Dive Deep into Survey Questions with Comprehensive Examples and Expert Assistance

A survey has been conducted to see how many people in a town of 40,000 people used Ebay to purchase a product last year. A simple random sample of 230 is taken and from this sample 52 people had used Ebay last year.
I'm asked to estimate the total number of people who used Ebay last year and compute the 95% confidence interval for the total (we are allowed to assum the 97.5% quantile of the normal distribution is 1.96).
The first part I believe is straightforward, I just find my ratio R and multiply by my population total t to get 52/230⋅40,000=9043.4783 ($\stackrel{^<!--\; ^\; -->}{t}$) but finding the confidence interval is a bit confusing because I'm hardly given any information here. I have formulas I would usually use to find the variance of R and then the variance of $\stackrel{^<!--\; ^\; -->}{t}$ and then find the confidence interval but all these formulas either require $\stackrel{¯<!--\; ¯\; -->}{y}$ (and sometimes $\stackrel{¯<!--\; ¯\; -->}{x}$) or they require $\sum <!--\; \sum \; -->y_i^2,\sum <!--\; \sum \; -->y_i{x}_i,\sum <!--\; \sum \; -->x_i^2$and I have none of these. I only have the total population, the sample size and the ratio. Any ideas how I would obtain my 95% confidence interval for the total?
Also, there is a second part that asks what sample size would be required for the total number to be within 940 units of the true value (I'm guessing they mean the total number of people in the town who used ebay last year), with confidence 95%? On this part of the question I'm just not sure what to do.
These are both low mark questions so I'm probably just forgetting a formula or missing something but I just can't see a way to get my answers with so little information given. I've double and triple checked and this is for sure the only information that is given about this survey.

A certain town has 25,000 families. These families own 1.6 cars, on the average; the SD is 0.90. And 10% of them have no cars at all. As part of an opinion survey, a simple random sample of 1,500 families is chosen. What is the chance that between 9% and 11% of the sample families will not own cars?
So I'm looking at the solution of this question it shows that to get the Standard Error of the Sum is SD= $\sqrt{0.9\cdot <!--\; \cdot \; -->0.1}$ However, I thought the formula to get SE is:
SE= SD of the Box $\sqrt{NumberofDraws}$
In this case, isn't the Box of the SD given at 0.9? Why do we still use the bootstrap method to get SD= $\sqrt{0.9\cdot <!--\; \cdot \; -->0.1}$=0.3 ??
Many thanks in advance!

Approximately 8.33% of men are colorblind. You survey 8 men from the population of a large city and count the number who are colorblind.
This question shows up in the binomial distribution chapter in my book.
g. How many men would you have to survey in order to be at least 95% sure you would find at least 1 who is colorblind?
$P(X\ge <!--\; \ge \; -->1)=1-<!--\; -\; -->P(X=0)=1-<!--\; -\; -->(\genfrac{}{}{0ex}{}{8}{0}){.0833}^0{.9167}^8=1-<!--\; -\; -->.4986752922=.5013247078$
But this is for a sample of 8 men. This question is asking me to find n with 95% confidence,
But this seems like a confidence interval question which I'm a little confused about since my book is about probability and not about statistics. Nothing like this shows up in my book, so I'm stuck. I feel like this is a "challenge" question.
Can someone give me a clue on what formula to use to deal with a question like this? Is this even a confidence interval question?

If I want to determine if people like tea or coffee more and conduct a simple random survey of about 50 people, I have some confusion about how to phrase the question to the people and how that affects the Null Hypothesis.
If I say, "Do you like tea or coffee more?" then I'm not sure how to state the null. If I say, "Do you like tea or coffee equally well?", then it is clear that the null would be mu = .5 and the alternative would be mu does not equal .5.
How would I set up the null for the first question of "Do you like tea or coffee?
I have read some articles that state journals don't like one sided tests (compared with two sided) unless the question is yes or no. So, I'm confused about how to set the null and alternative for this simple experiment.

A concert venue is located 4 minutes from a large university. A group of students are planning to attend the concert and think they can get there in less time. They test this hypothesis by using social media to survey a sample of other drivers to see how long this trip took them. The null and alternative hypotheses are given.
....
Here is the full question:
(a) what is the consequence of a Type I error in the context of this problem?
hey conclude it will average rewer than 4 minutes to get there, but the real average is more than 4 minutes.
They conclude that it will average more than 4 minutes, but the real average is 4 minutes to get there.
They conclude it will average 4 minutes to get there, but the real average is less than 4 minutes.
They conclude it will average rewer than 4 minutes, but the real average is 4 minutes to get there.
(6) What are the consequences of making a Type II error in the context of this problem?
They conclude it will average fewer than 4 minutes to get there, but the real average is more than 4 minutes.
They conclude that it will average more than 4 minutes, but the real average is 4 minutes to get there.
They conclude it will average 4 minutes to get there, but the real average is less than 4 minutes.
They conclude it will average fewer than 4 minutes, but the real average is 4 minutes to get there.

I got a question about the use of Chi Square test. Let's assume I am conducting a survey. And I have a question: "Have you ever heard of the Internet"?
The possible answers are: "Yes", "No", "Not sure" and I have 3 different age groups for my respondents: "<18", "18
When I want to do a Chi Square test to see if the age factor has an effect, can I just conduct this test on the "Yes" part? Or Do I need to include all the variables?

A question in GRE states:
In a survey of a town,it was found that 65% of the people surveyed watched the news on television,40% read newspaper, and 25% read a newspaper and watched the news on television. What percent of the people surveyed neither watched the news on television nor read a newspaper.

Question about random samplings and estimations.
100 students were selected to participate in a survey by the school in May 2013. The next month, another 100 students were randomly selected to participate in the same survey, out of which 2 had already participate in the survey the previous month. If the number of students did not change from May to June then what is the approximate number of students in the school?
Came across this question. Is the question weird or am I just dumb? Can anyone suggest a way to solve this or an approach on how it can be solved. Thank you.

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?

The following letters a,b,e,… denote Turing degrees.
We say $a\gg <!--\; \gg \; -->b$ if there exists a complete extension of degree a to the Peano Arithmetic augmented with axioms $\stackrel{\dot{}<!--\; \dot{}\; -->}{n}\in <!--\; \in \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{B}$ whenever $n\in <!--\; \in \; -->B$, and $\stackrel{\dot{}<!--\; \dot{}\; -->}{n}\notin \stackrel{\dot{}<!--\; \dot{}\; -->}{B}$whenever $n\notin B$, where n˙ are constant symbols for $n\in <!--\; \in \; -->\mathrm{\omega <!--\; \omega \; -->}$ and B is any given set of degree b.
Now in Volume 90, Pages ii-viii, 1-1165 (1977) HANDBOOK OF MATHEMATICAL LOGIC, Degrees of Unsolvability: A Survey of Results, Pages 631-652 by Stephen G. Simpson, Theorem 6.5 i), it is claimed that if a≫b then there exists another degree e such that $a\gg <!--\; \gg \; -->e\gg <!--\; \gg \; -->b$.
There was no proof there. I am wondering how to prove this?

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?

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.

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?

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?

Hello can someone please help me to answer this question it, it a binomial distribution question:
An email message advertises the chance to win a prize if the reader follows a link to an online survey. The probability that a recipient of the email clicks on the link to the survey is 0.0016. How many emails, to the nearest hundred, need to be sent out in order to have a 99% probability that at least 1000 will be answered?

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!

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?

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.

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?

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.