Recent questions in Survey Questions

Survey Questions
Answered

gaby131o
2022-09-20

The options on the ballot paper are "Yes" and "No", but voting is not compulsory.

The survey is being conducted over 8 weeks during which eligible voters can return their response.

At the end of the third week, 60% of voters have returned their response.

Exit-polling indicates of those who have already voted, 60% voted "Yes" and 40% voted "No".

Question: All other things being equal (e.g. early/late responders no indication of which way they vote) what % of the remaining 40% who haven't returned their response would need to vote "No" in order to achieve parity with the "Yes" vote (here's the kicker...) taking into account that not all eligible voters will ultimately return their ballots?

Is this work-outable?

Survey Questions
Answered

Melina Barber
2022-09-19

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?

One person provided an answer immediately and others suggested that more details were required before an answer could even be considered. But, there was a reason I asked this question the way that I did.

I was thinking about probabilities and I was coming up with a way to ask a more complicated question on math.stackexchange.com. I needed a basic example so I came up with the toys problem I posted here.

I wanted to run it by a friend of mine and I started by asking the above question the same way. When I thought of the problem, it seemed very clear to me that the question was "what is $\mathbb{P}(blue\cap ball)$." I thought the calculation was generally accepted to be

$$\mathbb{P}(blue\cap ball)=\mathbb{P}(blue)\cdot \mathbb{P}(ball)$$

When I asked my friend, he said, "it's impossible to know without more information." I was baffled because I thought this is what one would call "a priori probability."

I remember taking statistics tests in high school with questions like "if you roll two dice, what're the odds of rolling a 7," "what is the probability of flipping a coin 3 times and getting three heads," or "if you discard one card from the top of the deck, what is the probability that the next card is an ace?"

Then, I met math.stackexchange.com and found that people tend to talk about "fair dice," "fair coins," and "standard decks." I always thought that was pedantic so I tested my theory with the question above and it appears you really need to specify that "the toys are randomly painted blue."

It's clear now that I don't know how to ask a question about probability.

Why do you need to specify that a coin is fair?

Why would a problem like this be "unsolvable?"

If this isn't an example of a priori probability, can you give one or explain why?

Why doesn't the Principle of Indifference allow you to assume that the toys were randomly painted blue?

Why is it that on math tests, you don't have to specify that the coin is fair or ideal but in real life you do?

Why doesn't anybody at the craps table ask, "are these dice fair?"

If this were a casino game that paid out 100 to 1, would you play?

This comment has continued being relevant so I'll put it in the post:

Here's a probability question I found online on a math education site: "A city survey found that 47% of teenagers have a part time job. The same survey found that 78% plan to attend college. If a teenager is chosen at random, what is the probability that the teenager has a part time job and plans to attend college?" If that was on your test, would you answer "none of the above" because you know the coincident rate between part time job holders and kids with college aspirations is probably not negligible or would you answer, "about 37%?"

Survey Questions
Answered

tamnicufl
2022-09-18

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?

Survey Questions
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beobachtereb
2022-09-18

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?

Survey Questions
Answered

Kolby Castillo
2022-09-17

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?

Survey Questions
Answered

madeeha1d8
2022-09-17

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 ?

Survey Questions
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listgrein6u
2022-09-17

If I draw one toy at random, what're the odds I'll draw a blue ball?

Survey Questions
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nikakede
2022-09-16

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?

Survey Questions
Answered

obojeneqk
2022-09-15

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,

$$\mathbb{P}(L\mid O)=\frac{\mathbb{P}(O\mid L)\mathbb{P}(L)}{\mathbb{P}(O\mid L)\mathbb{P}(L)+\mathbb{P}(O\mid {L}^{C})\mathbb{P}({L}^{C})}.$$

We know

$$\mathbb{P}(O\mid L)=\frac{\mathbb{P}(O\cap L)}{\mathbb{P}(L)}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{6}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{5}}{\frac{1}{5}}\approx 1.7798$$

and

$$\mathbb{P}(O\mid {L}^{C})=\frac{\mathbb{P}(O\cap {L}^{C})}{\mathbb{P}({L}^{C})}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{10}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{9}}{\frac{4}{5}}\approx .2346.$$

I would plug this into the Bayes' Rule equation, but I got a probability above 1 in one of the cases. Why?

Survey Questions
Answered

Dulce Cantrell
2022-09-14

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?

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reinzogoq
2022-09-14

Let G be a topological group and X be a G-space (for a nice notion of "space"). He defines

$${X}_{hG}=EG{\times}_{G}X$$

the homotopy orbit space, and

$${X}^{hG}=F(EG,X{)}^{G}$$

the homotopy fixed point space.

He claims there are spectral sequences

$${E}_{p,q}^{2}={H}_{p}(G;{H}_{q}(X))\Rightarrow {H}_{p+q}({X}_{hG})$$

and

$${E}_{2}^{p,q}={H}^{-p}(G;{\pi}_{q}(X))\Rightarrow {\pi}_{p+q}({X}^{hG}).$$

Now, if I'm not mistaken the first one follows from the Serre spectral sequence applied to the Borel fibration $X\to {X}_{hG}\to BG$ (take the fiber bundle with fiber X associated to the action of G on X and to the G-principal bundle $EG\to BG$).

But where does the second one come from?

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Answered

kadirsmr9d
2022-09-14

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

my answer was 0.72 but it was wrong. Please, what am I missing?

Survey Questions
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Modelfino0g
2022-09-14

Then what I did was state:

$$n(U)=100\phantom{\rule{0ex}{0ex}}n(A)=48\phantom{\rule{0ex}{0ex}}n(B)=46\phantom{\rule{0ex}{0ex}}n(C)=55\phantom{\rule{0ex}{0ex}}n(AB)=18\phantom{\rule{0ex}{0ex}}n(AC)=20\phantom{\rule{0ex}{0ex}}n(BC)=23\phantom{\rule{0ex}{0ex}}n(ABC)=8.$$

Noting that $AB=A\cap B$

Then what I did was:

$$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(AC)-n(AB)-n(BC)+n(ABC)$$

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.

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Hrefnui9
2022-09-14

I know that 99% is 2.575 but i dont 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.

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equipokypip1
2022-09-13

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?

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tamolam8
2022-09-13

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?

Survey Questions
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nar6jetaime86
2022-09-13

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:

$$ME=(z\ast )(\sqrt{\frac{p(1-p)}{n}})$$

How can I determine a value for n?

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Answered

Pranav Ward
2022-09-13

When we have more than one regressor (a.k.a. multiple linear regression1), the model comes in its matrix form $y=X\beta +\u03f5$, (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), $\beta $ is the vector of regression coefficients, and $\u03f5$ is the residual vector distributing as a zero-mean multivariable Gaussian with a diagonal covariance matrix $\mathcal{N}\sim (0,{\sigma}^{2}{I}_{N})$, where ${I}_{N}$ is the $N\times N$ identity matrix. Therefore $y\sim \mathcal{N}(X\beta ,{\sigma}^{2}{I}_{N})$, (2)meaning that linear combination $X\beta $ explains (or predicts) response y with uncertainty characterized by a variance of ${\sigma}^{2}$.

Assume y,$\beta $, and $\u03f5\in {\mathbb{R}}^{\mathbb{n}}$ Under the model assumptions, we aim to estimate the unknown parameters ($\beta $ and ${\sigma}^{2}$) from the data available (X and y).

Maximum likelihood (ML) estimation is the most common estimator. We maximize the log-likelihood w.r.t. $\beta $ and ${\sigma}^{2}$ $\mathcal{L}(\beta ,{\sigma}^{2}|y,X)=-\frac{N}{2}\mathrm{log}2\pi -\frac{N}{2}log{\sigma}^{2}-\frac{1}{2{\sigma}^{2}}(y-X\beta {)}^{T}(y-X\beta )$

I am trying to understand that how the log-likelihood, $\mathcal{L}(\beta ,{\sigma}^{2}|y,X)$, 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

$\mathrm{ln}\prod _{i=1}^{N}\frac{1}{\sqrt{(2\pi {)}^{d}{\mathit{\sigma}}^{2}}}\mathrm{exp}(-\frac{1}{2{\sigma}^{2}}({\mathbf{x}}_{\mathbf{i}}-\mathit{\mu}{)}^{\mathrm{T}}\mathbf{(}{\mathbf{x}}_{i}-\mathit{\mu}))=\sum {}_{i}\mathrm{ln}\frac{1}{\sqrt{(2\pi {)}^{d}{\mathit{\sigma}}^{2}}}\mathrm{exp}(-\frac{1}{2{\sigma}^{2}}({\mathbf{x}}_{\mathbf{i}}-\mathit{\mu}{)}^{\mathrm{T}}\mathbf{(}{\mathbf{x}}_{i}-\mathit{\mu}))$. But in the case that is shown in this tutorial, there is no index I to apply summation.

Survey Questions
Answered

tashiiexb0o5c
2022-09-12

Let X be defined as the number of women in the sample never married

$P(2\le X\le 3)=p(2)+p(3)\phantom{\rule{0ex}{0ex}}=(202)(.18)2(.82)18+(203)(.18)3(.82)17\phantom{\rule{0ex}{0ex}}=.173+.228=.401\phantom{\rule{0ex}{0ex}}$

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 ?

Survey Questions
Answered

Gavyn Whitehead
2022-09-12

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.