VCE Methods Question Thread!

[nadiaaa]

Is the answer B? I think it should be, as the fact that we are generating 90% confidence intervals means that 90% of them would be expected to contain p.

Yep it is! Tbh i thought confidence intervals meant that like there's a 90% that the results sits within this range or something, so what does confidence intervals actually tell us??

[YellowTongue]

I thought confidence intervals meant that like there's a 90% that the results sits within this range or something, so what does confidence intervals actually tell us??

This question tests your understanding of the definition of what a confidence interval actually is. Someone else can probably explain this better than me, but I'll give it a try 

Basically, each confidence interval that you calculate is one of many possibilities. Each interval depends on the sample proportion that you find.

It is a common error to interpret that:

Pr(Lower boundary<p<Upper boundary)=0.90

However, this is incorrect. The correct interpretation of the confidence interval is that we expect 90% of intervals to contain the population proportion.

[Sine]

For methods and spec I think it's sufficient to say that an confidence interval is an interval that we are for example 95% sure the actual population proportion lies between however consider someone takes a sample size of n and constructs a C.I. (95% confidence) Now lets say he gets 99 of his friends to do the same with the same sample size n, now we have 100 different C.I. The 95% meaning is that 95% of the confidence intervals generated with the same sample size will contain the population proportion - 95 of the 100 C.I contains the population proportion - also meaning 5% of the C.I do not contain the population proportion -  5 of the 10 C.I don't contain the population proportion.

[nadiaaa]

Thanks Sine and YellowTongue i now get what confidence intervals mean !! Yay

[exit]

  yeah thats fine, in fact, all 1/2 is assumed knowledge, and they might throw in a 1/2 q to seem how rmbr what etc.

So I assume  it will only be allowed this year, since the new 1/2 doesn't have it.

Also, on the topic of confidence intervals, do i use square or round brackets when i find a 95% c.i.?

[Elizawei]

Also, on the topic of confidence intervals, do i use square or round brackets when i find a 95% c.i.?

Round brackets
(according to suggested answers for the sample exam)

[godsonsamuel888]

Hey guys, I need some help with the first question. What does it mean when it says the proportion is within: Standard Devaiation or possibly the Margin or error?

[LetsDoDis]

Thanks for the help on my last question guys! Appreciate it, now I have a new one sorry 

When is it necessary to divide by the coefficient of an x-value when differentiating or antidifferentiating?

Thank you! Just ensuring I know everything before tomorrows exam, still studying 

[Pineapple66]

I'm a bit confused about finding the anti-derivative using the rules 1/(ax + b) dx = 1/a loge (ax + b) + c when ax+b > 0 and 1/(ax+b) dx = 1/a loge (-ax - b) + c for ax + b < 0
how do you know if ax+b is positive or negative, if you don't know what x is? not sure if this is related but
 how would you work out the anti-derivative of (4-2x) ^-5 with respect to x?

And with circular functions for the question:
Solve the equation 2 sin (x/2) = -1/2 for [2pi, 4pi] would you find all the solutions between [4pi,8pi] or just keep to the domain in the q?
when you're solving for x in sin (x/2) = -1/2, and you get the base angle of -pi/6 (since sin(-number) = - number) WHY DO THEY FIND SOLUTIONS IN QUADRANTS WHERE SIN IS POSITIVE? / ?

ahhaha sorry for all the questions, at this point just trying to salvage my atar  thank you!

[bedigursimran]

 Hey guys. I have attached a graph from question 9 in the 2011 exam 1.

What I don't understand is why can you simply equate.

Despite g(x) being under the axis and above at another time. Also, why don't have to do it in parts like I was normally told to when this happens.
Thanks a lot, hope I didn't sound too confusing.

[P.GUAN]

Where can I find the 2016 Methods Sample Exam 1 answers?

[bedigursimran]

Where can I find the 2016 Methods Sample Exam 1 answers?

Here you go http://www.itute.com/download-free-vce-maths-resources/free-maths-exams/

[exit]

Hey guys. I have attached a graph from question 9 in the 2011 exam 1.

What I don't understand is why can you simply equate.

Despite g(x) being under the axis and above at another time. Also, why don't have to do it in parts like I was normally told to when this happens.
Thanks a lot, hope I didn't sound too confusing.

The location of the axis doesn't change the fact that you minus the lower graph from the upper to find the distance between. Parts is for when the upper and lower graph switch (when they intersect)

[bedigursimran]

The location of the axis doesn't change the fact that you minus the lower graph from the upper to find the distance between. Parts is for when the upper and lower graph switch (when they intersect)

Ah. That makes so much sense! Thanks

[MB_]

I'm a bit confused about finding the anti-derivative using the rules 1/(ax + b) dx = 1/a loge (ax + b) + c when ax+b > 0 and 1/(ax+b) dx = 1/a loge (-ax - b) + c for ax + b < 0
how do you know if ax+b is positive or negative, if you don't know what x is? not sure if this is related but
 how would you work out the anti-derivative of (4-2x) ^-5 with respect to x?

And with circular functions for the question:
Solve the equation 2 sin (x/2) = -1/2 for [2pi, 4pi] would you find all the solutions between [4pi,8pi] or just keep to the domain in the q?
when you're solving for x in sin (x/2) = -1/2, and you get the base angle of -pi/6 (since sin(-number) = - number) WHY DO THEY FIND SOLUTIONS IN QUADRANTS WHERE SIN IS POSITIVE? / ?

ahhaha sorry for all the questions, at this point just trying to salvage my atar  thank you!

For the circular functions questions:
1) You would find the solutions between [π,2π] because you change the domain to whatever is being done to x in the equation (in this case x is being divided by 2)
2) For negative values (eg. -1/2) I write the base angle as π/6 and the find solutions of sine where it is negative and I don't think they find solutions in quadrants where sine is negative, I would need to see an example