# Recent questions in Survey Questions

Recent questions in Survey Questions
gaby131o 2022-09-20

### Australia is having a national "postal survey" about same-sex marriage.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?

Melina Barber 2022-09-19

### I recently have this question: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 $\mathbb{P}\left(blue\cap ball\right)$." I thought the calculation was generally accepted to be$\mathbb{P}\left(blue\cap ball\right)=\mathbb{P}\left(blue\right)\cdot \mathbb{P}\left(ball\right)$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%?"

tamnicufl 2022-09-18

### 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?

beobachtereb 2022-09-18

### Statistical analysis of study with categorical and numerical variablesI 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?

Kolby Castillo 2022-09-17

### 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') = 0.28Would I be right in claiming that pr(d) = 0.65 therefore pr(d/l) = 0.65/0.72 ?

listgrein6u 2022-09-17

### 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?

nikakede 2022-09-16

obojeneqk 2022-09-15

### Bayes' Rule with a binomial distributionThe 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,$\mathbb{P}\left(L\mid O\right)=\frac{\mathbb{P}\left(O\mid L\right)\mathbb{P}\left(L\right)}{\mathbb{P}\left(O\mid L\right)\mathbb{P}\left(L\right)+\mathbb{P}\left(O\mid {L}^{C}\right)\mathbb{P}\left({L}^{C}\right)}.$We know$\mathbb{P}\left(O\mid L\right)=\frac{\mathbb{P}\left(O\cap L\right)}{\mathbb{P}\left(L\right)}=\frac{\left(\genfrac{}{}{0}{}{6}{1}\right)\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{5}}{\frac{1}{5}}\approx 1.7798$and$\mathbb{P}\left(O\mid {L}^{C}\right)=\frac{\mathbb{P}\left(O\cap {L}^{C}\right)}{\mathbb{P}\left({L}^{C}\right)}=\frac{\left(\genfrac{}{}{0}{}{10}{1}\right)\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{9}}{\frac{4}{5}}\approx .2346.$I would plug this into the Bayes' Rule equation, but I got a probability above 1 in one of the cases. Why?

Dulce Cantrell 2022-09-14

### 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?

reinzogoq 2022-09-14

### 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 $P\left(JoggingorSwimming\right)=P\left(Jogging\right)+P\left(Swimming\right)-P\left(Joggingandswimming\right)P\left(Jogging\right)=0.3P\left(Swimming\right)=0.54P\left(Joggingandswimming\right)=0.12$my answer was 0.72 but it was wrong. Please, what am I missing?

Modelfino0g 2022-09-14

### 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:$n\left(U\right)=100\phantom{\rule{0ex}{0ex}}n\left(A\right)=48\phantom{\rule{0ex}{0ex}}n\left(B\right)=46\phantom{\rule{0ex}{0ex}}n\left(C\right)=55\phantom{\rule{0ex}{0ex}}n\left(AB\right)=18\phantom{\rule{0ex}{0ex}}n\left(AC\right)=20\phantom{\rule{0ex}{0ex}}n\left(BC\right)=23\phantom{\rule{0ex}{0ex}}n\left(ABC\right)=8.$Noting that $AB=A\cap B$Then what I did was:$n\left(A\cup B\cup C\right)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(AC\right)-n\left(AB\right)-n\left(BC\right)+n\left(ABC\right)$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.

Hrefnui9 2022-09-14

### 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 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.

equipokypip1 2022-09-13

### 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?

tamolam8 2022-09-13

### Sample size requirements in surveyIf 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?

nar6jetaime86 2022-09-13