Dive Deep into Survey Questions with Comprehensive Examples and Expert Assistance

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!

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?

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?

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.

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)?

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?

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?

I am confused about creating a Venn diagram from the question below from one of the past papers as there is a negative number involved. Is this possible in Venn diagrams?
A survey of students in a lecture revealed that a total of 15 played rugby, 30 played football and 35 played cricket. 4 people played all three sports, while 2 played only football and rugby, 14 played only cricket and football, and 6 played only cricket and rugby.
If 4 people played all three sports and 2 played only football and rugby, how is it possible to create the Venn diagram? Are negative numbers allowed, as 2−4=−2?

Ice cream flavorsA survey was taken among 600 middle school students for their preferred ice cream flavours.- 250 like strawberry- 100 liked both strawberry and vanilla- 130 like strawberry but not chocolate- 120 like only vanilla- 300 like only one flavor- 270 students said they liked vanilla- 30 students like all three flavorsQuestions:- How many do not like any of the 3 flavors?- How many like at least one of the 3 flavors?- How many like at least 2 flavors?- How many like chocolate?