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keltingmeith

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A Week of Methods - VCE Preparation!
« on: November 01, 2017, 10:48:51 pm »
+15
Hey all - so something I realised just a couple of hours ago - the first methods exam is less than a week away! To help combat stress and help with all of your exam preparation, every day (with the exception of today) I'm going to put up one tip and go through some maths before 6:00 pm. Some of these will be massive, some will be small, and none of them have a specific audience. Some of these will definitely be more helpful to those who haven't even thought of revising, and others will be more helpful for people that started three years ago. Some of these will be tips for students who just want that 25 for their course, and some for the people that want a raw 50. I will also be doing this for specialist at the "1 week to go" mark, so keep an eye out.

Without further ado, let's start with today's topic - differentiation! For the last decade, the first question on EVERY methods exam 1 has been about differentiation - and on average has been worth 4 marks, which is a tenth of your paper. This should be a gimme question, but understandably will see not everyone getting full marks. There are lots of reasons for this - the three main ones are:

1. Not answering within the context of the question.
2. Choosing the "wrong" diffing rule.
3. Nerves

Now, the third one usually follows because it's the first question, the rest of the exam is really scary which you saw in reading time, or just general exam jitters. This whole series aims to stop the third one, so we'll ignore that one for now.

The first one is one you probably hadn't heard of before, as it's a small bit usually only mentioned at "meet the examiners" events. However, it's quite simple:

If the question is worded "if f(x)=x*sin(x), find f'(x)" DO NOT start off your answer by writing "dy/dx=...". The function was written in terms of f(x), NOT y, and so you're not answering within the context of the question. Yes, this is small. Yes, you will lose marks for it. If you really want, you can preface this with "Let y=f(x)", and then you're answering within the context of the question. HOWEVER, you may as well just get it right from the get go.

Next, choosing the right diffing rule. Now, I'm not going to go over when to choose each diffing rule - you've got your textbook for that, no point if me wasting both of our times by revising it. Instead, I'm going to show a few examples, and how your answer might look if you chose EACH rule. First, though, a little misnomer - EVERY RULE WILL ALWAYS WORK. If you use one rule weirdly, that doesn't immediately mean that you're wrong. I'm going to use a simple case, y=ln(x), to prove this to you.

First, the chain rule. If I set u=x, then the chain rule is dy/dx=dy/du * du/dx. Now, du/dx=1, and dy/du = 1/u. So, dy/dx = 1/u * 1 = 1/u=1/x.

Second, the product rule. Let's use u=ln(x) and v=1. Then, dy/dx = u * dv/dx + v * du/dx = ln(x) * 0 + 1 * 1/x = 1 * 1/x = 1/x.

Finally, the quotient rule. Same substitution as above, but dy/dx = (v * du/dx - u * dv/dx)/v^2 = 1/x / 1^2 = 1/x.


As you can see, ALL of the above got the same result. The problem is - if your function is a little more complicated, you might start seeing weird things that you're just not used to, and your nerves will make you continue on weird thoughts that will give you the wrong answer. There is, of course, a best method of the three for each one, so the idea is to figure out when each of these is not the best method, and when you should stop trying it. With that, let's start with the last example:

f(x)=x*sin(x)

Now, first, we'll try the chain rule. This is an easy one - the moment you try to pick your inner function, you'll immediately realise you only have one of two choices - u(x)=x or u(x)=sin(x). Doing this (remember, f'(x)=f'(u)*u'(x)), we immediately see choosing u(x)=x is pointless. In fact, this is a general rule - if you ever have to use the substitution u=x, you're on the wrong path, and should try a different route. Okay, so let's try u=sin(x). Using this, u'(x)=cos(x) - no problems. BUT, f'(x) cannot be calculated - there's still an x in the expression, as f(x)=x*u! The moment you see this, you should again know that the chain rule will not work.

Second, we'll try the product rule. This one's much easier - you can see that we have one thing multiplied by another thing! So, set u(x)=x and v(x)=sin(x). From here, we know that f'(x)=u*v' + v*u' = x*cos(x) + sin(x)*1 = xcos(x) + sin(x). DO NOT BE SCARED TO WRITE ALL OF THIS OUT. It will guarantee the examiners know exactly what you did, and you will get all four of those marks.

Finally, what if we tried the quotient rule? Using some algebra trickery, x*sin(x)=sin(x)/(1/x). Now, first trick - if you're every differentiating 1/x in a product or quotient rule, you're probably going to immediately struggle - you should stop, and use the other one instead. But, we can still make it work! From above, f'(x)=(v*u' - u*v')/v^2 = (cos(x)/x - -sin(x)/x^2)/(1/x^2) = x^2*(x*cos(x)/x^2 + sin(x)/x^2) = x*cos(x) + sin(x)

The beauty of the chain rule - you can immediately see it's a bad choice and move on. This is a bit more ambiguous in both the product and quotient rules, but once again, you can see that in the end you'll arrive at the right answer. Now, we'll do one more example.

y=x*e^(x^2)

This one's much more complicated - this time, I'll let you try each method individually, then look below for the answer!









Okay, so, the answer is - no one method is enough! In fact, some of you may have immediately noticed - if you try just the chain rule, you still have a product. If you try the product or quotient rule, you have something that's non-standard that you can't work with! So, if you try all three and run short, try combining, mixing, and matching. Practice makes perfect here if you're unsure, and so with that I recommend using this derivative calculator!

The derivative calculator not only calculates the answer for you, on any function you can imagine, but it will show you a step-by-step answer. And, if you don't understand one of the steps, hover your mouse over it, and it'll tell you what rule it used! Try it out on the last example I gave. Any questions on anything I've spoken about here, feel free to comment below, and I'll (or another great AN user) get to you before I post up tomorrow's tip and lesson!


Before I go, I urge you to bookmark this topic. All future posts in the series will be added here, so if you bookmark the topic, you just need to come back here each day to see the new tip.

keltingmeith

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Re: A Week of Methods - VCE Preparation!
« Reply #1 on: November 02, 2017, 12:18:46 pm »
+11
Alrighty, coming back in just over 12 hours later (definitely going to be on the order of 24 next time, sorry guys!), let's talk about every maths students' favourite topic: graphing.

This one will be shorter than the last one, but tbh the last one was a bit longer than most of these will hopefully be. So first, let's talk about graphs. It is inevitable that you will be asked to draw a graph - usually on exam 1 - and do something with that graph. But, here's the thing, graphing doesn't have to be your enemy. This episode will be dedicated to how you can use graphs to help yourself out, and potentially make some questions easier to answer! But first, every graph you draw needs these things:

1. A sense of scale. Honestly, if you just draw in an axes with a believable scale, and use two points to determine the shape of your graph, everything else can be fudged and hand-drawn.
2. Labelled axes. Show which one is y and which one is x. Again, answer in the context of the question - so if the equation is given as f(x), next to your curve you should also write "y=f(x)".
3. Shape. If you draw a parabola for a cubic, you done fucked up. On top of this, important points (particularly turning points and axes intercepts) should be labelled, PARTICULARLY if you were asked to calculate them in a previous part of the question, or you're trying to show something about that turning point. Don't forgot those labelled asymptotes, too!

Now, this may seem like a lot - but this one comes down to practice. When I did VCE, I used graphs to answer so many of my questions, that by the end of it I could cover all of these points in 5 seconds. (more complicated graphs usually involved my CAS finding points for me)


Okay, so now that that's done, question - how can you use graphs to help you answer some questions? Here's some examples to help you out:

First, there's the obvious "find the area between these curves". Tbh, this one is a no-brainer - you need to draw the graphs to figure out which one is on top. From there, you can properly use the right equation.

Another, less obvious, example: "show that for y=x^3-12x, x=2 corresponds to a minimum". Now, you could find the derivative, draw up your table, etc. You could do the double derivative test a la specialist maths. But, if you mind blank, you COULD just draw a graph. Not even kidding, and if you're quick at drawing graphs, it's as much effort as doing a slope table. Firstly, you need to find the derivative anyway, which should be quickly solved to get x=+/-2. Using this, all you need to do is make the claim that the graph is a positive cubic, and then you draw the right shape. Since x=+2 is the SECOND turning point, it HAS to be a minimum - otherwise the shape would be wrong. Point proven, move on. Obviously a slope table is a better way to show this - but if you're freaking out and forget how to do it, a graph is so much easier to remember.

Now for an example that's a bit more concrete: "prove that ln(x) is a constantly increasing function". There are two ways to do this - both of them first involve taking the derivative, which is 1/x. From here, you need  to prove that 1/x is always positive for x>0 (the domain that ln(x) is defined for).The first method, the one that most people will go to, is to do this algebraically - you make the argument that 1/x will never change the sign of x, and since x is always positive, 1/x has to always be positive as well. (alternatively, x>0 ==> 1/x > 0) This is likely straightforward for the top methods students, but most other students can stumble around it. The answer you want? DRAW THE GRAPH. It'll take you just a few seconds, but it can be clearly seen by the asymptotic behaviour of 1/x that it will always be positive for x>0 - so all you need to do is draw it and then right, "from the graph, it can be seen that f'(x)>0 for all x>0, therefore f(x) is a constantly increasing function".


Of course, to be able to use graphs like this in a question, you need to be able to draw a link from the question to the graph. For this reason, I recommend always thinking about what the graph will look like when answering a question. It'll give you an insight into what the scenario actually looks like - PARTICULARLY for those questions that go along the lines of "Buddy kicks a football that follows the trajectory of y=-x^2+x. If he's 50 metres from the post, will he score a goal?" The obvious answer is no, because the kid can only kick the ball 1 metre - which you'll know instantly if you can think of what the graph looks like off the top of your head!



Once again, any issues or concerns with this topic, make a post, ask me about it, I'll get back to you! Alternatively, if there's a topic you particularly want me to cover, let me know and I'll see what I can do to get it covered!

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Re: A Week of Methods - VCE Preparation!
« Reply #2 on: November 03, 2017, 04:18:01 pm »
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Hi! thank you for these posts, they are really helpful

Can u also please do a post on sampling (approximation)?
Thank you :)

keltingmeith

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Re: A Week of Methods - VCE Preparation!
« Reply #3 on: November 03, 2017, 10:05:40 pm »
+5
Hi! thank you for these posts, they are really helpful

Can u also please do a post on sampling (approximation)?
Thank you :)

Definitely! Circumstances intervened, so I'll be doing a twosome tomorrow, I'll see if I can squeeze it in either then or Sunday.

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Re: A Week of Methods - VCE Preparation!
« Reply #4 on: November 04, 2017, 06:45:27 pm »
+9
I set myself deadlines so that I have to follow them, unfortunately that didn't quite happen. :'( Nonetheless, here's lesson #3! Lesson #4 incoming in a few hours.


So, we know how to diff, we know how to graph, the next step - algebra. In particular, just basic algebra. Here's a common question example that comes up time and time again:

"If f(x)=x, show that f(x)-f(-x)=2f(x)"

See, the method to doing this question isn't difficult, but it seems to very quickly throw people off. Let's work through it one step at a time - firstly:

f(x)=x
f(-x)=-x
2f(x)=2(x)=2x

Putting these together:

f(x)-f(-x)=x - (-x) = x+x = 2x = 2f(x), QED

See, not hard! All you need to do is follow basic algebra, start with one side, and watch it move to the next. Let's go over two more examples - the first, from last year's exam 1:



Seems much trickier at first glance - and this definitely has a few more steps. But, it's only worth 2 marks - because all you need to do it is basic algebra! So, we start by figuring out each bit:



Now, just plug things together! Note the use of the log law in the second line:


See, easy! Little bit more effort, but it all comes down to basic algebra.

On a side-note, these types of questions like to pop-up in the exam a lot. In particular, they'll do it in this style:

If f(x)=ln(x), which of the following are true?


The answer is, again, just a matter of writing them all out - however, don't be scared to do this on your calculator! Simply define f(x)=ln(x), and then write all of those expressions into your calculator. If it's true, the calculator will tell you "TRUE" - but if it's not, it'll say "FALSE" - it could end up being a much quicker way to answer the question!

*NOTE: This is based on my experience with the TI-nspire, Classpad users may need to consult their teachers on how to do this*

keltingmeith

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Re: A Week of Methods - VCE Preparation!
« Reply #5 on: November 05, 2017, 01:02:12 am »
+8
Alrighty, lesson number 4, let's get back into this groove!

Now, I'm going to be clear, since this was requested - I'm going to very specifically start with sampling tomorrow. The reason for this, is that I want to go over integration by recognition, and it makes more sense to do that after covering that algebra lesson. Furthermore, sampling is going to go over two lessons - one where I talk about the probability side, and another where I talk about the statistics side, so I just need a little more time to go over what to cover in each part.

So, with that, integration:

Before we talk about about difficult integration, we need to talk about simple integration. Integration is easy to understand when you think of it as its other name - antidifferentiation. It's the opposite of differentiation - if you diff sin(x), you get cos(x), so if you anti-diff cos(x), you get sin(x). Simple, right?

Okay, so let's put some symbology in there. If we have:



And we take the integral of both sides:



We can now do something cool - we know that the integral of cos(x) is sin(x), right? So, if we remember our +c, then the equation starts to look like this:



Now, if this looks like it should be obvious, then good news - you've a great mathematical mind! If it doesn't seem obvious, then that's okay, now you know. But, importantly, this makes a very important point - if you have a derivative, and take an integral, you get what's on the inside. So, let's do a practice question with this. Now, integration by recognition ALWAYS follows in two parts. The first part: you differentiate something. So, first example, let's differentiate x*e^-x:



Second part: using this information, find an anti-derivative for x*e^-x. Now, if you DON'T know what you're doing, remember that you have to use the previous part. So, just start IMMEDIATELY by integrating both sides:



From here, remember that the integral and the derivative cancel each other out. We can also split up the integral on the other side - so let's do that. Also, don't forget - anytime you lost an integral from one side, you need to include a +c:



Now, remember our basic algebra and what we're trying to find. We want an antiderivative for xe^{-x}, so let's solve for that:



And to finish this one off - we wanted to find AN antiderivative, so we should let that pesky c=0:



And we're done!

So let's up the ante a little bit - instead of working with two known functions, let's make them unknown. As they say, if you can work with the general case, you can work with every case. Now, the way these questions unfold, they ALMOST ALWAYS involve you starting with the product rule. So, let's find an antiderivative of u'(x)v(x) (note the dash - I've done this on purpose!). I'm going to start by diffing u(x)v(x), and all the other steps without explanation, see if you can follow:



Now, from here, we don't know the integral of u(x)v'(x), so we can't go any further. BUT, usually this will be an integral that's very straight-forward. HSC students may recognise this - it's called integration by parts, and is a really useful technique. Feel free to practice this for VCE, but make sure NOT to use it at any point, because your examiners might not actually know what it is.


---

Specialist series starts tomorrow (errr, today, this took two hours to write, I thought it'd only take half O.o), as well, so if you're doing spec, stay tuned for that!

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Re: A Week of Methods - VCE Preparation!
« Reply #6 on: November 05, 2017, 04:04:26 pm »
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Alrighty - today, it's probability, and specifically in reference to sampling. But first, let's try and figure out what probability exactly represents in real life, because that's the part that's important to statistics.

So, let's say I want to buy some milk - but, I don't know if I want chocolate milk or vanilla milk. They cost the same, I don't have a taste preference, so I decide to pick one "fairly" - by tossing a coin. So, I flip a coin - if it's heads, I get chocolate milk, if it's tails, I get vanilla milk. Easy, fine, so we have two events - heads (or chocolate), or tails (or vanilla). Easy, fine.

But what if NOW I also have the choice of banana? Here's the thing, if I wanted to do the same test to pick one, I can't - a fair coin is always going to be fifty-fifty. So, what if instead I chose to roll a die - if I get a 1 or 2, I'll chocolate, 3 or 4, vanilla, 5 or 6, banana. Now, I have three events, which can all happen equally - either I get chocolate milk (1 or 2), vanilla milk (3 or 4), or banana milk (5 or 6). Even better: we can now write things in terms of numbers. So, let's say that X=the type of milk I get. If I were to get chocolate, then X=1 or X=2. From here, we can actually define some type of function that allows us to put numbers to everything:

X=1,2,3,4,5,6
P(X)=1/6,1/6,1/6,1/6,1/6,1/6

Now, here's the thing - you can think of this as X is your x variable and P(X) is your y-variable, and make a graph of this. The beauty of this is that we like graphs - we know how to work with graphs, and from here we can use all the skills we've learned previously to analyse this graph. The problem with this is that it doesn't work with the coin scenario, because the coin doesn't work in numbers! So, instead, we make a decision - let's treat heads as the number 1, and tails as the number 0. All of a sudden, immediately, we know have a function for the scenario with just chocolate and vanilla milk:

X=0,1
P(X)=1/2,1/2

Which you can actually think of as also looking like this:

X=Vanilla, Chocolate
P(X)=1/2,1/2

So, now that we're comfortable with this X, here's the thing - let's say I go to the shop again tomorrow, and after getting chocolate today, I flip a tails, and so buy vanilla tomorrow. This is the nature of a probabilistic event - it can change every time. So, even though I had chocolate milk today, I might get vanilla the next. And even if it's the three case scenario, I might never roll a 5 or 6, and so I might never buy banana milk, even though there's always the possibility that I might.

This is where we hit a nice little distinction - since X is always random, we don't like to say "X=1", or "X=2", or stuff like that - because that makes it sound like X is always going to be one of those numbers, or at some point it will be one of those, but the whole point of probability is that anything can happen and it's only predictable up to a certain point. So instead, what we like to say is "X=x, x=1,2", which instead says that big X is the same as little x, but little x can take on a range of values. It's a small point, but valuable.

Now, the reason we work with this weirdness all comes back to numbers. Because we now have numbers we can work with, we can now do maths with it, which is the important part. The reason why we want to be able to do maths on probability? Because then we can model statistics by using probability, which is what I'll get to tomorrow.

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Re: A Week of Methods - VCE Preparation!
« Reply #7 on: November 06, 2017, 09:14:15 pm »
+7
Only two days until exam 1! Tomorrow is the last topic before it, and the day after I'll be writing about tips for your CAS calculator - so if anybody wants a particular topic discussed, now's your chance to ask!! (and I promise tomorrow and wednesday will actually be before 6...)


Okay, so sampling. Yesterday we discussed probability in terms of trying to assign numbers to things, and we also discussed that if you try to repeat a probabilistic experiment, you won't get the same thing each time. This should be common sense, but if you're confused by it, just spend an afternoon flipping a coin.

So, what does this have to do with statistics? Well, as with all maths, the answer to more complicated questions (such as who voted for which political group) can often be solved by first understanding much easier questions.

Let's say I wanted to know the proportion of times I bought chocolate milk in my coin example from yesterday. Over a period of 10 days, I got fat 7 cartons of chocolate milk. This would mean that the proportion (think percentage, but out of 1) of milk that was chocolate is 7/10. This is easy for a human to do, but what about a computer? We know that if we were to get a computer to do this, it wouldn't be able to do much harder questions unless we personally figured out all the proportions ourselves and fed them to the computer. So, instead, we'll approach the problem this way:

Just like yesterday, where we said chocolate was 1 and vanilla was 0, let's do that again. We can actually now model this as a random variable - even better, we can model it as a binomial random variable, where n is the amount of times we buy milk. Here's the thing, though - we don't actually know what the probability of a success is! If our coin was fair, it would be 0.5 - but we're no longer talking about a made-up, probabilistic scenario. We're talking about a real, statistical scenario. The difference is this:

In probability, you start with a distribution, and use that to see what might actually occur in real life.
In statistics, you see what actually occurred in real life, and you try to find the distribution it came from.

Since we only have two possible outcomes (either I get chocolate or vanilla milk), we know the type of distribution - binomial. To find out what the distribution is, we need to know its two parameters - n (number of trials) and p (probability of success). In probability, this is known - and we'll often refer to the made-up, probability scenario as the population. The statistical experiment is then referred to as the sample. The aim of sampling is simple - can we predict the population parameters from the sample statistics?

Okay, so back to our example - we know the type of distribution, we know the first parameter is n=10. So, can we predict the probability of success? One guess you might make is that the probability of success is the same as the proportion, which we could then calculate like so:



Now, there's a lot in this equation, believe it or not. So, some points: firstly, P-hat is capital because it's ALSO random. We call this a sampling distribution, which is called that way because P can change every time you do an experiment. For example, let's say while watching myself buy milk, you also buy milk yourself using the same method (I even give the you my own coin when I'm done, to make things equally fair). You might buy chocolate milk 4/10 times instead of 7/10 - in this case, when we both calculate P-hat, even though we've used the same formula, we got a different result. The next point - the hat means that P is an estimator of the population parameter p. This is how we differentiate between a population parameter and a sample statistic. Finally, note that X is a random variable, and corresponds to the amount of successes we have. This goes back to remembering that the capital X, after an experiment, will itself be a number - which is a point that's often forgotten.


Now, I mentioned that P-hat is a sampling distribution - which should mean that it can also be modeled with a probabilistic distribution, right? Because it can - in fact, using some scary maths from specialist, I can tell you that P-hat has a normal distribution, with the following parameters:



You can prove this yourself if you want - it's not actually that hard, you just gotta remember that X is binomial. If you do specialist, I recommend giving the proof a shot - it's not hard, and very much within the realm of things you could be asked.


Okay, so now that we know this, let's get back to trying to predict the population parameter, p. Before we got two estimates, 0.7 (from my experiment) and 0.4 (from your experiment). These are both wildly different - which is annoying. We can make them get closer by doing more experiments - let's say we do 100. This time, you might get 0.48 and I might get 0.61. This is still wildly different, but much closer. We could up this to 1000 experiments, and get 0.499 and 0.564, but as you can see, we very quickly get to a point where we have to do way too many experiments to get these values to be the same. However, they are still useful to us - we often call them point estimates, and give them the notation of little p-hat. We'll come back to them in a second.

So, to get some useful numbers, what we do is we consider the interval of points it's likely to be. The reason we might want to do this depends on the application - for example, let's say you're digging for gold, and due to budgetary reasons, you should only dig in an area in which 20% of it consists of gold. Even more, let's say we do an experiment (that doesn't involve digging, but more complicated things), and it gives us a point estimate of little p-hat=0.4 after 25 replications. This point-estimate looks like it'll be enough, but what if another experiment gave little p-hat=0.1? We need to know with more certainty that the first experiment is reliable, and we've very well established that we can't trust point estimates alone.


Well, let's think of it this way, instead - these values came from a probability of a continuous random variable (big P-hat). So, what if we instead consider a probability - in particular:



What this equation tells us is what the probability (alpha) is of big P-hat being in the range of a and b. So, if big P-hat is between 0.2 and 1 95% of the time, this might be enough for us to decide to dig in that area. Yes, this IS what people do in real-life - countless experiments have demonstrated that this is actually not a bad way of trying to make these kinds of decisions. However, normally, we don't put in values for a and b and then find alpha - the reasons as for why, I'm not entirely sure, probably a historical thing - although, there are definitely situations where starting from alpha is more beneficial.

Okay, so, let's think about this - we know that big P-hat is normally distributed. So, to figure out what those bounds could be (and find this magical interval that I keep talking about), let's try starting with the standard normal distribution and seeing what I can get to with some algebraic changes:



An explanation: in the second line, I multiplied by that bit square-root thing to make the standard deviation the same as big P-hat, and then added p to make the mean the same as big P-hat, and the last line just makes the middle bit big P-hat. THIS IS BEYOND METHODS. DON'T WORRY IF YOU DON'T UNDERSTAND IT.

I just want to get to the point that we now have an interval in a probability - that interval is:



This, mathematically, is the interval of big P-hat that gives a probability of alpha - and we call this a confidence interval. Normally, all we care about is the 95% confidence interval, which comes from when z=1.96. If you wanted to do another interval, say a 68% confidence interval, you'd have to pick the z-values from the standard normal distribution that would give this percentage. Now, we don't have a lot of precedent to go off of, but it looks like VCAA will always just tell you the z-value straight up. If you want to cover your bases, you can use the following formula in your CAS calculator to find the z-value:



where invnorm is the inverse normal function.

So, back to our confidence interval. We're almost done, but there's one small problem with it - it uses a population parameter, which is what we've been trying to estimate this whole time! This is where the point estimate comes in - because we're now dealing with an interval, even though the point estimates gave a margin of error, that margin is reduced. In fact, it's reduced to the point that it's not as significant - so, what we do, is substitute our population parameter with our sample statistic. So, if we were to calculate the confidence interval from our milk buying situation from earlier, we'd get:



Now, you should be asked how many decimal points to put this to - if not, just pick something reasonable. I've gone with three.

Now, this results makes a lot more sense all of a sudden - and it very obviously explains why our coin gave such a wide variety of results - from only 10 tosses, 95% of the time our proportion is going to be in this range. The range gets smaller if we do more tosses - feel free to calculate it and get a feel from it all.

And let's also consider our mining example earlier, just so we can see how to make a conclusion from this interval. We ran our experiment 25 times and got a little p-hat of 0.4, and we want it to be above 0.2 95% of the time. So, doing our calculation gives the 95% confidence interval to be:



Since numbers below 0.2 are not in this interval, we know that it's safe to dig for gold in this area. HOWEVER, if we had instead gotten 0.1 as our point estimate:



Now, we did get a negative number - leave it in there, otherwise the interval doesn't correspond to 95% probability. The reason we got a negative number is because we're basing it off a normal distribution - even though it's nonsensical for p-hat to take on a value smaller than 0 or greater than 1. It seems weird, but trust me - the maths is sound.

This time, numbers less than 0.2 are in our interval - so, it's not safe to dig for gold.

There's only final point I want to make - you'll notice that the first gold example has a range smaller than the milk example. Hopefully this can help you appreciate just how big of a difference just a few more experiments can make. Since n is under a square-root, this difference becomes much smaller every unit as n get massive (as an example, the range decrease from n=1 to n=4 is the same range decrease from n=4 to n=9, even though we've gone up an extra two experiments). The other thing you'll notice is that the second gold example has an even smaller range than the first gold example. This isn't because our point estimate is smaller - in fact, the range for p-hat=0.1 is the same as the range for p-hat=0.9. The key here is that point estimates further from 0.5 give smaller ranges than point estimates closer to 0.5

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Reminder, tomorrow's the last lesson before exam 1, so if you want something covered, get it in quick!

Elegant_Chaos

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Re: A Week of Methods - VCE Preparation!
« Reply #8 on: November 07, 2017, 01:46:03 am »
+1
After reviewing NHT exam 2 questions 15 and 20, anything seems fair game.
So... Have a crack then at this fresh logs test; I'm sure now that the difficulty is within the realm of "reason" according to what was assessed this year.

keltingmeith

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Re: A Week of Methods - VCE Preparation!
« Reply #9 on: November 07, 2017, 01:05:13 pm »
+7
I honestly had no idea what I wanted to put here. At all. So, I thought about it, and realised that you're all guaranteed to get a graphing question in exam 1 - and that graph could be ANYTHING that's been taught in the year. So, here's a list of all the possible graphs you could be given:

First, the linear graph:




Easy, you've been doing this for years. Notice the extra two equations people often forget - the first allows you to find the equation from just the gradient and a point (also useful when asked to find tangent lines!), and the second is my personal favourite - allows you to construct the equation from just the y-intercept (y_0) and the x-intercept (x_0). Next:




Quadratic. To find the equation, you always need three points. No ifs, whys, or buts. Preferably when you graph it, you use the second equation (where you draw it based off of the two intercepts, and know that the turning point is in the middle) or the third (which is straight up just turning point form).




Cubics! This time, you need FOUR points - unless you know for sure that the graph can be put into turning point form, in which case you can get away with three. But tbh, if you know it's in turning point form, you probably don't need to find the equation. You can also be expected to do any higher polynomial graph, for example:



But, it's unlikely that VCAA will ask for anything higher than a quartic, just because the time-cost is too much for you guys.

Finally, the other two you need to be aware of are reciprocal and root-graphs:




In methods, we only deal with basic reciprocals. Above is n=1, the hyperbola. If n is even, you both tails are positive, giving a truncus shape. If n is is odd, you keep the hyperbola shape. Not too difficult.




a is a dilation factor, h and k are translation factors - you should be comfortable with this, if not, consult your text book. The real kicker here is the p and q. If q is even, you only get one tail of the graph (see the square root and quartic root above) - if q is odd, you get two tails (see the cube root above). p can also have an effect - if p>q, then the graph will curve upwards (not shown, but thing somewhere between a quadratic and linear graph, really), but if p<q, then the graph will curve downwards (as shown). If p=q, you have a linear graph. Rejoice. These graphs are worth looking over if you haven't already - but, not so worth it to know them inside and out, because they care much more about transcendent graphs:




These are a lot more tricky - exponential graphs. Lots of people like to make points such as "if b is less than 1, then the graph flips" - I prefer to just use the following log rule:



So that b is ALWAYS greater than 1. Makes life just a little bit easier, and then because the negative ends up in front of the x, you know it's flipped. You'll never need to deal with the situation that b is negative, it will always be positive. Note that if you have a number in front of the b, you can just do some exponential law abuse to turn it into the dilation factor a above.




Logarithms! These are much easier to work with, as we only care about b=e,10. Just like with the exponentials, you can use some log-law trickery to get rid of anything inside the brackets if something ends up in-front of the x.

Finally, circular functions:





These are much more complicated, and tbh require their own full review. Any questions, go to the question thread.

---

And that's the last of our revision until exam 1! Good luck tomorrow guys, and make sure to come back for a calculator lesson!

keltingmeith

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Re: A Week of Methods - VCE Preparation!
« Reply #10 on: November 08, 2017, 03:39:44 pm »
+4
Alright everyone - this is the last one. It's been a hard week, but nothing compared to you guys, so let's power-through!

The point of this post is to just go over all the functions in your calculator you might not be aware of - some of them are incredibly simple, some slightly more complicated, and you may need to check your manual for details on how to do it.

So, by area of study:

Functions
-The obviously helpful feature: graphing. Your calculator can make graphs. What you might not know is that it might actually have the functionality to make piece-wise graphs and graphs with restricted domains, as well! Not range, though, unless there's been an update since I did VCE in 2013.

-You can define functions, so that if f(x)=x^2+x+1, and then on your graph page write =f(x), you should see that function appear on the graphing page. You can then write f(2), and the calculate will return with 7. Really useful for those extended response questions which revolve around a single equation.

-You can also do the above to calculate sums, differences, composites, etc. of functions.

Algebra
-Your calculator has a solve function - and it can also do simultaneous equations! Up to any amount of variables, too.

-Above function has a literal myriad of uses, I'll let you guys go through and find them all.

-Your calculator can factorise and expand for you, as well. It can also find greatest/lowest multiples if you're into that.

Calculus
-Your calculator can differentiate things. It can also find gradients at a point if you use the given command (for TI-nspire, it's the | button, followed by x=blah). You can also define a function that is equal to derivative of something else and find point-gradients that way.

-Your calculator can find tangent line equations.

-Your calculator can find anti-derivatives. The TI-nspire even remembers to include the +c! It can also do anti-derivatives that you can't, if you get asked another integration by recognition question, you can proof-check your answer on your calculator.

-Your calculator can also do definite integrals, and if you draw graphs, you can get a numerical estimation of the area under them or between two graphs.

Probability
-There's a function to evaluate the binomial distribution! To memory, it can find the mean, variance, and I think even the median and mode. More importantly, the pdf function can find P(X=x), and the cdf can find P(X<x).

-See above, but for the normal distribution. Ignore the pdf function, it's useless in this case. DON'T try and use the integration functions for the normal distribution, you'll break your calculator.

-The normal distribution also has an inverse button - in which it essentially solves the equation P(Z<z)=alpha, where you tell it alpha, and it tells you z.


Now, I cannot help anyone with the classpad, and I barely remember the TI-nspire anymore, so I might not be able to help with that anymore (and definitely can't tell you exact buttons to push). After high school, you don't tend to use these calculators again, instead opting for computer programs. If you want to know specifically how to do any of these functions, you'll have to ask someone else, your teacher, or consult the internet or your calculator manual.

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My final piece of advice - you're not going to learn anything new between now and your exam tomorrow. The best thing you can do now is consolidate your bound reference, make sure it's got everything in it, and make sure you know as much about your calculator as you can.

Joseph41

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Re: A Week of Methods - VCE Preparation!
« Reply #11 on: November 08, 2017, 04:28:40 pm »
+1
Nice work, Meithy!

Oxford comma, Garamond, Avett Brothers, Orla Gartland enthusiast.