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February 21

Browse History

Is it possible that every Wikia contributor has browse history, which shows every page on Wikipedia they have visited? If there is, how do I get to it? I want to see all of the pages I've visited on this site.--68.186.238.19 (talk) 00:27, 21 February 2013 (UTC)[reply]

Note that this is Wikipedia, not Wikia. That said, no, the MediaWiki software that powers Wikipedia does not retain a record of pages you visit. Your browser history will, however (subject to its settings). — Lomn 01:24, 21 February 2013 (UTC)[reply]

Bank of America Humane Society credit card

This card formerly supported the Humane Society instead of giving cash back to cardholders. However, they've recently announced plans to discontinued this practice. Why ? StuRat (talk) 00:34, 21 February 2013 (UTC)[reply]

The obvious answer is because they are evil, the common sense answer is because there is no return in it, and the psychic answer I will leave to someone else. But I am surprised it's not obvious that businesses keep business strategies like this secret unless there's a return in publicizing the reason (e.g., he was charged with murdering his girlfriend) for it. μηδείς (talk) 02:51, 21 February 2013 (UTC)[reply]

There's nothing on Google News about it that I can quickly find. I like Medeis' answer. Shadowjams (talk) 08:56, 21 February 2013 (UTC)[reply]

Grading system C being considered below average

I've studies Academic grading in China, Japan and Canada their academic grading scale 80-100% is an A, 70-79% is a B, 60-69% is a C, 0-59% is a F. In Japan and China is their C being considered as average or is it below average. I heard in many countries they avoided D as part of academic grading scale. In Canada is C being considered as average or below average. I have studied some territory 55% would be a D, and 64% would be a moderate C, 67% is a C+, 62% is a C-. They said C is below government standard. Is E passing or is it failing. Because USA never use E, is there such thing as E+ and E-, because failing grade usually don't put pluses or minuses. It is because when I was a middle school student I always get in trouble by contradicting 60% is C, and I would usually identify 61% as a C-, 65% as a C, 65% as a C+, that was how I upset my teachers. if I consider 60% as a C can I classify C as below average.--69.226.39.147 (talk) 04:44, 21 February 2013 (UTC)[reply]

In a U.S. graduate school I briefly attended a C grade was essentially failure. You wouldn't have to retake the class but you would not get your degree unless you did much better in your remaining classes. When I was in elementary/high school an A was usually 90 -94% depending on the teacher. F (failure) was 65%-68% usually, sometimes 70%. But the difficulty of the material is just as variable between countries as the grading system is. Rmhermen (talk) 06:25, 21 February 2013 (UTC)[reply]

See academic grades and Grade inflation. DOR (HK) (talk) 07:17, 21 February 2013 (UTC)[reply]

As DOR says, part of the answer is no doubt covered at grade inflation. In the United States, one significant cause was the Vietnam War. Students had to keep up a certain grade average to keep their student exemptions to the military draft; sympathetic instructors were willing to help them out.

Addressing the supposed relationship between "percentages" and grades, one has to ask, percentage of what? A lot of people seem to feel that this is somehow well-defined, that there is some standard of "90% of the material" that can be compared cross-course and cross-exam. But in my opinion that's the most obvious sort of complete crap.

How much a "percentage" on an exam means — and it's hard to imagine any statement more obviously true than this — depends on how hard the questions are. Where I did my undergrad, at Caltech, it was not unusual for an A grade to correspond to less than 50% on an exam, because the questions were really that hard. In my view this is an excellent approach, "hard questions and lots of partial credit", because it turns the exam itself into a learning experience where the student discovers new things not explicitly taught in class, and remembers them very well because of the extra mental focus generated by the exam situation. --Trovatore (talk) 07:47, 21 February 2013 (UTC)[reply]

My unreferenced opinion is that very few universities (with professional schools like law, medicine, some business schools being an exception) maintain GPA curves. The old letter grades are meaningless in their "average" or whatever meaning. It's all relative to others within the same general classes. Shadowjams (talk) 08:30, 21 February 2013 (UTC)[reply]

I question your premise. I doubt many schools in China use "A, B, C" etc. for grading. When I went to school in China (a couple of decades ago) grades were given as "优、良、中、可、差", a descriptive system originally translated from the (English?) "high distinction, distinction, credit, pass and fail" system. Generally, 90-100% was an HD, 80-90% was D, 70-80% was C, 60-70% was P, and 0-60% was F, though with variations between places and between schools.

While you could say HD = A, D = B, etc, the key difference is that this sytem has five grades, whereas your "A-B-C-F" system only has four grades. If your source has somehow matched up a five grade system to a four grade system, there must have been a conflation somewhere, and it is not obvious, given what you described in your question, that the conflation was necessarily "HD+D" -> "A". --PalaceGuard008 (Talk) 11:29, 21 February 2013 (UTC)[reply]

Again, "percent" of what? This is one of my peeves. It's just so obvious that "percent" is one hundred percent meaningless without saying something about how hard the questions are, and that there is no canonical standard for that. People talk about mastering such and such a percentage of the "material", but that's also meaningless in any decent class. The only time it could be anything other than meaningless is in a class where all you have to do is memorize discrete facts. --Trovatore (talk) 17:08, 21 February 2013 (UTC)[reply]

Trovatore, was that meant to be a response to my post? If so I don't understand your point. I said nothing about how hard the tests were or not. My point was that the OP's description of the Chinese grading system is not correct. If it was addressed to me, well, it is 100% of the marks allocated in a test. Isn't that obvious? --PalaceGuard008 (Talk) 17:40, 21 February 2013 (UTC)[reply]

It's complete nonsense. There is no exam-to-exam comparability between the percentage scored and how well you did. --Trovatore (talk) 17:42, 21 February 2013 (UTC)[reply]

Right, I see you have a chip on your shoulder against exam-based testing. That's irrelevant to my post, which is only about the grading system, and arguably to the OP's question. Please do not read in a moral judgment where none was intended. --PalaceGuard008 (Talk) 17:45, 21 February 2013 (UTC)[reply]

Huh? Not at all. I'm a big believer in exams. The chip on my shoulder relates to the idea that there is some uniform meaning to a percentage on an exam. --Trovatore (talk) 17:47, 21 February 2013 (UTC)[reply]

Without suggesting that a percentage on one exam is comparable to the same percentage on a different exam, I believe that the traditional percentage-to-letter-grade conversion in the United States is 90-100% = A, 80-89% = B, 70-79% = C, 60-69% or 65-69% = D, and below 60% or 65% = F. There is some variation in the cutoff point for F. See Academic grading in the United States. Marco polo (talk) 23:51, 21 February 2013 (UTC)[reply]

It's "traditional" in some sense, but it's totally meaningless, so luckily it's mostly ignored, at least at the post-secondary level.

Unfortunately it lingers in students' head at some level, and this is a very bad thing, because it means they panic unnecessarily when exam questions are difficult. Exam questions should be difficult, and then they should be graded in such a way that the difficulty is taken into account. --Trovatore (talk) 00:19, 22 February 2013 (UTC)[reply]

How is "6 months' passport validity" calculated?

Hi there, everyone:

It's often the case that countries require an individual to have six months' validity in his or her passport to allow them into the country or to process a visa for them. But how is this calculated? Do they take the date a person's passport expires and subtract 180 days and that's the latest date they're allowed to enter, or do they use some other process? I'm curious about the whole thing.

Say someone had a passport that was due to expire on the 23rd August, 2013. What would you say would be the latest day that they could enter a country with a six months validity requirement?

All the best!

--95.141.31.4 (talk) 08:42, 21 February 2013 (UTC)[reply]

Well, the obvious interpretation is to subtract 6 from the month (add 12 if you are below zero). So the latest day the passport is good for entry is 23rd of February (typically, the passport is still valid on the day of expiry). --Stephan Schulz (talk) 12:40, 21 February 2013 (UTC)[reply]

Thanks a lot for your advice, Stephan. This would certainly make sense and I hope you're right! Am I right in thinking that if you enter in this case on or before the 23rd of February, you could even get a six months' visa in many countries? Cheers for your input! --95.141.31.4 (talk) 15:07, 21 February 2013 (UTC)[reply]

I suspect so, but I'm not quite sure. And this may well differ from country to country. Your best bet is to check with the embassy in question. --Stephan Schulz (talk) 15:44, 21 February 2013 (UTC)[reply]

Party listing

Why wasn't the Independent Party listed along with the other political parties? My search question was how many people are registered in each political party in the USA? — Preceding unsigned comment added by 173.55.112.225 (talk) 09:53, 21 February 2013 (UTC)[reply]

I've added a section title for your question. List of political parties in the United States has Independent American Party, regional parties, parties formerly named Independent and Independence Party of America. If it's not one of those, you'll have to be more specific. Clarityfiend (talk) 10:02, 21 February 2013 (UTC)[reply]

The OP is a little more likely to mean the longer-established American Independent Party, which has been around since 1968 (it won five states in the '68 presidential election) and was for a long time one of the five parties having automatic ballot access in California. (Registration numbers were generally considered to be inflated by voters who thought they were claiming themselves to be independent of any party, but nevertheless I believe the AIP has or had a small but substantial base of committed members. I seem to recall they may have merged at some point with the Constitution Party but I'm not sure about that.)

Just to add further confusion, there's also America's Independent Party. --Trovatore (talk) 03:59, 22 February 2013 (UTC)[reply]

People in the United States do not generally register with a party. Our systems are different. And that is systems as it varies by state. See Primary election#Types for descriptions of some of the ways. Some states have parties require registration some don't, some do sometimes. Rmhermen (talk) 18:23, 21 February 2013 (UTC)[reply]

What a ridiculously wrong statement. Most people, at least most voters, in the U.S. do register with a party. HOwever how states treat party affiliation, especially in primaries, is HIGHLY dependent on state legislatures. But Rmhermen gives an entirely wrong impression, as is typical here at the reference desk. Shadowjams (talk) 13:52, 24 February 2013 (UTC)[reply]

References please. "most voters do register"? See also Voter Registration#United States: "Declaring a party affiliation is never required, however, and some states, including Georgia, Michigan, Virginia, Minnesota, Wisconsin, and Washington, do not offer the option of declaring party affiliation with registration at all." and from Political party strength in U.S. states:" As of 2010, 28 states and the District of Columbia allow registered voters to indicate a party preference when registering to vote; the following 22 states (mostly in the South and the Midwest) do not provide for party preferences in voter registration:..." Also "Illinois and Ohio are the two states which don’t have registration by party on the voter registration, and yet which try to classify voters as party members depending on what actions they take. They results of the Illinois/Ohio system create many ambiguous legal problems."[1] Rmhermen (talk) 20:00, 24 February 2013 (UTC)[reply]

I've never heard a weaker response. You mention a few states that have differing party affiliation statues. So what. You don't answer the simple and only fact I challenged you on, and that on which you're wrong; that is, the vast majority of Americans who vote are affiliated with a party. Let me quote you: "People in the United States do not generally register with a party" > WRONG, [2] at least 70%+ of American voters are registered with a party, and I seriously doubt that independent figure in that poll. You can spew whatever response you want, and the fact you linked to something unrelated is meaningless. Even our own article says that the supermajority of people are affiliated with a political party: Political_party_strength_in_U.S._states#Gallup. Your early statement, the one I challenged, is obviously wrong. Why are you digging in your heels? Shadowjams (talk) 05:39, 25 February 2013 (UTC)[reply]

Your response is as puzzling as your intensity over this matter. Your source does not mention anything about number or percentages of Americans who are registered to political parties. Likely voters and affiliated does not equate to registered party members. 22 states as I have shown do not have party registration as part of voter registration. Some might register on their own but is likely small numbers. In Illinois a Republican candidate got dropped from the ballot for signing a petition for a Democrat which meant to the state that he was a Democrat despite the state not registering voter's parties. Sounds complex as I indicated before. Rmhermen (talk) 05:58, 25 February 2013 (UTC)[reply]

Are you just ignoring the giant chart in the middle? Here, i'll link it: [3]. It lists a Republican + Democrat percentage at over 70%. Insofar as you're surprised at my intensity of calling you out on being wrong, I"m surprised at your intensity of defending an obviously wrong position. Maybe you misspoke (I've certainly done that here) and maybe I've misinterpreted your comments, but your very straight forward comment that I quoted above, is obviously wrong. That's what I'm correcting you on. Shadowjams (talk) 06:20, 25 February 2013 (UTC)[reply]

No I think those things mean something other than you think they do. (The chart which says REGISTERED & LIKELY does not say registered) I grew up in Illinois and live in Michigan now. In my life I think I have known one single person who registered with a political party, paid dues and carried a membership card. The question we are trying to answer here is "how many people are registered in each political party in the USA" not how many people voted in the Democratic primary+Republican party or how many people are associated in some way with the Independence Party. It is not unheard of for people here to cross-party vote in open primaries to try to get a weaker candidate to run against your real candidate. It doesn't mean you have stopped considering yourself associated with your original party (except sometimes in Illinois and Ohio, where it does mean that) What I am trying to point out is the U.S. situation is complex and varies by state and times. Unlike in some foreign countries, to be a party member we do not sign up to a membership and pay yearly dues to our party (although in some places you do). If you want to consider "registered" as the sum of registered, LIKELY, affiliated and perhaps other terms you have mentioned to reach your totals, you are, of course, allowed to keep changing the bar. But it goes to illustrate my point of the complexities. Rmhermen (talk) 15:45, 25 February 2013 (UTC)[reply]

Indeed, the linked chart says literally nothing about how many people in the US register a party affiliation. It charts, instead, party 'identification' as recorded in responses to opinion polls. Pollsters will separate this from registration; that may be asked about elsewhere in the survey but the question behind this chart would be something along the lines of, "Regardless of which party you plan to vote for, which party do you currently identify most closely with?" That's not to say that Rmhermen is correct in saying that the majority of people (or voters) do not register with a party. But neither of you has so far presented any evidence either way, which is kind of the point of the Reference Desk. OpenToppedBus - Talk to the driver 18:34, 25 February 2013 (UTC)[reply]

Edible Oil

Dear Wikipedia, I need some information on the nutritional values and health related information of edible oils in India. — Preceding unsigned comment added by Jaikishanm (talk • contribs) 12:29, 21 February 2013 (UTC)[reply]

You might want to start with our List of vegetable oils. Itsmejudith (talk) 15:34, 21 February 2013 (UTC)[reply]

For nutrition info, I recommended the site www.nutritiondata.com. Here's their list of oils: [4]. Pick on each for full nutritional info. Note that there are multiple pages in the main list. StuRat (talk) 16:09, 21 February 2013 (UTC)[reply]

how to prove stability

First, I am NOT asking anyone to do this. I am asking for references so I can learn how to do this...

I did an experiment and I've been asked to prove stability. The experiment is rather simple. I have a population of students. Currently, there is a test to decide which students are best to take a certain class (it is only a 5 day class). If the students do well on the test, the county pays for the class. Then, the students either get certified or fail. The test is there to weed out those who won't certify, wasting county money. My experiment was to take students at random and give them the class (a fake one that doesn't give certification, but the same material), but give them a test each day. If they passed the test, they get to continue in the class. If they fail any day, they are kicked out and can't pass. They are replaced with different student who starts at day 1. I found that by doing this, I was able to consistently meet (or exceed) the pass rate of the real class. Also, I changed the class to a completely different one without warning and found that the random set of students very quickly reformed into a set that could pass the new class.

Now, my reference point is the real class with a real test and real pass/fail statistics. My experiment is my fake class with the same material and a random set of students who might pass or fail (I consider a fail on day 1 to be the same as a fail throughout any day of the class). I've been reading about stability and I think that I need to do something like show the probability that the random set of students will be good as taking the class and show that the probability increases over time. But, I cannot find any examples of this to work from and I'm having a huge mental block trying to just get started. Any pointers in the right direction will be greatly appreciated. — Preceding unsigned comment added by 128.23.113.249 (talk) 14:59, 21 February 2013 (UTC)[reply]

Are you trying to prove that your sample is representative of the target population ? The number of students in your sample is also critical, as low numbers of students will not predict the behavior of the larger population as well as large groups. And, if the students know the difference between the fake class and real class, or even you do, then this brings in the possibility of bias. I would expect that students wouldn't try as hard in a fake class, but you never know. StuRat (talk) 16:15, 21 February 2013 (UTC)[reply]

Also, of course, the fake class will at least prepare them about what kind of material to expect. So even if selection isn't helping, there is a training effect. --Stephan Schulz (talk) 16:28, 21 February 2013 (UTC)[reply]

As far as I can see, you haven't said anything about how it is determined whether a student passes the class (the real class). Do they take a test at the end? Is the test fully objective? Those are really essential to know. Looie496 (talk) 16:38, 21 February 2013 (UTC)[reply]

Let's just see if I've got this right. You conducted an experiment where you wasted quite a bit of peoples' time without their consent, and made a mess of the experiment because you didn't ask someone with a clue to help you design the experiment so it did something halfway meaningful. I feel no great inclination to help you progress along this path. Dmcq (talk) 16:46, 21 February 2013 (UTC)[reply]

Let's assume good faith. The OP was probably assigned this task in a middle-school statistics class, and they may well be lax on the standards, just wanting to get the students involved. StuRat (talk) 16:51, 21 February 2013 (UTC)[reply]

BTW, I assume this is an online class, as vacating a seat and putting a new student in his place doesn't work in a brick-and-mortar classroom. Also, allowing students to continue, at their own expense, might be a good option to consider. If they do manage to pass, they could then be reimbursed. StuRat (talk) 16:53, 21 February 2013 (UTC)[reply]

OK. This is how I tried to do this. There are N students. Only P of them will pass the test at the end of the real class (also used as the test at the end of the fake class). The control is the current system of trying to identify members of P with a pre-test. I pick people who I call F. There is a probability of |P|/|N| that a member of F will be in P. I remove members of F who don't pass a daily quiz. The same quiz is given in the real class, but is not graded. Replacing the removed member of F who probably won't pass the test and replacing him with someone else again gives me the change of picking someone in P. So, to prove stability, I have to prove that the intersection of F and P doesn't get worse, correct? — Preceding unsigned comment added by 128.23.113.249 (talk) 17:42, 21 February 2013 (UTC)[reply]

You need to define stability. You seem uncertain that stability occurs if and only if the intersection of F and P doesn't get worse over time, but I'll assume that that property is equivalent to stability.

The intersection of F and P won't get worse over time if the probability p that the new replacement member of F is in P exceeds the probability q that the student who fails a quiz is in P. Two things here: (1) How do you select the initial members of F, and how do you select the replacement members? If it's purely random, then p remains constant over time; but if you have some prior information about the students and you start with the best ones, then p decreases over time. (2) The probability q that the student who fails the quiz is in P obviously increases the more quizzes that student has already passed. For example, if a student has already passed all but the last quiz and then fails the last quiz, then q is probably pretty high, and likely is higher than p. So maybe the optimal procedure is to stop replacing people after they've passed a certain number of quizzes. Duoduoduo (talk) 00:30, 22 February 2013 (UTC)[reply]