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# Edexcel S3 - Wednesday 25th May AM 2016 watch

1. (Original post by anndz3007)
Chi square for the doctor question ?
Do you remember what the degrees of freedom were etc?
2. What did people put for q1?

For 1a I put if the sample required is large, and if the population divides naturally into mutually exclusive strata

For 1b, I put if sampling errors need to be estimated (can't be done for quota as its non-random), and if the characteristics being sampled are hard to judge
3. Messed up a few questions I should have got right which is annoying. Somehow didn't get the last q
4. (Original post by Armpits)
Do you remember what the degrees of freedom were etc?
2 i think, and s.l is 10%
5. Answer for part c lung que?

Posted from TSR Mobile

1a) More representative, less chance of bias
1b) it's random, unlikely to have interview bias

2) 5.614>4.605 so reject H0

3a) Want to check correlation not linear ability
3b) 0.75
3c) 0.75>0.7143 so reject H0
3D) 0.65<0.6694 so accept H0
3e) positively correlation but non-linear

4a) 0.7794
4b) X-N(760, 302.25)
4c) 0.0107

5a) 1.954>1.6449 so reject H0
5b) CLT, means approx normal, bar sample = var population
5c) 12.7

6a) 4.1<5.991 so binomial suitable
6b) r= 26.78, s=16.07
6c) 14.65>11.345 so poison not suitable

7a) [19.15,19.85]
7b) reduce 20% value as outside interval
7c) n=44
7. (Original post by arks_007)
Answer for part c lung que?

Posted from TSR Mobile
Remind me what it was roughly?
8. For the Po question a lot of people seem to have assumed that lambda is 1.8 by taking the mean of the sample. I personally took the first expected value and equated it to e^-lambda, getting lambda as being 1.78. Not a major difference but it means I didn't subtract an additional degree of freedom in my Chi squared test. Any opinions?
9. (Original post by lai812matthew)
got question 1 wrong :/
i only list the advantages of stratified to simple random sampling at a and quota at b will i get any marks......
10. (Original post by ZeusOfScience)
For the Po question a lot of people seem to have assumed that lambda is 1.8 by taking the mean of the sample. I personally took the first expected value and equated it to e^-lambda, getting lambda as being 1.78. Not a major difference but it means I didn't subtract an additional degree of freedom in my Chi squared test. Any opinions?
Yeah I did the same, got lambda as a natural log and used that. I did subtract a degree of freedom for it though, as it was calculated from the data afterall?
11. My answers agree with yours - so i am liking that!!!!
(Original post by SHJBHB)

1a) More representative, less chance of bias
1b) it's random, unlikely to have interview bias

2) 5.614>4.605 so reject H0

3a) Want to check correlation not linear ability
3b) 0.75
3c) 0.75>0.7143 so reject H0
3D) 0.65<0.6694 so accept H0
3e) positively correlation but non-linear

4a) 0.7794
4b) X-N(760, 302.25)
4c) 0.0107

5a) 1.954>1.6449 so reject H0
5b) CLT, means approx normal, bar sample = var population
5c) 12.7

6a) 4.1<5.991 so binomial suitable
6b) r= 26.78, s=16.07
6c) 14.65>11.345 so poison not suitable

7a) [19.15,19.85]
7b) reduce 20% value as outside interval
7c) n=44
12. (Original post by SHJBHB)

1a) More representative, less chance of bias
1b) it's random, unlikely to have interview bias

2) 5.614>4.605 so reject H0

3a) Want to check correlation not linear ability
3b) 0.75
3c) 0.75>0.7143 so reject H0
3D) 0.65<0.6694 so accept H0
3e) positively correlation but non-linear

4a) 0.7794
4b) X-N(760, 302.25)
4c) 0.0107

5a) 1.954>1.6449 so reject H0
5b) CLT, means approx normal, bar sample = var population
5c) 12.7

6a) 4.1<5.991 so binomial suitable
6b) r= 26.78, s=16.07
6c) 14.65>11.345 so poison not suitable

7a) [19.15,19.85]
7b) reduce 20% value as outside interval
7c) n=44
Got same for all. Im safe now
13. (Original post by SHJBHB)

1a) More representative, less chance of bias
1b) it's random, unlikely to have interview bias

2) 5.614>4.605 so reject H0

3a) Want to check correlation not linear ability
3b) 0.75
3c) 0.75>0.7143 so reject H0
3D) 0.65<0.6694 so accept H0
3e) positively correlation but non-linear

4a) 0.7794
4b) X-N(760, 302.25)
4c) 0.0107

5a) 1.954>1.6449 so reject H0
5b) CLT, means approx normal, bar sample = var population
5c) 12.7

6a) 4.1<5.991 so binomial suitable
6b) r= 26.78, s=16.07
6c) 14.65>11.345 so poison not suitable

7a) [19.15,19.85]
7b) reduce 20% value as outside interval
7c) n=44
1a) Gives fair representation of strata, more suitable for larger samples
1b)Can estimate sampling error, no interview bias

2)agree.

3)a)When data is already ranked or variables are not jointly normally distributed
b)agree
c)agree
d)not sure

4)agree

5a)agree
5b) Not sure you 'assume' CLT but this might be correct, I answered that the people were selected independently.
5c) can't remember

6)agree

7)agree
14. (Original post by cjlh)
Yeah I did the same, got lambda as a natural log and used that. I did subtract a degree of freedom for it though, as it was calculated from the data afterall?
We can't assume they calculated their lambda from the data, they could've completely made up 1.78 - a very poorly phrased question
15. (Original post by SHJBHB)

1a) More representative, less chance of bias
1b) it's random, unlikely to have interview bias

2) 5.614>4.605 so reject H0

3a) Want to check correlation not linear ability
3b) 0.75
3c) 0.75>0.7143 so reject H0
3D) 0.65<0.6694 so accept H0
3e) positively correlation but non-linear

4a) 0.7794
4b) X-N(760, 302.25)
4c) 0.0107

5a) 1.954>1.6449 so reject H0
5b) CLT, means approx normal, bar sample = var population
5c) 12.7

6a) 4.1<5.991 so binomial suitable
6b) r= 26.78, s=16.07
6c) 14.65>11.345 so poison not suitable

7a) [19.15,19.85]
7b) reduce 20% value as outside interval
7c) n=44
I got virtually the same for the calculations and for the worded questions:

1a) More representative as reflects population structure, gives estimates for individual strata
1b) Random process so able to estimate sampling errors, interviewer could introduce bias in quota sampling due to choice of sample

3a) Use Spearman's when the two variables cannot be assumed to be normally distributed / when the data is already ranked

5b) Assumptions are sample variance is equal to population variance and that the means (I used Xbar and Ybar) are independent

7b) 20% is above (I believe they didn't allow 'outside' in one of the past papers I'd seen) confidence interval so he should recommend them to lower their stated fat content
16. (Original post by cjlh)
Yeah I did the same, got lambda as a natural log and used that. I did subtract a degree of freedom for it though, as it was calculated from the data afterall?
I did this initially, however estimated 1.8 correctly afterwards. I believe the 1.8 answer is probably correct (it is nicer). In any case you should have subtracted an extra degree of freedom as the mean has still been calculated from observed data.
17. Hella pissed I didn't say sample size is large for 1a.
Can you still get anything if you write along the lines of its results best reflecting the population as proportion of stratum in sample equals proportion of that strata in population. And for point 2 stating population divides into mutually exclusive groups ??
18. (Original post by Mav1)
1a) Gives fair representation of strata, more suitable for larger samples
1b)Can estimate sampling error, no interview bias

2)agree.

3)a)When data is already ranked or variables are not jointly normally distributed
b)agree
c)agree
d)not sure

4)agree

5a)agree
5b) Not sure you 'assume' CLT but this might be correct, I answered that the people were selected independently.
5c) can't remember

6)agree

7)agree
Yeah, my definitions aren't brilliant but I'm pretty sure all my maths is right
19. (Original post by ZeusOfScience)
We can't assume they calculated their lambda from the data, they could've completely made up 1.78 - a very poorly phrased question
Very good point, I think it was poorly worded too. Oh well, suppose we'll have to see haha. Hopefully just one lost mark in this case
20. My answers - I thought t was fairly standard

1a) More representative, individual strata estimates availiable
1b) random process, no interviewer bias

2) 5.61 and I think you rejected

3a) When the data isn't bivariate noral
b) 0.75 and I think you rejected
You accepted the pmcc question
e) You HAD TO say about ranks - days where sales of ice cream ranked highly, ranked highly for sunglasses

4a) 0.7794
b) 760,302.25
c) 0.0107

5a) 1.95 and you rejected
b) Var of sample = var of population. Samples are independent of each other. CANT MENTION CLT - as its not an assumption, it's a thorem
c) tricky - 12.7

6a) Test statistic was 4.1 and you accepted
b) r= 26.78 s= 16.07
c) Test statistic was 14.65 you reject

7a) 19.15%,19.85%
b) Paul shoud tell them to lover stated weight
c) n=44

Who agrees?

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