, how many standard errors the difference between sample mean and population mean comprise of. For example: If t was calculated to be 2, then the sample mean in question would be 2 standard errors away from the mean of the sample distribution.
1.) Phew, ok. So question 1: Let's just say a t-score of 1 was calculated for a sample mean and since a distribution of sample means is normal according to the CLT, does that mean that the sample mean in question is part of the 68% (because of the $68-<!--\; -\; -->95$ rule)of all sample means that are within 1 standard error of the sample mean distribution?
2.) Let's say we have a distribution of sample means of sample size 15. Is this distribution equivalent for a t-distribution of degrees of freedom 14? Or more importantly: Is the t-value from a t-table for 14 degrees of freedom and 95 confidence EQUIVALENT to a calculated t-value using a sample mean that is 2 standard errors away from the mean of a distribution of sample means with sample size 15?

Which of the following is not a form of bias?
Multiple Choice
Those responding to a survey or poll differ systematically from the nonrespondents.
Information from the sample overemphasizes a particular stratum of the population.
Portions of the population are excluded or underrepresented from the sample.
Certain groups in the population are systematically underrepresented in the sample.

Which of the following is not an example of probability sampling?
simple random sampling
stratified random sampling
proportionate stratified random
sampling quota sampling

Researchers wanted to know about the alcohol drinking habits of high school students in the United States that vape. In order to do so, they randomly selected 10 high schools in the United States and then sampled all of the students from the school by asking them if they had ever vaped. Of the students that were surveyed, 67% said that they frequently vaped.
a.)Does the 67% represent a statistic or a parameter? Explain your reasoning.
b.) What type of sample was obtained? (Random, systematic, convenience, stratified, or cluster). Explain your reasoning

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 would $T=XY+Z$ not be a sufficient statistic for $X,Y,Z{\sim <!--\; \sim \; -->}_{i.i.d}Bern(p)$

In a certain village, 20% of the population has some disease. A test is administered which has the property that if a person is sick, the test will be positive 90% of the time and if the person is not sick, then the test will still be positive 30% of the time. All people tested positive are prescribed a drug which always cures the disease but produces a rash 25% of the time. Given that a random person has the rash, what is the probability that this person had the disease to start with?
I am looking for P(S|R) given that a person tested positive where S denotes a sick person R denotes a person with a rash, given that they tested positive. If + denotes a person who tested positive and I use Baye's formula and the data to calculate $P(S|+)$ would $P(S|R)=\frac{P(R|+)P(S|+)}{P(R|+)}=P(S|+)$? Or would the answer be $P(S|R)=P(R)P(S|+)$? Or are both of these answers wrong? Also, I cannot tell if in the problem statement $P(R)=.25$ or $P(R|+)=.25$

A sampling frame should have the following characteristics:
i) It is not necessary to include all members of the population on the list
ii. The information should be correct and up to date.
iii. Each member of the population appears on the list only once. Which of the above statement/s is /are incorrect?

$f_{\mathrm{\theta <!--\; \theta \; -->}}(x)$, then $T$ is sufficient for $\mathrm{\theta <!--\; \theta \; -->}$ if and only if nonnegative functions $g$ and $h$ can be found such that
$f_{\mathrm{\theta <!--\; \theta \; -->}}=h(x)g_{\mathrm{\theta <!--\; \theta \; -->}}(T(x))$

How to identify and distinguish a sample and population mean?
In a village the mean rent was 1830; a rental company comes and surveys with a sample size of 25 for a hypothetical testing to test if the μ is equals 1830 or not.
This new survey gets a mean value of 1700 with a standard deviation of 200.
For the company to deduce t-value(for hypothesis testing) which is sample mean and which will be population mean? I assumed 1830 to be the population mean but it turned out to be the sample mean and 1700 to be the population mean.

Classify each of the following random variables as discrete or continuous. Y = the number correct on a recent multiple-choice test for a randomly selected student in your class.

How to calculate the diversity index, the probability that two people chosen at random are of different races or ethnicities?
The United States Census Bureau started releasing the results of the latest census. Among many other pieces of information, the bureau recoded the race or ethnicity of the residents of every county in every state. From these results the bureau calculated a 'diversity index,' which measures the probability that two people chosen at random are of different races or ethnicities. The census determined that in a county in Wisconsin, 83% of its residents are white, 8% are black, and 9% are Asian. Calculate the diversity index for this county.

When performing a log-likelihood ratio test, do I take log to both of the likelihood first or find the ratio first and take the log?

A man tests for HIV. What is the predictive probability that his second test is negative?
In a population, it is estimated HIV prevalence to be $\mathrm{\lambda <!--\; \lambda \; -->}$. For a new test for HIV:
- $\mathrm{\theta <!--\; \theta \; -->}$ is the probability of an HIV positive person to test positive
- $\mathrm{\eta <!--\; \eta \; -->}$ is the probability an HIV negative person tests positive in this test.
A person takes the test to check whether they have HIV, he tests positive.
What is the predictive probability he tests negative on the second test?
Assumption: Repeat tests on the same person are conditionally independent.
From my notes predictive probability is given as:
$P(\stackrel{~<!--\; ~\; -->}{Y}=\stackrel{~<!--\; ~\; -->}{y}|Y=y)=\int <!--\; \int \; -->p(\stackrel{~<!--\; ~\; -->}{y}|\mathrm{\tau <!--\; \tau \; -->})p(\mathrm{\theta <!--\; \theta \; -->}|\mathrm{\tau <!--\; \tau \; -->})$ here $\stackrel{~<!--\; ~\; -->}{Y}$ is the unknown observable, y is the observed data and η the unknown.
I am interested in the probability of the second test is negative, given that the first test is positive,without knowing if the man really has HIV or not.
To facilitate this I define:
$y_1$ as the event of the first test being positive and
$\stackrel{~<!--\; ~\; -->}{y_2}$ as the second test being negative
Would this adaption to the formula given above be the correct/best approach to this problem ?
$p(\stackrel{~<!--\; ~\; -->}{y_2},y_1|\mathrm{\tau <!--\; \tau \; -->})=p(\stackrel{~<!--\; ~\; -->}{y_2}|\mathrm{\tau <!--\; \tau \; -->})p(y_1|\mathrm{\tau <!--\; \tau \; -->})p(\mathrm{\tau <!--\; \tau \; -->})$ and this is really $\propto <!--\; \propto \; -->p(\stackrel{~<!--\; ~\; -->}{y_2}|\mathrm{\tau <!--\; \tau \; -->})p(\mathrm{\tau <!--\; \tau \; -->}|y_1)$
I've gotten for the $p(\mathrm{\tau <!--\; \tau \; -->}|y_1)$ from Bayes' theorem:
$$\begin{gathered} p(\mathrm{\tau <!--\; \tau \; -->}|y_1)=\frac{p(\mathrm{\tau <!--\; \tau \; -->})p(y_1|\mathrm{\tau <!--\; \tau \; -->})}{p(y_1)} \\ =\frac{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\theta <!--\; \theta \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\eta <!--\; \eta \; -->}(1-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->})} \end{gathered}$$
How could I then find $p(\stackrel{~<!--\; ~\; -->}{y_2}|\mathrm{\tau <!--\; \tau \; -->})$? Is this the correct approach?

A coach has made a statement that his players have bigger lung capacity than the average of the population of the same age which is 3.4. (Normal distribution)
The measurements yield the following data: 3.4, 3.6, 3.8, 3.3, 3.4, 3.5, 3.7, 3.6, 3.7, 3.4 and 3.6.
$n=11$
$\stackrel{¯<!--\; ¯\; -->}{X}=3.545$
$S=0.157$
Find the required sample size, which lung capacity should be measured, so coach can state his statement with 99% confidence. (assume ${\mathrm{\sigma <!--\; \sigma \; -->}}^2=0.09$)
I don't even know how should I start. My initial thought was to use the U statistics $U=\frac{\stackrel{¯<!--\; ¯\; -->}{X}-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}\sqrt{n}$~$N(0,1)$. But I don't know the U.