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

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?

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?

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?

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.

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 ?

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.

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.

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.

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 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 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.

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?

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?