Improve Your Understanding of College Level Statistics Problems

$(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$ 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?

Combinatorics - How many ways are there to arrange the string of letters AAAAAABBBBBB such that each A is next to at least one other A?
I found a problem in my counting textbook, which is stated above. It gives the string AAAAAABBBBBB, and asks for how many arrangements (using all of the letters) there are such that every A is next to at least one other A.
I calculated and then looked into the back for the answer, and the answer appears to be 105. My answer fell short of that by quite a bit. I broke down the string into various cases, and then used Stars and Bars to find how many possibilities there are for each. Now here's what I have got so far. First case would be all As are right next to each other, leaving 2 spots for the 6 Bs. That gives $(\genfrac{}{}{0ex}{}{7}{1})$ from Stars and Bars, 7 possibilities. Second case was dividing the As into 2 groups of 3 As. There would have to be 1 B between the two, which leaves 5 Bs that can be moved. Using Stars and Bars, there are 3 possible places to place a B and 5 Bs in total, so $(\genfrac{}{}{0ex}{}{6}{2})$, 15 possibilities. Then there's a group of 4 As and another group of 2 As. 1 B would be placed inbetween, and then the calculation would be the same as the second case, except it would have to be doubled to account that the groups of As can be swapped and it would be distinct. That gives 30 possibilities. Then I found one final case of dividing the As into 3 groups of 2 As. 2 Bs would immediately be placed between the 3 groups, leaving 4 Bs to move between the 4 possible locations. I got $(\genfrac{}{}{0ex}{}{5}{3})$ for that, which adds 10 possibilities. Summing it up, I only have 62 possibilities, which is quite far from the 105 answer. Any ideas where I might have miscalculated or missed a potential case? Additionally, are there any better ways to calculate this compared to this method of casework?

Equations of significance probabilities
Consider a population of independent light bulbs with an exponential lifetime distribution with mean $\mathrm{\mu <!--\; \mu \; -->}$. It is claimed that their expected lifetime is 1000 hours. A definition of a 100(1−$\mathrm{\alpha <!--\; \alpha \; -->}$)% confidence interval obtained from an observation to is the set of all ${\mathrm{\mu <!--\; \mu \; -->}}_0$ which are not rejected in a test of a null hypothesis ${\mathrm{\mu <!--\; \mu \; -->}}_0$ against an alternative hypothesis $\ne <!--\; \ne \; -->$.
One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T "" 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$) = 0.05 (for a test of ${\mathrm{\mu <!--\; \mu \; -->}}_0$ versus ${\mathrm{\mu <!--\; \mu \; -->}}_0$) and Pr(T "" 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$ = 0.05 (for a test of ${\mathrm{\mu <!--\; \mu \; -->}}_0$ versus ${\mathrm{\mu <!--\; \mu \; -->}}_0$) for $\mathrm{\mu <!--\; \mu \; -->}$. Determine the range of values of $\mathrm{\mu <!--\; \mu \; -->}$ such that both of the probabilities Pr(T "" 622 | $\mathrm{\mu <!--\; \mu \; -->}$) and Pr(T "" 622 |$\mathrm{\mu <!--\; \mu \; -->}$) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for $\mathrm{\mu <!--\; \mu \; -->}$.)
I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T $\ge <!--\; \ge \; -->$ 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$), Pr(T $\le <!--\; \le \; -->$ 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$) and compare with 0.05? I believe I will get the second part after I understand this bit.

What is needed to specify a group?
I have come across several groups, some of which have the same number of generating elements and of the same orders. Take, for instance, $D_{2n}$ and $S_n$. I have never seen it read explicitly, but it seems to me that many groups have some number of generating elements of a given order, and then also have some additional structure on top of this, such as the requirement that $sr=r^{-<!--\; -\; -->1}s$ for the Dihedral group, which allows the "object" to still satisfy the group axioms but it has a restricted set of allowed elements compared with the group generated by the elements with no restriction alone. Another example coming to mind is any Abelian group, which has this additional structure on it.
Are these properties (i.e. the number of generating elements, their orders, and any additional properties/constraints) necessary and sufficient to specify a group? I am trying to think in terms of isomorphisms; I would try to show two groups are 'the same' by showing that an isomorphism exists between them. onsidering this, it certainly seems sufficient that two groups are matched with these properties. The isomorphism can then simply map the corresponding elements on to each other. But is it necessary? Or, in other words, if two groups are not exactly matched in these properties, is there no isomorphism between them? Perhaps it also makes a difference if the groups are of infinite order. Then reerring to 'the number of generating elements' and 'their orders' seems a bit suspect...

Classify each of the following random variables as discrete or continuous. X = the reported score of a randomly selected senior at your school on the SAT Math test.

Finding confidence interval
A random sample of 700 units from a large consignment showed that 200 were damaged.how can we find the 95% confidence interval for the proportion of damaged unit in the consignment.

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?

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?

Given the problem of a patient taking a test for a disease where having the disease is denoted by X and the a positive test is denoted by Y,
- the rate of occurrence of the disease in the general population is 1%
- the probability of a false positive is 3%
- the odds of getting tested positive is 90% if you have the disease
is it appropriate to solve the following through rearranging the total probability
$P(A)=P(A|B)P(B)+P(A|NotB)P(NotB)$
Into
$P(A|B)=(P(A)-<!--\; -\; -->P(A|B)P(B))/P(NotB)$
and then getting P(B) from the total probability
$P(B)=P(B|A)P(A)+P(B|NotA)P(NotA)$
then given B is binary P(Not B) from
$1=P(B)+P(NotB)$
and getting P(A|B) from Bayes
$P(A|B)==P(A)P(B|A)/P(B)$
and then substituting that all back into the first equation to get the result

A population includes a fraction m of individuals carrying a disease exists and has the following characteristics:
P(positive test | individual with the disease) = p
P(positive test | individual without the disease) = r
What is the probability of a false negative: P(individual with the disease | negative test)?
My try
Let W = with the disease, ~W = without the disease;
Let + = positive test, - = negative test
From the given, we have: P(W) = m, P(~W) = 1-m
We want P(W | -) = (P(- |W) P(W))/(P(-)) = (P(- | W)P(W)/(P(- | W)P(W) + P(- ~W)P(~W))
We have: P (+) = P(+|W)P(W) + P(+|~W)P(~W) = p * P(W) + r * P(~W)

Classify each of the following random variables as discrete or continuous. W = the exact amount of sleep that a randomly selected student from your school got last night.

How to check whether a lifting between given mappings is possible?
Suppose mappings
$$M\to <!--\; \to \; -->P_1,M\to <!--\; \to \; -->P_2,$$
where M is some manifold, while $P_{1,2}$ are in general different spaces. I want to clarify if there exists a lifting between these two mappings.
Suppose $M=S^n,n>1$. Then have I to compare the homotopy groups
$${\mathrm{\pi <!--\; \pi \; -->}}_i(P_1)\text{with}{\mathrm{\pi <!--\; \pi \; -->}}_i(P_2),i=1,...,n,$$
or only
$${\mathrm{\pi <!--\; \pi \; -->}}_n(P_1)\text{with}{\mathrm{\pi <!--\; \pi \; -->}}_n(P_2)?$$

Suppose two random variables have standard deviations of 0.10 and 0.23, respectively. What does this tell you about their distributions?

Finding confidence level given confidence interval and sample size
The Austin-American Statesman asked an SRS of 100 Austinites for their opinion about Mayor Adler's performance. Using the results of the sample, the C% confidence interval for the proportion of all Austinites who approve of the mayor's job performance is .565 to .695. What is the value of C?
A) 82
B) 86
C) 90
D) 95
E) 99
To do this, I set up the equation to get the confidence interval:
Confidence interval =$.63\pm <!--\; \pm \; -->z\cdot <!--\; \cdot \; -->\sqrt{{\displaystyle \frac{0.63\cdot <!--\; \cdot \; -->0.37}{100}}}$
Solving for z gives around 91%...
Am I doing this right?

I'm struggling to compare the %increase/decrease in the average homicide rate of a select group of cities to the state homicide rate.
For example, if the state saw a 4% decrease (-4) in the homicide rate over a two-year period but the average homicide rate in my group of comparison cities increased by 2%(+2), how much bigger was the increase in my group of comparison cities compared to the state? Is it possible to make a statement along these lines: "The average homicide rate of the comparison group increased X times more than the state rate over the same period."
It can't be this $|-<!--\; -\; -->4|/2=2$ because it's definitely more than double.

A clinical researcher has developed a new test for measuring impulsiveness and would like to establish the validity of the test. The new test and an established measure of impulsiveness are both administered to a sample of participants. Describe the pattern of results that would establish concurrent validity for the new test.

Generation of unique normalized and weight value from a set with different number of elements
To generate a specific formula for my research, I need to define some steps, and for this, I need to take a group of elements, specifically, three elements and have as return a single normalized value.
For example:
$f_1=(105,312,75)$. These elements can range from zero to values greater than one million.
I tried to use fnorm $=(X-<!--\; -\; -->min(X))/(max(X)-<!--\; -\; -->min(X))$ and calculated the arithmetic mean for the group of elements, but the representation of the result does not agree with what I would like.
Comparing two sets of these elements, for example:
$f_1=(105,312,75)$
$f_2=(22,23,1)$
When I apply fnorm and the arithmetic mean of the values, I get the final result $\to <!--\; \to \; -->f_1=0.3755$ and $f2=0.65$.
These results are not complete for me, because even though they are normalized, the elements of f1 are greater than f2. Therefore, I need f1 to be more representative, have greater weight than f2. Another point is that if there is a value 0 within the group, it cannot zero the whole result the other different elements of zero are relevant to the outcome.

Size of automorphism group of element of $Y_0(5)$
The modular curve $Y_0(5)$ parametrizes elliptic curves E with an isogeny of degree five. So an element of $Y_0(5)$ can be interpreted as
$E\stackrel{\mathrm{\varphi <!--\; \varphi \; -->}}{\to }{E}^{\prime }$
Suppose we are working over a finite field. An automorphism of these curves is an automorphism of the curve E, that sends the kernel G of the map $\mathrm{\varphi <!--\; \varphi \; -->}$ to itself, over some algebraic closure. The elements of $Y_0(5)$ do admit automorphisms, because any curve has the hyperelliptic involution.
Because G is of order five, it can have at most four automorphisms. If the curve E has six automorphisms, it is clear, by comparing the sizes of the groups, that only the identity and the hyperelliptic involution respect the subgroup G. If E has four automorphisms, this kind of reasoning doesn't work. Is there a similar result if E has four automorphisms? To me it is not clear if there are always only two automorphisms that map G to itself, or that there could be four.

Confidence Intervals that Contain the Mean: Designing an Activity
Roll a fair 6 sided dice 25 times. Take the sample mean of the face value. Using the standard deviation of the uniform probability distribution, construct confidence intervals with $C=.90,.95,.99$.
I had 16 groups total doing this exercise. All 16 groups found their 90% confidence intervals containing the the mean of the probability distribution (3.5.).
I thought I would have at least one confidence intervals that wouldn't contain the mean... considering that the probability that all 16 confidence intervals contains $\mathrm{\mu <!--\; \mu \; -->}$ would be ${.90}^{16}\simeq <!--\; \simeq \; -->.1853$.
Is my prediction completely flawed? Or was I just unlucky? Or is there something inherently wrong with my experiment?

Determine hollow marble with least number of weighings
We have 135 boxes each containing 100 marbles which look identical and weigh 10 grams, with the exception of one box, which contains hollow marbles of 9 grams each. By using a scale that can weight up to 999 grams, what is the least number of weightings required to determine the box with the hollow marbles?
Well, I know the method of numbering the boxes and then taking 1 marble from box 1, 2 from box 2 etc and then compare the result with $\underset{k=1}{\overset{135}{\sum <!--\; \sum \; -->}}k\ast <!--\; \ast \; -->10$, but this is 91800, so we would roughly need 100 weighings! Maybe we can split the 135 boxes into two groups, $68+67$, then from the first group weigh one of each, to see if the scale displays 680 or 679, then continue with either group which contains the hollow marble and do the same?
2nd weighing: 34 or 33
3rd: 17 or 17
And then we can apply the $1+2+3$… method to determine in which of the two groups 9 or 8 are the hollow marbles.
But this way we will need 5 weighings, which I doubt is the optimum method.