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S4 A-Level Maths Edexcel June 26 2018 Unofficial Markscheme

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Original post by k92e67
putting this here in advance

UNOFFICIAL MARK SCHEME

Q1) test statistic = 3.33 > 2.821 reject h0

Q2a)

b)

Q3) a) differences are normally distributed

b) 0.044<m<0.377. ( something like this)


There was a second range for m as well I think
I got two ranges for m. As it was two tailed I think you had to take into account the positive and negative tails
Original post by Pepperpeople
I got two ranges for m. As it was two tailed I think you had to take into account the positive and negative tails


Yes, you did have to find two ranges for m, but then if you bring them together, it will simplify to just one range (with two tails). Look for the overlap of the two ranges.
Original post by mupsman2312
Yes, you did have to find two ranges for m, but then if you bring them together, it will simplify to just one range (with two tails). Look for the overlap of the two ranges.


The question said find the 'ranges' (plural). Would that mean anything here?
Original post by Redcoats
The question said find the 'ranges' (plural). Would that mean anything here?


Well, we all would have put both ranges in our working, so they should see that anyway. Just giving it as one range is only a simplification of these two ranges combined. It shouldn't matter too much either way.
Original post by Redcoats
The question said find the 'ranges' (plural). Would that mean anything here?


I know I definitely got one of the ranges wrong anyway but the way the question was phrased made me think it would be two separate ones
(edited 5 years ago)
Original post by mupsman2312
Well, we all would have put both ranges in our working, so they should see that anyway. Just giving it as one range is only a simplification of these two ranges combined.


Agreed, you'll still get the full marks. I was wondering if the simplification was necessary and if one would be penalised for merely giving two separate ranges. Thoughts?
Original post by Redcoats
Agreed, you'll still get the full marks. I was wondering if the simplification was necessary and if one would be penalised for merely giving two separate ranges. Thoughts?


It may not have been necessary, so don't worry about it. If they're REALLY fussy, then you might drop one mark, but the main part of the question would just have been to find the two ranges, so they shouldn't mind too much about how you write them.
Reply 28
Original post by Nyathigiz
Ye


Because it was consistent unlike s2, and because for s1 n has to be at least 2 and so was smaller also? (S2 had 1/11 as its fraction of theta)
Reply 29
Wtf did we have to do for the last 2 parts of the glue question? I didn't manage to do much with those questions. I wrote no, I tried finding the new s^2 and mean of the glue that had the errors. Also, what did everyone do for the last question, for a and b, and for showing the minimum variance was a given fraction. I think my a's and b's (had 2 values for each were 1 and -1 and the other set were 2 fractions, 11 was in the number)
Reply 30
Original post by Evil Homer
If you guys want to set up an unofficial markscheme I could promote it around the site and get it filled in as soon as possible :smile:

Hope the exams went okay today! :biggrin:


Please :smile:
Guys do you think 55/75 would be an A this year ?
So hopeful that I would get an A* this year after FP2 and S3 but this S4 exam dashed all hopes . What can you do that’s life I suppose .
My ranges were m less than 0.37 and m is greater or equal to 0.04 and some long decimals after that.
Reply 34
Original post by mupsman2312
These are some of the answers which I remember getting:

P(Type I error) = 0.05
Minimum value of n is 35

E(X) = (4/3)theta
E(X^2) = 2theta^2 and Var(X) = (2/9)theta^2
Var(S1) = (theta^2)/(8n)
a = 6/11 and b = 9/11 (4a + b = 3)
S1 is the better estimator

Differences are normally distributed
0.0490 &lt;= m &lt;= 0.3774

Confidence interval for mean: [47.93, 56.07]
Confidence interval for sd: [3.57, 10.14]
Don't change to Zesty, as 3.57 &lt; 5.5 (&lt; 10.14) - I.e. Jeremiah can't be sure that sigma &lt; 5.5

F-test: 4.53 &lt; 5.19 ==&gt; not significant
Pooled estimate of variance = 0.24698...
t-test: 2.825 &gt; 1.833 ==&gt; significant

Q1: 3.333 &gt; 2.821 ==&gt; significant

Can anyone please confirm? Sorry for going in such a weird - almost reverse - order!


YESSS. THIS IS ALL GOOD. Although I think I got slightly (Ever so slightly different a's and b's, i have that same equation, but also a quadratic for a and b) there was an 11 in there though.
Reply 35
Original post by Lifeofwind88
Guys do you think 55/75 would be an A this year ?


Unlikely to be that low because of the calibre of students who do S4, but I hadn't gotten less than like 68/75 in all the past papers I did and I'm pretty sure I lost around 7-8 marks. I'm thinking this gives me a comfortable A* though. Perhaps 61 for an A?
Original post by dugdg
Unlikely to be that low because of the calibre of students who do S4, but I hadn't gotten less than like 68/75 in all the past papers I did and I'm pretty sure I lost around 7-8 marks. I'm thinking this gives me a comfortable A* though. Perhaps 61 for an A?



As long as I get into UCL I’m happy , best of luck to you man as well
Reply 37
@mupsman2312 There were two separate ranges, and they couldn't be combined. a < m < b and c < m < d where c > b, in other words, there's a bunch of values there in between which m can not be.

Would you mind showing me how you worked out the value for n? I got a really weird answer.
Original post by CornOnTheCog
I think I got like 35?


me too
Original post by EdwardG
@mupsman2312 There were two separate ranges, and they couldn't be combined. a < m < b and c < m < d where c > b, in other words, there's a bunch of values there in between which m can not be.

Would you mind showing me how you worked out the value for n? I got a really weird answer.


Oh, really? I only did a one-tailed inequality for one of the significance levels (I kept on making mistakes and having to correct stuff on that question, so I wouldn't have had time to do the other tail). I had something like 0.0490 < m < 2-point-something and 0.3774 <= m, so my ranges combined quite nicely. I did keep on getting all of the signs around the wrong way, however (I was getting mixed-up which test was significant and what it meant), so chances are that I made some sort of error to end-up with this.

[EDIT: Wait, didn't the value of m have to satisfy both ranges simultaneously? So, if the two ranges didn't overlap, then no values of m would be possible, would they?]


For the question with finding the minimum value of n, I started by finding the 5% one-tail critical value of X-bar in terms of n:

x-bar = 150 + 1.6449{4/sqrt(n)} = 150 + 6.5796/sqrt(n).

Then, using mu = 152, the probability of a type II error was:

P[X-bar < 150 + 6.5796/sqrt(n) | X-bar ~ N{152, 4/sqrt(n)}] < 0.1 (the 0.1 was given in the question).

After standardising, I had:

P[Z < 1.6449 - sqrt(n)/2] < 0.1
From tables: 1.6449 - sqrt(n)/2 < -1.2816
sqrt(n)/2 > 2.9265
sqrt(n) > 5.853
n > (5.853)^2 = 34.257609

But, n is an integer, so n >= 35. Thus, the minimum possible value of n is 35.
(edited 5 years ago)

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