Poll: Grade boundary prediction
65 (5)
35.71%
64 (3)
21.43%
63 (2)
14.29%
62 (1)
7.14%
(3)
21.43%
Acey110
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#1
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#1
So how did you all find it? Found it pretty straightforward how did you all do the last question?
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Casio123
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#2
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#2
I got 3.18... some people got 3.5-3.18 =0.32. Not sure whether it was the discount or the discounted price.
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Acey110
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#3
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#3
(Original post by Casio123)
I got 3.18... some people got 3.5-3.18 =0.32. Not sure whether it was the discount or the discounted price.
Yeah same I got 3.18 not sure either, we'll have only lost a mark ah well, any other tricky questions?
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lilrabbits
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#4
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So the values in my calculator are as follows:
Test statistic for the t-test? 1.539
Confidence interval (29.7115,...)
Test statistic for the Wilcoxon paired test: W=15, v=8-1=7
Test statistic for the goodness of fit: 12.6
P(mean of x>4)=0.00921
0.3165
0.2621
0.3629
Mean for D 4.68w
S.d. For D 0.2219w
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Acey110
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#5
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Yeah I agree with all of the above except wasn't the last one 0.212w, root(0.067^2x10) . And if I left the confidence interval as a bracket like (x,y) would I get all the marks?
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Casio123
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#6
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Nope, found it alright.
Here are my answers (from my calc) except for the worded ones
1. T- test test statistic = 0.75/1.9494/4 =1.5390
Cv= 1.753 so insignificant, reject h1.

Confidence interval 30.75 +- 2.131 x square root of 3.8/16

Wilcoxin test statistic 15 cv 5 so insignificant

2. Chi square sum = 12.6
Cv = 12.02 (v=7 10%)

3. Prob (x>4) = 0.00921
A bunch of proves and CLT

4. Prob (x>0.5) = 0.3165

Prob of mackerel smaller than trout = 0.2621

Prob of one macker plus two trout > 5 pounds = 0.3629

W=3.18

I have the working if you want it as well.

Do any of these numbers look familiar to you?
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Acey110
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#7
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#7
What did you put for the central limit theorem
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Casio123
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#8
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(Original post by Acey110)
Yeah I agree with all of the above except wasn't the last one 0.212w, root(0.067^2x10) . And if I left the confidence interval as a bracket like (x,y) would I get all the marks?
I got same as him
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jn998
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#9
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(Original post by Casio123)
Nope, found it alright.
Here are my answers (from my calc) except for the worded ones
1. T- test test statistic = 0.75/1.9494/4 =1.5390
Cv= 1.753 so insignificant, reject h1.

Confidence interval 30.75 +- 2.131 x square root of 3.8/16

Wilcoxin test statistic 15 cv 5 so insignificant

2. Chi square sum = 12.6
Cv = 12.02 (v=7 10%)

3. Prob (x>4) = 0.00921
A bunch of proves and CLT

4. Prob (x>0.5) = 0.3165

Prob of mackerel smaller than trout = 0.2621

Prob of one macker plus two trout > 5 pounds = 0.3629

W=3.18

I have the working if you want it as well.

Do any of these numbers look familiar to you?
I got the same as you apart from the first critical value in Q1 which I got to be 2.131
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Acey110
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#10
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(Original post by Casio123)
I got same as him
How did you get it?
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Casio123
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#11
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#11
(Original post by Acey110)
What did you put for the central limit theorem
It was the 0.00921 one
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Acey110
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#12
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#12
(Original post by Casio123)
It was the 0.00921 one
I mean the reason it
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Casio123
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#13
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(Original post by Acey110)
How did you get it?
I did 10x0.067^2 x w^2 for the variance

Square root that to get 0.2119
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Casio123
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#14
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(Original post by Acey110)
I mean the reason it
I said n is large enough to assume that the mean of the population is same as the mean of the sample

Don't know if that's right
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Acey110
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#15
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(Original post by Casio123)
I said n is large enough to assume that the mean of the population is same as the mean of the sample

Don't know if that's right
Yeah makes sense, I wrote so you could approximate it to the normal :/
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Casio123
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#16
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(Original post by Acey110)
Yeah makes sense, I wrote so you could approximate it to the normal :/
Your answer makes sense to me too
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Casio123
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#17
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(Original post by jn998)
I got the same as you apart from the first critical value in Q1 which I got to be 2.131
Wasn't it one tail tho? So you have to take 10% instead of 5... not sure
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Ellis.B.Redding
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#18
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#18
so my values are as follows:
test statistic = 1.5894 (may be slightly off) < critical value = 1.753 (not significant)
Interval [29.74,31.76]
W = 15 > Critical value = 5 (not significant)
Chi-squared = 12.6 > 12.02 (significant)
mode = 4
E(X) = 3.6
0.00921
0.3165
0.2621
0.3629
4.68w 0.212w
w = £3.05 (may be wrong)
i think some of my answers will have a small error becauseif the difference between calculator and table values
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jn998
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#19
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#19
(Original post by Casio123)
Wasn't it one tail tho? So you have to take 10% instead of 5... not sure
Damn, you're right. I forgot about that. How many out of the 10 do you think I'd lose on that question?
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Casio123
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#20
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#20
(Original post by jn998)
Damn, you're right. I forgot about that. How many out of the 10 do you think I'd lose on that question?
3 marks according to previous mark schemes
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