Competition - post a photo you've taken of nature >>

2017 mei s3

Poll

Grade boundary prediction

Acey110

So how did you all find it? Found it pretty straightforward how did you all do the last question?

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Reply 1

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.

Reply 2

Acey110

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?

Reply 3

lilrabbits

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

(edited 6 years ago)

Reply 4

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?

Reply 5

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?

Reply 6

Acey110

What did you put for the central limit theorem

Reply 7

Casio123

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

Reply 8

jn998

Original post by Casio123

I got the same as you apart from the first critical value in Q1 which I got to be 2.131

Reply 9

Acey110

Original post by Casio123

I got same as him

How did you get it?

Reply 10

Casio123

Original post by Acey110

What did you put for the central limit theorem

It was the 0.00921 one

Reply 11

Acey110

Original post by Casio123

It was the 0.00921 one

I mean the reason it

Reply 12

Casio123

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

Reply 13

Casio123

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

Reply 14

Acey110

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 :/

Reply 15

Casio123

Original post by Acey110

Yeah makes sense, I wrote so you could approximate it to the normal :/

Your answer makes sense to me too

Reply 16

Casio123

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

Reply 17

Ellis.B.Redding

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