August 27

I often face problems while finding angles as in case of inverse trigonometric functions e.g. sin ^ -1 = 0.4 and cos ^ -1 = 0.67. How can I find the vale of such angles without using calculator? Publisher54321 (talk) 01:47, 27 August 2013 (UTC)[reply]

The old way was a slide rule. A table also works. Otherwise you will spend a lot more time to solve it mathematically. StuRat (talk) 01:50, 27 August 2013 (UTC)[reply]

If you can approx square roots fairly decently, then for the sin case you can use (-3 + sqrt(192x - 15)) / 12, where x is your inverse sin (and it is > 15 / 192). But, according to excel, for arcsin between 0 and 0.5, just taking whatever value you start with as the solution will be within .02 of the correct answer (for the 0.4 case, sin 0.4 is 0.389..., quite close). Then, using the formula for 0.5 to 1.0 will net you an error of under .02 until about x = 0.86, at which point it starts going up (but it's still only about 0.06 around x = 0.96, so not super horrid- and you can always just remember this and subtract off 0.05 when in this area). If you're wondering where the formula comes from: since arcsin .5 is close to .5, I just pretend it is (gives nicer coeffs.), take the taylor expansion of arcsin around 0.5, discard all cubic and higher order terms, then solve the quadratic equation you get setting it equal to arcsin (,which becomes the input x).
Honestly, though, you'd be better served by a calculator (especially if this is too much a margin of error), but if this is somehow better for you, I can work out the cosine case. Note: I got all the numeric values plugging into excel, so I'm just going with what it spits out as arcsin for comparisons.Phoenixia1177 (talk) 08:11, 27 August 2013 (UTC)[reply]

Actually, arcsin = pi/2 - arccos; so you can just use the above method for the cosine case as well.Phoenixia1177 (talk) 09:40, 27 August 2013 (UTC)[reply]

The method explained above is of very high level for me. Isn't there any other method for doing the same? Publisher54321 (talk) 12:05, 27 August 2013 (UTC)[reply]

Maybe, it depends on what counts as high level. Is it the explanation of where it comes from that's difficult, or the formula? For example, are you comfortable with taking roots and aren't familiar with taylor series, etc? Or do roots and such present a problem? I could also try to come up with an easy trick (but more memorizing) to track a "rolling correction", like if sin^-1 is bigger than .5, add y to it, that's a close answer; if it's bigger than .75, then add y2, that's a close answer; etc. However, if you're looking to get better than within .1 of the answer, that's going to require memorizing a lot of numbers.
The better question is: what is your motivation? If it is just to calculate, why not use a calculator? And, if for some reason that's not viable, how close do you need your result to be? Is it okay if you're off by like .3; or do you need it accurate to 5 decimal places? The more error you're willing to accept, the simpler a scheme we can use; but you won't find something simple, easy to do in your head/paper, and highly accurate- you're going to have to take a hit somewhere, just tell me where and I'll cook something up.Phoenixia1177 (talk) 12:24, 27 August 2013 (UTC)[reply]

First of all, I would like to thank you for helping me. Now, it is still not completely clear to me how can I compute Sin^-1 (0.89). Please,if possible, show me the steps to compute Sin^-1 (0.89). I promise this is my last comment. Publisher54321 (talk) 14:50, 27 August 2013 (UTC)[reply]

No problem at all:-) Also, this is fun, so you don't have to make this your last comment:-)

1. Since 0.89 > 0.5, we're going to use the formula, here's a step by step (I assume you have pen and paper).
2. We figure out the number under the root, which is 192x - 15. Plugging in x = 0.89, we get 155.88.
3. Since we are taking a root, then dividing by 12; we can just use 155. sqrt(155) is going to be between 12 and 13, taking a few guesses and trying some numbers, 12.45 comes out pretty close.
4. The final step, is (-3 + 12.45) / 12 = 0.7875. The correct answer by calculator is 0.77707...; we're close and took some extra shortcuts.
   

If you wanted to get closer, you could have kept the 15.88, but this would take longer. On the other hand, high x has higher error, and even with some shortcuts, we're pretty close for 0.89. So, if this is tolerable, if you'd just guessed 12 for the root, you'd get (-3 + 12) / 12 = .75, which is still pretty close. If anything is still unclear, I'd be willing to work out a few more cases on your/my talkpage, just let me know:-) By the way, are these answers close enough for your purposes?Phoenixia1177 (talk) 03:53, 28 August 2013 (UTC)[reply]

(unindenting) I just realized: are you trying to compute sin^-1(0.89) or are you trying to find the number y so sin^-1(y) = 0.89? My answer is for the latter case (which is what your initial question looks like). We can do the other version too, it's just a different formula.Phoenixia1177 (talk) 05:50, 28 August 2013 (UTC)[reply]

for small x: ${\displaystyle \arcsin(x)=x+{\frac {1}{6}}x^{3}+{\frac {3}{40}}x^{5}+{\frac {5}{112}}x^{7}+\ldots }$ Count Iblis (talk) 16:31, 30 August 2013 (UTC)[reply]

How would I find the only solution where a and b are integers? Thank you in advance for any useful advice! — Preceding unsigned comment added by 220.255.230.150 (talk) 13:07, 27 August 2013 (UTC)[reply]

One integer solution will occur when 1000 - b = 1. Thus, b = 999, a = -499 000. But, since the numerator is even for integers b <> 500, another pair of integers will happen at 1000 - b = 2, another at 1000 - b = 4, another at 1000 - b = 8, and there are also integer pairs at 1000 - b = -1, -2, -4, -8. So, there is a problem with your "given". EdChem (talk) 13:21, 27 August 2013 (UTC)[reply]

In fact, 1000 - b can be any divisor of 1000, so we have 1000 - b = +/-1, +/-2, +/-4, +/-5, +/-8, +/-10, +/-20, +/-25, +/-40, +/-50 etc. Gandalf61 (talk) 13:45, 27 August 2013 (UTC)[reply]

Maybe the question was supposed to restrict a and b to being positive integers. But then there are still multiple solutions, such as (a, b) = (1500, 2000) and (1250, 3000). Duoduoduo (talk) 14:34, 27 August 2013 (UTC)[reply]

I count 84 solutions, 63 of which are positive. --RDBury (talk) 15:21, 27 August 2013 (UTC)[reply]

On a mathematics forum that I visit, a student asked for advice (not DYOH - they had arrived at an answer, but were not sure if it was correct) on the following question:

A statistician believes that in her town each child born has a 50% chance of being a girl, independently of all other children. There are 281 three-child families in her town; 39 of these families have no girls, 94 have one girl, 115 have two girls, and the remaining 33 have three girls. If the statistician’s belief is correct,what is the probability that a three-child family in the town will have exactly one girl ?

I think the question as posed is ambiguous, since it is not clear whether it is asking about a three-child family drawn at random from the current population in the town (in which case the answer is 94/281 and the statistician's beliefs are irrelevant) or a hypothetical three-child family drawn at random from the town's past or future population (in which case the statistics of the current town population are irrelevant and the answer is 3/8). Possibly the ambiguity is deliberate, and the question is intended to motivate a discussion about interpretations of probability.

So ... is the question ambiguous ? Or am I missing something ? Gandalf61 (talk) 13:36, 27 August 2013 (UTC)[reply]

I agree with your analysis -- whatever the intent, there is some irrelevant information in the question. Since the actual question says "If the statistician’s belief is correct,...", the town's current stats are the irrelevant part. But the last part of the question is ambiguous regarding whether it asks about the current sample or about the hypothetical population. Duoduoduo (talk) 13:56, 27 August 2013 (UTC)[reply]

Does the question have followups?

When I read this what I see is a question about a hypothetical family (3/8) which should then be followed by questions that contrast the statistician's belief with the observed reality. -- Meni Rosenfeld (talk) 18:25, 27 August 2013 (UTC)[reply]

Agreed. Either that or they are just trying to teach the student to ignore irrelevant info. Also note that it says "will have", which to me means they are asking about a future 3-child family, and the current families therefore play no roll. On the other hand, if it asked "What is the probability that a three-child family in the town does have exactly one girl ?", then we would use those numbers to do the math. StuRat (talk) 04:52, 28 August 2013 (UTC)[reply]

The full question appears to be:

A statistician believes that in her town each child born has a 50% chance of being a girl, independently of all other children. There are 281 three-child families in her town; 39 of these families have no girls, 94 have one girl, 115 have two girls, and the remaining 33 have three girls.

If the statistician’s belief is correct, the probability that a three-child family in the town will have exactly one girl is ______%. If the statistician’s belief is correct, then among the three-child families in the town what is the expected number of families that have exactly one girl? If the statistician’s belief is correct, then among the three-child families in the town what is the expected number of families that have no girls? To test whether the data support the statistician’s belief, the appropriate test statistic to use roughly follows a ___________ distribution. The P-value of the appropriate test is about ________%.

The conclusion of the test is that the data support the statistician’s belief or not?

A full copy appears on [1], [2], and [3]. I'm guessing either the teacher used a limited version, poorly, for a more elementary class; or the person asking omitted the rest of it.Phoenixia1177 (talk) 05:16, 28 August 2013 (UTC)[reply]

Good find - thank you. Yes, when you see the complete question then it all makes more sense. As usual, context is key ! Gandalf61 (talk) 08:25, 28 August 2013 (UTC)[reply]

Hello,

I recently read a paper by R.S. Nickerson (Null hypothesis significance testing: a review of an old and continuing controversy) in which one of the many fallacious beliefs discussed is the "belief that the value at which alpha is set for a given experiment is the probability that a type I error will be made in interpreting the results of that experiment" (this section starts on page 258 of the issue in which this article appeared). I was not previously aware that this belief is incorrect, particularly since many sources (including statistical textbooks I've read) state that ${\displaystyle \alpha }$ is the probability of making a type I error.

Nickerson states that alpha is the probability of rejecting the null hypothesis given that it is in fact true (i.e., ${\displaystyle \alpha =\mathrm {P} (\mathrm {reject} \;H_{0}\mid H_{0})}$), which is a common assertion. However, he also states that the probability of making a type I error is ${\displaystyle \mathrm {P} (H_{0}\mid \mathrm {reject} \;H_{0})}$, which he emphasizes is not equal to ${\displaystyle \alpha }$.

Has anyone else come across this definition of the probability of making a type I error? If the definition is valid, Nickerson's point that this probability is different from the probability represented by ${\displaystyle \alpha }$ is obvious, but I find the definition itself to be a bit odd. I suppose it's equivalent to saying, "given that you've rejected the null hypothesis, what is the chance that you've made a type I error?", but since a type I error can only occur when the null hypothesis is true to begin with it seems to me that ${\displaystyle \mathrm {P} (\mathrm {reject} \;H_{0}\mid H_{0})}$ is a more reasonable definition.

Any thoughts on this (or references to articles/sites/etc. that discuss this definition) would be welcome. Most sites on the internet discussing type I error represent the view that ${\displaystyle \alpha }$ is indeed the probability of making a type I error; the one site I found that agrees with Nickerson's view doesn't offer a definition of the probability of making a type I error but merely states that it is proportional to ${\displaystyle \alpha }$ (which admittedly would seem to imply the same definition used by Nickerson via Bayes' theorem).

142.20.133.199 (talk) 15:03, 27 August 2013 (UTC)[reply]

Alpha is the probability of rejecting the null given that it is true. Therefore it is the probability of making a type I error given that the null is true. The phrase "the probability of making a type I error" is an unconditional phrase, not really of much interest in my opinion, but one which would have to be interpreted as the probability that the null is true times the probability (alpha) of making a type I error given that the null is true. Unless you have a Bayesian prior, this cannot be computed. As for his expression ${\displaystyle \mathrm {P} (H_{0}\mid \mathrm {reject} \;H_{0})}$, I think this would have to be described with the conditional wording "the probability of a type I error given that the null is rejected", which is not the same as his unconditional wording. Duoduoduo (talk) 15:37, 27 August 2013 (UTC)[reply]

In plain English, it would seem that he is talking about the probability that you have made a type 1 error (i.e. I know that I rejected the null, what are the chances that I'm wrong?), which is arguably more useful by itself than knowing how likely a type 1 error is if the null hypothesis is true, and probably what the intuitive meaning of "probability of a type 1 error" is. MChesterMC (talk) 08:50, 28 August 2013 (UTC)[reply]