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1. ocr s2 hypothesis testing..
stuck on jan 2006, Q3 and statistics 2 in general..

The manufacturers of a brand of chocolates claim that, on average, 30% of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly. Show the values of any relevant probabilities

so

(8, 0.3)
and because its H0 : p = 0.3 and H1 : p is not 0.3 than the significance level is now 0.025?

i don't really understand what to do next... am i trying to find out p ( x = 5 ) ? and then if the answer is smaller than 0.025 then we reject it.. ?

i am very confused.
2. Re: ocr s2 hypothesis testing..
Also where po (1.6) find the smallest value of r for which P (R > or equal to r) < 0.01..... what??
3. Re: ocr s2 hypothesis testing..
help anyone...
4. Re: ocr s2 hypothesis testing..
(Original post by peachesandcream77)
stuck on jan 2006, Q3 and statistics 2 in general..

The manufacturers of a brand of chocolates claim that, on average, 30% of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly. Show the values of any relevant probabilities

so

(8, 0.3)
and because its H0 : p = 0.3 and H1 : p is not 0.3 than the significance level is now 0.025?

i don't really understand what to do next... am i trying to find out p ( x = 5 ) ? and then if the answer is smaller than 0.025 then we reject it.. ?

i am very confused.
You will work on the assumption that p=0.3 On this assumption, test the evidence.

So if your hypotheses is two-tailed (i.e. "0.3 or not 0.3") then you will be calculating 2p(X>=5) on Bi(8,0.3) Once you find this probability, you must evaluate what it means for this situation.
5. Re: ocr s2 hypothesis testing..
(Original post by peachesandcream77)
Also where po (1.6) find the smallest value of r for which P (R > or equal to r) < 0.01..... what??
Use your list of tables for poison distribution (of 1.6).

From S1, you will have learned that P(R>=n) = 1- P (R<n). Knowing this, and knowing the probability you're working with is 0.01, what can you deduce?
6. Re: ocr s2 hypothesis testing..
(Original post by Llewellyn)
You will work on the assumption that p=0.3 On this assumption, test the evidence.

So if your hypotheses is two-tailed (i.e. "0.3 or not 0.3") then you will be calculating 2p(X>=5) on Bi(8,0.3) Once you find this probability, you must evaluate what it means for this situation.
ok so that means the significance level is now 0.025% ?

how do i actually evaluate what it means for this situation though, the question says : ' In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly'

So does that mean I'm testing P ( x = 5 ) or when x is smaller than 5 .. ? sorry, very confused
7. Re: ocr s2 hypothesis testing..
The probability of x chocolates with hard centres, assuming the null hypothesis is true, is P(x=T) such that T~B(8,0.3). The critical region is defined such that the probability that a value of T in the distribution T~B(8,0.3) falls within the critical region is 5%. Is the probability for the observed value of T (5) within the critical region? The result should imply whether to accept/reject the null hypothesis.
8. Re: ocr s2 hypothesis testing..
(Original post by aznkid66)
The probability of x chocolates with hard centres, assuming the null hypothesis is true, is P(x=T) such that T~B(8,0.3). The critical region is defined such that the probability that a value of T in the distribution T~B(8,0.3) falls within the critical region is 5%. Is the probability for the observed value of T (5) within the critical region? The result should imply whether to accept/reject the null hypothesis.
ok but i dont actually understand what the next step is i have to do.....
how do i found out if the probability for the observed value of t 5 is within the critical reigon?
would that be p (x = 5 ) using binomial theorem?
what do i actually have to do
9. Re: ocr s2 hypothesis testing..
(Original post by peachesandcream77)
The manufacturers of a brand of chocolates claim that, on average, 30% of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly. Show the values of any relevant probabilities

As you have correctly identified, .

The purpose of the hypothesis test is to prove whether 5 out of 8 chocolates with hard centres is enough to conclude that the percentage of chocolates with hard centres is not 30%.

Someone could just say, "well 5/8 is 62.5%, so clearly their claim is wrong." However, this is not good enough in statistics. We have to prove that the chance that the percentage with hard centres in this sample is not 30%, with a probability of less than or equal to 0.05 that this alternative percentage has occurred by chance. This is because if you took the sample again, the percentage may be around 30%.

So, the next thing we do is find the probability that 5 or more of these chocolates have hard centres. That's

(using tables).

Now, this means that the probability, from a sample of size 8, that 5 or more had hard centres is 0.058. Since this is a two-tail test (meaning that we're interested in two rejection regions - one above the mean and one below the mean), for us to reject the null hypothesis and accept that there are more chocolates than 30% on average with hard centres, the test statistic (0.058) must be less than 0.025 (half the significance level of the test).

, so do not reject .

Make sure to follow this up with a written conclusion to ensure you get those last marks, acknowledging uncertainty with the phrase 'There is insufficient evidence to conclude that...'.

Ask for any more help, I'll be around tomorrow.
Last edited by CraigKirk; 18-06-2012 at 08:21.
10. Re: ocr s2 hypothesis testing..
(Original post by peachesandcream77)
ok so that means the significance level is now 0.025% ?

how do i actually evaluate what it means for this situation though, the question says : ' In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly'

So does that mean I'm testing P ( x = 5 ) or when x is smaller than 5 .. ? sorry, very confused
To address your confusion regarding the values you are testing for , consider that 5 out of 8 chocolates is actually above the proportion stated in the claim. With a bit of thought, this should help you realise that you must be testing for values which include 5 or more. It certainly isn't less than or equal to 5, since that would include the expected proportion if the claim is correct, which makes no sense.

You then have to realise that you have to test for values of 5 and above, because if it happened that in this sample there were 6, 7, or 8 chocolates with hard samples, you would be moving further away from the expected number of chocolates with hard centres according to the claim. As such, you want to include these possibilities in your probability. This is best understood by looking at the critical values on a graph of the distribution, but I don't have a facility to draw this for you.

Maybe it would just be best, in this case so close to the exam, that in hypothesis testing, you NEVER use P(X=x); it's always less than/greater than or equal to depending on which side of the mean you're on.

Make any sense?
11. Re: ocr s2 hypothesis testing..
(Original post by CraigKirk)
To address your confusion regarding the values you are testing for , consider that 5 out of 8 chocolates is actually above the proportion stated in the claim. With a bit of thought, this should help you realise that you must be testing for values which include 5 or more. It certainly isn't less than or equal to 5, since that would include the expected proportion if the claim is correct, which makes no sense.

You then have to realise that you have to test for values of 5 and above, because if it happened that in this sample there were 6, 7, or 8 chocolates with hard samples, you would be moving further away from the expected number of chocolates with hard centres according to the claim. As such, you want to include these possibilities in your probability. This is best understood by looking at the critical values on a graph of the distribution, but I don't have a facility to draw this for you.

Maybe it would just be best, in this case so close to the exam, that in hypothesis testing, you NEVER use P(X=x); it's always less than/greater than or equal to depending on which side of the mean you're on.

Make any sense?
ok thank you for such a thorough and step by step guide. I find statistics very confusing but that really helped. The only bit I dont quite understand still is whether to test for larger or less than 5. When you say 5 out of 8 chocolates is above the proportion stated, do you mean above 30%? For the sake of the exam, shall I assume that because its 5 out of 8 which is roughly 62.5%, this is larger than 30% therefore I will find values for x larger or equal to 5?
Not the most mathematical and statistical reasoning but I dont think in these next few days i'll be able to properly understand it!
12. Re: ocr s2 hypothesis testing..
And additionally, am I right in saying if it is less than the significance level than we reject, if it is greater than we accept. And writing do not reject is the same as writing accept right? or are these two different concepts?
13. Re: ocr s2 hypothesis testing..
(Original post by Llewellyn)
Use your list of tables for poison distribution (of 1.6).

From S1, you will have learned that P(R>=n) = 1- P (R<n). Knowing this, and knowing the probability you're working with is 0.01, what can you deduce?
My ability at S1 was also questionable. So, sorry I really don't know what you're talking about.

Does the formula you've written above mean in relation to the question: 'where po (1.6) find the smallest value of r for which P (R > or equal to r) < 0.01.'

mean P(R > r ) = 1 - P(R < r - 1 )
1 - P(R < r - 1 ) < 0.01
- P(R < r - 1 ) < -0.99
P(R < r - 1 ) < 0.99
So looking at the Poisson where the mean is 1.6, 5 = 0.9940, 6 = 0.9987, 7 = 0.9997 .. so which value would I take? Then I would add 1 to it, to get the least possible value.. ?
14. Re: ocr s2 hypothesis testing..
I'm trying to do the June 2006 paper, and just thought I'd write out step by step each question and its answer and why it is that answer.. because I really have no clue. If someone could confirm what I'm doing is correct or not then I'd really appreciate it.

In a study of urban foxes it is found that on average there are 2 foxes in every 3 acres.
(i) Use a Poisson distribution to ﬁnd the probability that, at a given moment, (a) in a randomly chosen area of 3 acres there are at least 4 foxes,

a.) So this means that firstly the Poisson distribution is (2). So for question a it would be P (x > or equal to 4 ) = 1 - P(x is smaller or equal to 3). Looking up poisson distribution where its 2, and under 3 would be 1 - 0.9473 = 0.1429

(b) in a randomly chosen area of 1 acre there are exactly 2 foxes. [3]

b.) its in 1 acre now compared to 3, which means divide poisson mean of 2 by 3, the mean is now Po (2/3). Finding it exactly 2 is P (x = 2). 2/3 isn't in the table, so using that e lambda formula.. is e to the power of 2/3 times 2/3^2 over 2! = 0.114

(ii) Explain brieﬂy why a Poisson distribution might not be a suitable model. [2]

ii.) explain why poisson might not be suitable, poisson distribution is required when it is in a fixed interval, known average rate and independently. So it could not be possibe because, foxes could come at different times? foxes could travel together so not independent?
Last edited by peachesandcream77; 18-06-2012 at 12:16.
15. Re: ocr s2 hypothesis testing..
Q2. The random variable W has the distribution B (40, 2/7)
Use an appropriate approximation to ﬁnd P(W > 13).

Binomial can take two approximations.. the poisson and normal ?

Seeing N is not larger than 50, than the approximation must be normal.
As np = 80.7 > 5
npq = 400/49 > 5

(np, npq) = (80/7, 400/49) this now as a normal approximation means (u, o^2)?
Therefore to standardise we have

x - u
____
o

Applying a continuity correction, seeing as we have to find P, W > 13, we add 0.5
so its 13.5 - 80/7 divided by the square root of 400/49. (It's square root because before o was squared because it approximated to normal right?)
That gives us 0.725. Seeing as its bigger than 13, that means 1 - 0.725, which looking in the tables is 1 - 0.7657 = 0.2343
16. Re: ocr s2 hypothesis testing..
Q3. The manufacturers of a brand of chocolates claim that, on average, 30% of their chocolates have hard centres. In a random sample of 8 chocolates from this manufacturer, 5 had hard centres. Test, at the 5% signiﬁcance level, whether there is evidence that the population proportion of chocolates with hard centres is not 30%, stating your hypotheses clearly. Show the values of any relevant probabilities.

So with this one, its in binomial form (8, 0.3). As the hypothesis is testing whether it is 0.3 or not 0.3 then that means the significance level is split into 0.025. (I'm not sure I understand exactly why but for the sake of the exam if its a question of it is the average or not, then you always split the sig level). So, as Craigkirk explained to me earlier, because its 5 out of 8 chocolates, which is around 62.5 %, this is larger than 30% so we take x is larger or equal to 5 which is 1- P(x smaller or equal to 4) which is 1 - 0.9420 which is 0.058.
0.058 is bigger than 0.025 meaning we accept Ho.
However how do I conclude in relation to the question.... I've found out we accept the original hypothesis meaning ..
17. Re: ocr s2 hypothesis testing..
DVD players are tested after manufacture. The probability that a randomly chosen DVD player is defective is 0.02. The number of defective players in a random sample of size 80 is denoted by R.
(i) Use an appropriate approximation to ﬁnd P(R ≥ 2). [4]
(ii) Find the smallest value of r for which P(R ≥ r) < 0.01.

i.) Write this in binomial form
(80, 0.02)
I can see that N > 50 meaning we could do a poisson distribution. as np is 1.6, smaller than 5. Therefore it is Po(1.6)

To find P(R ≥ 2), that is 1 - P (R< or equal to 1) , 1 - 0.5249 = 0.4571

ii) P(R > r ) = 1 - P(R < r - 1 )
1 - P(R < r - 1 ) < 0.01
- P(R < r - 1 ) < -0.99
P(R < r - 1 ) < 0.99
So looking at the Poisson where the mean is 1.6, 5 = 0.9940, 6 = 0.9987, 7 = 0.9997 .. so which value would I take? Then I would add 1 to it, to get the least possible value.. ?
18. Re: ocr s2 hypothesis testing..

Q5. In an investment model the increase, Y%, in the value of an investment in one year is modelled as a continuous random variable with the distribution N (µ, 1/4 u^2)
The value of µ depends on the type of investment chosen.
(i) Find P(Y < 0), showing that it is independent of the value of µ.
(ii) Given that µ = 6, ﬁnd the probability that Y < 9 in each of three randomly chosen years.
(iii) Explain why the calculation in part (ii) might not be valid if applied to three consecutive years.

I really have no idea what this question is asking.. Any direction?
19. (Original post by peachesandcream77)
ok thank you for such a thorough and step by step guide. I find statistics very confusing but that really helped. The only bit I dont quite understand still is whether to test for larger or less than 5. When you say 5 out of 8 chocolates is above the proportion stated, do you mean above 30%? For the sake of the exam, shall I assume that because its 5 out of 8 which is roughly 62.5%, this is larger than 30% therefore I will find values for x larger or equal to 5?
Not the most mathematical and statistical reasoning but I dont think in these next few days i'll be able to properly understand it!
Yes, I mean exactly that. This reasoning is fine and perfectly statistical. To summarise, if the sample mean is above the claim, use greater than or equal to, and vice versa.
(Original post by peachesandcream77)
And additionally, am I right in saying if it is less than the significance level than we reject, if it is greater than we accept. And writing do not reject is the same as writing accept right? or are these two different concepts?
When we talk of rejection or acceptance, we mean with regards to the null hypothesis. So yes, this is correct.

I'll do some of your other questions later as my revision!
20. Re: ocr s2 hypothesis testing..
(Original post by peachesandcream77)
Also where po (1.6) find the smallest value of r for which P (R > or equal to r) < 0.01..... what??
I'll take this through from the start, using spoilers. If, along the way, you can guess what to do next, DON'T open the spoiler, do it first, then check the spoiler.

We want to find, for ~ , the value of such that

This means that we want to find the smallest value of so that the probability is less than 0.01.

Spoiler:
Show

First, take note that , since we are dealing with discrete variables in a Poisson distribution.
Spoiler:
Show
(Be careful with the direction of your inequality sign here!).
Spoiler:
Show

Now recall that since this is a discrete distribution, your value of is going to take an integer value; this means that we can not use the Poisson equation to calculate an exact value for , as this value be a decimal (not to mention that it's either impossible or very difficult to calculate!).

We know that we are dealing with distribution , so in the tables check the column for .

Spoiler:
Show

Now you need to find the lowest value in the x-row for which the probability is greater than 0.99, as we can deduce from . Therefore, . To check this, calculate . The answer is 0.006; less than 0.01.

Last edited by CraigKirk; 18-06-2012 at 16:41.