Master the Art of Research Methodology: Comprehensive Examples and Expert Advice

First off, this is a vague question about a survey which is, I guess, meant to be vague. So bear with me
In Morel's "Motivic Homotopy Theory" survey he mentioned the following fact in motivating the Tate circles. One notes that ${\mathbb{P}}^1$ is equivalent to $S^1\wedge <!--\; \wedge \; -->{\mathbb{G}}_m$ while, if we follow our topological intuition, ${\mathbb{P}}^1$ ought to be $S^2$ which is equivalent $S^1\wedge <!--\; \wedge \; -->S^1$. So we need to keep track of this difference between the "usual" topological situation and count the number of ${\mathbb{G}}_m$'s.
Now Morel justified the first equivalence by saying that ${\mathbb{A}}^1$ is invertible in the the unstable homotopy category. With just this information, is there an "intuitive", or maybe "geometric" (i.e. without going to the details of the motivic unstable category) way to see how this fact leads to the equivalence above?

Is there a way to test the "accuracy of a binomial survey"?
I've been away from mathematics for a while and forgotten almost everything. It doesn't come from a text book; I was given an assignment in my training for a job and wondered if I can use my mathematical knowledge. All I have to do is actually interpret the data and say "more than 50 percent of the people surveyed thinks "yes" to the question" but I'm taking a step further and trying to say how "accurate" is the result? Basically, here's what I am given
There's a supermarket that is experiencing a fall in revenues. A survey was conducted and it asked whether "the customer thinks the workers are unfriendly/unhelpful." out of a 100 randomly chosen customers on the same day (100 different customers) 51% answered "yes."
However, the total number of customers that visited the supermarket is expected to be around 485. The total that visited the supermarket that month is 19700. How confident are we to say that more than 50% are not happy with the workers among all of those who visited the store a. that day b. that month?
I vaguely recalled Chi-squared ad z-test but I wasn't so sure; I tried the z-test with
$$z=\frac{p-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}{\sqrt{\frac{\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->})}{n}}}$$
where p=0.51,π=0.5,n=100. Thing is, I get z=0.2 and the z table seems to tell me this is a very inaccurate result. In any case, my data and the question I ultimately want to answer is as above. Along the process of doing so, if no one would want to actually show me how to do this, can you please answer
What's the most apt test to answer a question like this? And why?
I think, as people start writing some answers, my senses will come back, some words and terms ringing a bell, reminding me of certain formulas, rules etc.

A survey shows that 63% of kids like brand A shoes and 76% like brand B shoes. brand B. Each boy in the group surveyed expresses at least one preference. Calculate the percentage of the boys who like both brand A and B and the percentage of those who like only the A.
My attemps: I have thought that the total number nb of boys cannot obviously exceed 100%; but I have instead
$$|A\cup <!--\; \cup \; -->B|=63\mathrm{\%<!--\; \%\; -->}+76\mathrm{\%<!--\; \%\; -->}=139\mathrm{\%<!--\; \%\; -->}$$
Subtracting the percentage of boys who like both brand A and B that is 100% I will have the percentage n′:
$$n^{\prime }=139\mathrm{\%<!--\; \%\; -->}-<!--\; -\; -->100\mathrm{\%<!--\; \%\; -->}=39\mathrm{\%<!--\; \%\; -->}$$
After the percentage of those who like only the A is:
$$|A-<!--\; -\; -->B|=|A|-<!--\; -\; -->|A\cap <!--\; \cap \; -->B|=63\mathrm{\%<!--\; \%\; -->}-<!--\; -\; -->39\mathrm{\%<!--\; \%\; -->}=24\mathrm{\%<!--\; \%\; -->}$$
Question: I have used the logic: surely the percentage of kids can't be more than 100%. Is there another motivation or is that the only one?

Maclane gives a definition (in a survey on categorical algebra) of category in the following way (in my own words), a category is a two-sorted system the sorts being called the objects and the arrows and that satisfies certain properties...
My questions are
What is this system, is a logic, a theory? Why a category is a two-sorted system? Then, where are the categories, in another system?

You wish to estimate,with 99% confidence, the proportion of Canadian drivers who want the speed limit raised to 130 kph. Your estimate must be accurate to within 5%. How many drivers must you survey,if your initial estimate of the proportion is 0.60?
I know that 99% is 2.575 but I don't know how to set up the problem. I don't think that 130 kph even has anything to do with the problem. I think i am over thinking this question.
I think it would be about 93 people. Is that right?

Averaging Whole Numbers
I'm curious which way is correct when rounding whole numbers. My issue is that the customer wants decimal places included in the average when the User only ever will select whole numbers in a range from 1-5. In my head it makes sense to return a whole number instead of a rational or irrational number.
Results out of 5 questions:
Question1 = 4
Question2 = 3
Question3 = 5
Question4 = 3
Question5 = 2
SUM = 17
which is correct for the average 3 or 3.4?

A survey of adults found that 34% say their favorite sport is professional football. You randomly select 130 adults and ask them if their favorite sport is professional football.
a) Find the probability that at most 66 people say their favorite sport is professional football.
The steps I have taken to solve this include:
$$n=130,p=.34,q=.66$$
mean=np=44.2;standard deviation=$$\text{mean}=np=44.2;\text{standard deviation}=\sqrt{npq}=5.4$$
So
$$z=\frac{66-<!--\; -\; -->44.2}{5.4}=\mathrm{4.03.}$$
This z-score is too high to use on my chart. I don't know where to go from here.
The next question I have is how I would find the probability that MORE than 31 people say their favorite sport is professional football. I know I am doing something wrong, can anyone help? Thank you!

truggling with these problems, which I believe are both cases of pigeonhole principle but I am struggling to get the right answer.
1) An IT survey asks desktop computer users the following questions: -- which OS type they are using (Windows, macOS, Linux, other) -- if Windows: Windows 10, 8.1, 7 or other -- if macOS: 10.12, 10.11, lower -- if Linux: Mint, Debian, Ubuntu, other -- screen resolution used (lower, equal or higher than 1920x1080) -- whether they play games on their desktop (yes/no) How many people have to be asked to guarantee that you will find 10 with identical answers?
2) How many subsets are there of a set of 100 elements that contain at least 4 elements? You must simplify your answer until it contains only unresolved large exponential expressions of the form 𝑎 𝑏 with a and b integers.

I just began filling out a survey for a contest that Walmart is hosting. Since I live in Canada there is a skill testing question which was as follows:
$$(4\times 2)+(6/3)?5=$$
The answer to the equation was omitted.
What kind of mathematical significance does the question mark hold?
I have tried Googling the equation for an explanation, but to no avail.
Is the equation bunk?
FWIW, the answer turned out to be 5... so obviously the question mark represented a subtraction symbol, but how was I ever supposed to induce that if the answer is/was omitted?

What is the most Most relevant statistic method to compare those samples?
One person is answering a survey. There are two type of questions :
4 questions about intellectual work affinity.
4 questions about manual work affinity.
For each of these question, the person gives an aswer from 1 (do not agree) to 5 (totally agree).
I want to know if there is a significant difference between answers about intellectual work and manual work, and eventually be able to say "this individual is more an intellectual / a manual" or "the difference is not significant.

Testing a hypothesys about a survey?
A survey of 61,647 people including questions about office relationships. Of the respondents, 26% reported that bosses scream at employees. Use a 05 significance level to test the claim that 1/4 of the people say that bosses scream at employees.
Please keep things as simple as possible.
My work: $H_0:p=.25$
$H_1:p><!--\; >\; -->.25$
P is given as 26
Using the formula $z={\displaystyle \frac{\stackrel{^<!--\; ^\; -->}{P}}{\sqrt{\frac{pq}{n}}}}$ with q=75, I get a z-score of around 5.73. However, this is an extreme value but percent is very close to percent., so I was expecting a z score of almost zero. What did I do wrong?

Suppose that in a city of 100 people, a survey conclude that 30 of them do not agree (says 'no') with the building of a new luxury apartment. If you randomly chose 12 people in the city what is the probability that 2 to 6 of them are those who disagree with the building of the new luxury apartment?
This seemed to be a binomial probability question.... though I fail to recall how to do so.
bonus point if you know how to solve this in minitab!

How is this the right answer?
In a survey of customer satisfaction, participants are asked to give a score of 1,2,3 or 4 to each of the 6 questions. If participants are instructed not to give the same numerical score to more than 4 questions, how many responses are possible?? I know the answer is 4,020 from the solutions, however I don't understand why. The solution says:
$$4\ast <!--\; \ast \; -->3\ast <!--\; \ast \; -->(\genfrac{}{}{0ex}{}{6}{5})=72$$
Hence, the number of possible responses is:
$$4,096-<!--\; -\; -->4-<!--\; -\; -->72=4,020$$
Why and how is this the answer? I don't understand where the −4 comes from.

Why am I not allowed to use $P(A\wedge <!--\; \wedge \; -->B)=P(A)P(B)$ formula?
Question: A survey of 1000 people determines that 80% like walking and 60% like biking, and all like at least one of the two activities. What is the probability that a randomly chosen person in this survey likes biking but not walking?
What I did was use the formula:
$$P(A\wedge <!--\; \wedge \; -->B)=P(A)P(B)=0.8\cdot <!--\; \cdot \; -->0.6=0.48$$
$$P(B)-<!--\; -\; -->P(A\cap B)=P(A^{\prime }\cap B)=0.6-<!--\; -\; -->0.48=0.12$$
but the answer should be 0.2.
In the answer key, $P(A\cup B)=P(A)+P(B)-<!--\; -\; -->P(A^{\prime }\cap B)$ is used instead. Why?

Rings for which the Köthe conjecture holds
Is there an overview of rings for which the Köthe conjecture is known to hold? In particular, I am interested in endomorphism rings of graded modules over multivariate polynomial rings. I read one survey states that the conjecture holds for left artinian rings (alas, without a reference). However, in the notion of that survey, to hold means that "the ideal generated by every nil left ideal of R is nil."
So my explicit question is: In any left artinian ring, or in a particular special case thereof, is it true that the sum of two left nil ideals is nil again?

Survey of the tax increase
In a survey of 1,000 people, 420 are opposed to the tax increase.
a) Construct a 95% confidence interval for the proportion of those people opposed to the tax increase.
b) Interpret the CI in terms of the question.
c) Is the estimate in part (a) valid? Explain.
My work:
I was able to solve a) and get the answer of [0.4205,0.4195].
For b) I am confused as to how to interpret the CI in terms of the question... could it be that its the percentage range of the possible number of people opposed to the tax increase?
For c) I think that it is valid because the number of people opposed to the tax increase is 420, which lies in the range of the confidence interval.
EDIT:
To solve for a) I did $$p\pm <!--\; \pm \; -->1.96x[\sqrt{\frac{0.42x0\times <!--\; \times \; -->0.58}{1000}}$$, which equalled 0.42 $$\pm <!--\; \pm \; -->$$ 0.0005...
I now see that I forgot to apply the sqrt before multiplying by 1.96. The correct answer should then be: 0.42 $$\pm <!--\; \pm \; -->$$ 1.96x0.0156 = 0.42 $$\pm <!--\; \pm \; -->$$ 0.0306, which results in the range of [0.4506, 0.3894]

Logical disjunction with sets
I have a question involving sets and logical disjunction, I have no idea of how to go about solving it. The question:
A survey of 40 IT users established that:
All of them used at least one of Word, Excel, Access
34 used Word or Access
38 used Word or Excel
30 used Word
26 used Excel
17 used Access
10 used all three packages
With reference to a Venn diagram, or otherwise, calculate the number of IT users in the survey who:
Used only Excel
Used only Access

In a past paper question for maths, I have encountered the following problem which is confusing me. The question states that:
When drivers approach crossroads, a survey shows that:
40% went straight on;
35% turned left;
25% turned right.
Assume that the drivers choose direction independently.
Find the probability that the next three vehicles approaching the crossroads will all turn left given that none turns right.
The answer provided is:
$$P(\text{all left}\mid <!--\; \mid \; -->\text{none turn right})=\frac{{0.35}^3}{(1-<!--\; -\; -->0.25{)}^3}=0.102\text{(3 s.f)}$$
This is where I'm confused. For independent events should it not be the following:
$$P(\text{all left}\mid <!--\; \mid \; -->\text{none turn right})=P(\text{all left})={0.35}^3=0.0429\text{ (3 s.f)?}$$?
Edit:
I have realized I misunderstood the independence part of the question, but I'm still left confused over the answer. From the equation
$P(A\mid <!--\; \mid \; -->B)=\frac{P(A\cap <!--\; \cap \; -->B)}{P(B)}$, how does $P(A\cap <!--\; \cap \; -->B)$ equal P(all left)?

Use of the Bonferroni correction
I am conducting an analysis of survey data.
I have carried out sign tests comparing the answers provided by two different groups to each question. There are several questions for which I am making this analysis.
Am I right in thinking that I do not need to use a Bonferroni (or any other type) of correction?

Sorry if my title sounds vague and inaccurate. I can't think of a better way to put it lol. Anyway, I stumbled upon this problem today while preparing for Oxford tsa:
A survey of households in a town showed that (allowing for sampling errors) between 75% and 85% owned a dishwasher, between 35% and 40% owned a tumble dryer and less than 5% owned neither.
How many people own both a tumble dryer and a dishwasher?
In which the answer is: "between 10% and 30%"
I spent lots of time trying to figure out how to solve it but to no avail. The inclusion of the ranges in a typical Venn diagram question totally throws me off. Can anyone help me with this?