Dive Deep into Survey Questions with Comprehensive Examples and Expert Assistance

A survey was conducted of graduates from 30 MBA programs. On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is $168,000 and $117,000, respectively. Assume the standard deviation for the male graduates is $40,000, and for the female graduates it is $25,000.
Question 1: What is the probability that a simple random sample of 40 male graduates will provide a sample mean within $10,000 of the population mean, $168,000?
Question 2: What is the probability that a simple random sample of 40 female graduates will provide a sample mean within $10,000 of the population mean of $117,000?

Statistical analysis of study with categorical and numerical variables
I am researching the effect of a certain innovation type on firm performance. The innovation type is measured through a 6-item survey with nominal answers (yes/no; 1/0) and is retrospective (e.g. Did you introduce XY in the last 5 years). For firm performance I have financial data for the 5 year period I'm interested in. Now, there are two possible approaches I could take:
1) I compute an "innovation" variable from the survey answers, to distinguish between adopters and non-adopters and I examine whether the adopter-group shows to have better firm performance than the non-adopter group. Which type of analysis would this be? And how would I control for firm size and time effects?
2) I investigate whether firms that answered more questions with "yes" perform better than firms that answered with fewer "yes". Which type of analysis would this then be? Regression?

Just want to check if my answer and reasoning is correct for the following problem (Not a homework problem - it is a sample question for a test I'm preparing for)
In a survey, viewers were given a list of 20 TV Shows and are asked to label 3 favourites not in any order. Then they must tick the ones that they have heard of before, if any. How many ways can the form be filled, assuming everyone has 3 favourites?
My reasoning:
1) Choose 3 shows out of 20: c(20,3)
2) Choosing 0-17 shows from 17 choices: $c(17,0)+c(17,1)+c(17,2)+...+c(17,16)+c(17,17)$
and add 1) and 2) together for the final answer.
Would this be correct? Is there a better way of doing the second part that doesn't involve so many calculations?

I currently have this question..
A survey was conducted that found 72% of respondents liked the new motorway. Of all respondents, 65% intend to drive more. Suppose that 81% of those who like the new motorway intend to drive more.
I get rather confused with how the 65% and 81% intertwine. I assume I'm working backwards to find out the percentage of those who don't like the new motorway but intend to drive more.
Let l = like, d = drive..
pr(l) = 0.72, pr(l') = 0.28
Would I be right in claiming that pr(d) = 0.65 therefore pr(d/l) = 0.65/0.72 ?

I have a bag of toys. 10% of the toys are balls. 10% of the toys are blue.
If I draw one toy at random, what're the odds I'll draw a blue ball?

I have a result that uses ordered fields. However, I am ignorant of the literature surrounding ordered fields. I have read the basic facts about them, such as those contained in the wikipedia page. However, in the introduction, I would like to provide some motivation.
My result has the property that if it holds for some ordered field k, then it holds for any ordered subfield of k. Does that mean it is enough to prove my result for real-closed ordered fields?
I also read the statement that to prove a 1st order logic statement for a real-closed ordered field, then it is enough to prove it for one real-closed ordered field, such as R for instance. Can someone please provide a reference for that?
A mathematician mentioned to me a classical link between iterated quadratic extensions and ordered fields. What is the precise statement please?
I would also like a number of interesting examples of ordered fields. I know for example of an interesting non-archimedean example using rational functions (that I have learned about from the wikipedia page on ordered fields).
Does anyone know of a survey on ordered fields, or a reference containing answers to my questions above?

Bayes' Rule with a binomial distribution
The question is as follows:
Each morning, an intern is supposed to call 10 people and ask them to take part in a survey. Every person called has a 1/4 probability of agreeing to take the survey, independent of any other people’s decisions. Eighty percent of the days, the intern does what they are supposed to; however, the other twenty percent of the time, the intern is lazy and only calls 6 people. One day, exactly 1 person agree to take the survey. What is the probability the intern was lazy that day?
My solution is partially shown using Bayes' Rule with the binomial distribution. Let L be the event he is lazy and let O be the event of one person agreeing to the survey. Then,
$$\mathbb{P}(L\mid <!--\; \mid \; -->O)=\frac{\mathbb{P}(O\mid <!--\; \mid \; -->L)\mathbb{P}(L)}{\mathbb{P}(O\mid <!--\; \mid \; -->L)\mathbb{P}(L)+\mathbb{P}(O\mid <!--\; \mid \; -->L^C)\mathbb{P}(L^C)}.$$
We know
$$\mathbb{P}(O\mid <!--\; \mid \; -->L)=\frac{\mathbb{P}(O\cap <!--\; \cap \; -->L)}{\mathbb{P}(L)}=\frac{(\genfrac{}{}{0ex}{}{6}{1})\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^5}{\frac{1}{5}}\approx <!--\; \approx \; -->1.7798$$
and
$$\mathbb{P}(O\mid <!--\; \mid \; -->L^C)=\frac{\mathbb{P}(O\cap <!--\; \cap \; -->L^C)}{\mathbb{P}(L^C)}=\frac{(\genfrac{}{}{0ex}{}{10}{1})\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^9}{\frac{4}{5}}\approx <!--\; \approx \; -->.2346.$$
I would plug this into the Bayes' Rule equation, but I got a probability above 1 in one of the cases. Why?

In a survey searching at reading conduct at a college, it become observed that 48% study mag A, 46% read mag B, 55% examine magazine C, 18% examine magazines A and B, 20% examine A and C, 23% examine B and C, ad 8% examine all three. what number read atleast one of the three magazines.
Then what I did was state:
$$\begin{gathered} n(U)=100 \\ n(A)=48 \\ n(B)=46 \\ n(C)=55 \\ n(AB)=18 \\ n(AC)=20 \\ n(BC)=23 \\ n(ABC)=8. \end{gathered}$$
Noting that $AB=A\cap <!--\; \cap \; -->B$
Then what I did was:
$$n(A\cup <!--\; \cup \; -->B\cup <!--\; \cup \; -->C)=n(A)+n(B)+n(C)-<!--\; -\; -->n(AC)-<!--\; -\; -->n(AB)-<!--\; -\; -->n(BC)+n(ABC)$$
when I did this I got that 96% read either A, B, or C and I was just wondering if I had made an error because it seemed a bit high when I first did it.

So the question looks like this A survey finds that 30% of all Canadians enjoy jogging. Of those Canadians who enjoy jogging, it is known that 40% of them enjoy swimming. For the 70% of Canadians who do not enjoy jogging, it is known that 60% of them enjoy swimming.
Suppose a Canadian is randomly selected. What is the probability that they enjoy exactly one of the two activities? Select the answer closest to yours.
I solved this using $$P(JoggingorSwimming)=P(Jogging)+P(Swimming)-P(Joggingandswimming)P(Jogging)=0.3P(Swimming)=0.54P(Joggingandswimming)=0.12$$
my answer was 0.72 but it was wrong. Please, what am I missing?

A survey of students revealed that 30% of them have a part-time job. If students are chosen at random what is the probability of the following events: That less than 2 have part time work? More than 3 are working part-time? None have a part-time job?
This is a question from a mock exam paper as I prepare for exams.
I guess I could start by imagining a number of 1000 college students surveyed with 300 having part-time jobs.
so for part (i) -> (30/100)(70/10)(70/10)(70/10)(70/10)=7%
That ain't right. Can someone shed some light on this?

I'm reading Carlsson's "A survey of equivariant homotopy theory" and I have a question.
Let G be a topological group and X be a G-space (for a nice notion of "space"). He defines
$$X_{hG}=EG{\times <!--\; \times \; -->}_{G}X$$
the homotopy orbit space, and
$$X^{hG}=F(EG,X{)}^G$$
the homotopy fixed point space.
He claims there are spectral sequences
$$E_{p,q}^2=H_p(G;H_q(X))\Rightarrow <!--\; \Rightarrow \; -->H_{p+q}(X_{hG})$$
and
$$E_2^{p,q}=H^{-<!--\; -\; -->p}(G;{\mathrm{\pi <!--\; \pi \; -->}}_q(X))\Rightarrow <!--\; \Rightarrow \; -->{\mathrm{\pi <!--\; \pi \; -->}}_{p+q}(X^{hG}).$$
Now, if I'm not mistaken the first one follows from the Serre spectral sequence applied to the Borel fibration $X\to <!--\; \to \; -->X_{hG}\to <!--\; \to \; -->BG$ (take the fiber bundle with fiber X associated to the action of G on X and to the G-principal bundle $EG\to <!--\; \to \; -->BG$).
But where does the second one come from?

Sample size requirements in survey
If am doing some market research and want to answer the question "What percentage of the users of a service, searched for the given service online?". Lets say I go out and get people to take a survey.
How do I calculate the required sample size that would create the correct distribution for a given country or region?

I'm trying to study for a test in my AP Statistics course. My lecturer spent the majority of the unit going over the various proportion tests. On my review, I was presented with the following question:
Suppose that you wanted to estimate p, the true proportion of students at your school who have a tattoo with 98% confidence and a margin of error no more than 0.10. How many students should you survey?
What I'm not understanding is what should be substituted for p. In the given problem, no value for p is given, but yet I need to find n using the following formula:
$$ME=(z\ast <!--\; \ast \; -->)(\sqrt{\frac{p(1-<!--\; -\; -->p)}{n}})$$
How can I determine a value for n?

I am studying a Tutorial on Maximum Likelihood Estimation in Linear Regression and I have a question.
When we have more than one regressor (a.k.a. multiple linear regression1), the model comes in its matrix form $y=X\mathrm{\beta <!--\; \beta \; -->}+\mathrm{ϵ<!--\; ϵ\; -->}$, (1)where y is the response vector, X is the design matrix with each its row specifying under what design or conditions the corresponding response is observed (hence the name), $\mathrm{\beta <!--\; \beta \; -->}$ is the vector of regression coefficients, and $\mathrm{ϵ<!--\; ϵ\; -->}$ is the residual vector distributing as a zero-mean multivariable Gaussian with a diagonal covariance matrix $\mathcal{N}\sim <!--\; \sim \; -->(0,{\mathrm{\sigma <!--\; \sigma \; -->}}^2{I}_N)$, where $I_N$ is the $N\times <!--\; \times \; -->N$ identity matrix. Therefore $y\sim <!--\; \sim \; -->\mathcal{N}(X\mathrm{\beta <!--\; \beta \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2{I}_N)$, (2)meaning that linear combination $X\mathrm{\beta <!--\; \beta \; -->}$ explains (or predicts) response y with uncertainty characterized by a variance of ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$.
Assume y,$\mathrm{\beta <!--\; \beta \; -->}$, and $\mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{\mathbb{n}}$ Under the model assumptions, we aim to estimate the unknown parameters ($\mathrm{\beta <!--\; \beta \; -->}$ and ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$) from the data available (X and y).
Maximum likelihood (ML) estimation is the most common estimator. We maximize the log-likelihood w.r.t. $\mathrm{\beta <!--\; \beta \; -->}$ and ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$ $\mathcal{L}(\mathrm{\beta <!--\; \beta \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2|y,X)=-\frac{N}{2}\log <!--\; \; -->2\mathrm{\pi <!--\; \pi \; -->}-\frac{N}{2}log{\mathrm{\sigma <!--\; \sigma \; -->}}^2-\frac{1}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^2}(y-X\mathrm{\beta <!--\; \beta \; -->}{)}^T(y-X\mathrm{\beta <!--\; \beta \; -->})$
I am trying to understand that how the log-likelihood, $\mathcal{L}(\mathrm{\beta <!--\; \beta \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2|y,X)$, is formed. Normally, I saw these problems when we have ${\mathbf{x}}_{\mathbf{i}}$ as vector of size d(d is number of parameter for each data). specifically, when xi is a vector, I wrote is as
$\ln <!--\; \; -->\underset{i=1}{\overset{N}{\prod <!--\; \prod \; -->}}\frac{1}{\sqrt{(2\mathrm{\pi <!--\; \pi \; -->}{)}^d{\mathit{\sigma <!--\; \sigma \; -->}}^2}}\exp <!--\; \; -->(-<!--\; -\; -->\frac{1}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^2}({\mathbf{x}}_{\mathbf{i}}-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}{)}^{\mathrm{T}}\mathbf{(}{\mathbf{x}}_i-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}))=\sum <!--\; \sum \; -->{}_i\ln <!--\; \; -->\frac{1}{\sqrt{(2\mathrm{\pi <!--\; \pi \; -->}{)}^d{\mathit{\sigma <!--\; \sigma \; -->}}^2}}\exp <!--\; \; -->(-<!--\; -\; -->\frac{1}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^2}({\mathbf{x}}_{\mathbf{i}}-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}{)}^{\mathrm{T}}\mathbf{(}{\mathbf{x}}_i-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}))$. But in the case that is shown in this tutorial, there is no index I to apply summation.

If 20 items have 95% chance for X, can X be expected to happen for 19?
When conducting a survey, confidence level of 95% means (I know it is not exact interpretation) that I can be that much sure about the interval covering the correct value/mean.
Therefore I have 20 questions, each with 95% probability of including the correct value within the margin of error and also 5% probability of being out of margin of error. Can I then say that 1 (5% of 20) can be expected as out of margin of error?

Tensor product of varieties : What's this notation $V_1\otimes <!--\; \otimes \; -->V_2$ ?
I saw this notation $V=V_1\otimes <!--\; \otimes \; -->V_2$ in a survey on universal algebra, where was a variety, but the survey in question didn't define this notation. Could anyone explain what it means ?

A recent survey indicated that 82% of single women aged 25 years old will be married in their lifetime. Using the binomial distribution, find the probability that two or three women in a sample of twenty will never be married.
Let X be defined as the number of women in the sample never married
$$\begin{gathered} P(2\le <!--\; \le \; -->X\le <!--\; \le \; -->3)=p(2)+p(3) \\ =(202)(.18)2(.82)18+(203)(.18)3(.82)17 \\ =.173+.228=.401 \end{gathered}$$
My Question
If I recognize it effectively, the binomial distribution is a discrete chance distribution of a number of successes in a sequence of n unbiased sure/no experiments.
But choosing 2 (or 3) women from a 20-women sample is not independent experiments, because choosing the first woman will affect the probability for the coming experiments.
Why the binomial distribution was used here ?

Hi I am stuck on this probability question and don't know what to do.
Q. A survey is carried out on a computer network. The probability that a log on to the network is successful is 0.92. Find the probability that exactly five out of nine users that attempt to log on will do so successfully.

I'm really good at probability, but this time I seems like I'm not.
My friends asked me a very tricky question, and I want to see if there's anyone who can find out the answer.
Here's the question:
In a survey of 1,000 randomly selected people:
756 believed that the price of the particular product of is reasonable compared to the last year increase,
487 believed that the quality of that product is worse than the last year, and
363 believed both
From here we know:
Product is reasonable + Product is worse = 363
But how about
Product is reasonable + Product is not worse
Product is not reasonable + Product is worse
Product is not reasonable + Product is not worse
I'm trying to figure it out using set like $P(A\cup <!--\; \cup \; -->B\cup <!--\; \cup \; -->C\cup <!--\; \cup \; -->D)$ but seems like hard for me.

Survey Results Computation Problem Solving
There is a survey for a product for A number of people regarding its quality and price. B number of people are satisfied both with the quality and price. Find the number of people who are both dissatisfied with quality and price.
Here’s the result of survey: Price:
– Satisfied: C number of people
– Dissatisfied: D number of people
Quality:
– Satisfied: E number of people
– Dissatisfied: F number of people
To be clear, C + D = A and E + F = A.
A, B, C, D, E and F are actual numbers that are given in the problem but forgot them.