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s3 help

for the june 2002 paper q6, why is t=43.76 and not 43.74? 43.76 doesn't make any sense as the expected frequencies won't add to 200?
thanks


http://www.thestudentroom.co.uk/attachment.php?attachmentid=86889&d=1276342376
(edited 9 years ago)
Original post by MrB24
for the june 2002 paper q6, why is t=43.76 and not 43.74? 43.76 doesn't make any sense as the expected frequencies won't add to 200?
thanks


http://www.thestudentroom.co.uk/attachment.php?attachmentid=86889&d=1276342376


Can you post the question please?
Reply 2
Avatar for MrB24
MrB24
OP
3
.Data were collected on the number of female puppies born in 200 litters of size 8. It was decided to test whether or not a binomial model with parameters n = 8 and p = 0.5 is a suitable model for these data. The following table shows the observed frequencies and the expected frequencies, to 2 decimal places, obtained in order to carry out this test.

Number of females
Observed number of litters
Expected number of litters
0
1
0.78
1
9
6.25
2
27
21.88
3
46
R
4
49
S
5
35
T
6
26
21.88
7
5
6.25
8
2
0.78



(a) Find the values of R, S and T.
(4)
Original post by rayquaza17
Can you post the question please?
Original post by MrB24
for the june 2002 paper q6, why is t=43.76 and not 43.74? 43.76 doesn't make any sense as the expected frequencies won't add to 200?
thanks


http://www.thestudentroom.co.uk/attachment.php?attachmentid=86889&d=1276342376



How did you get 43.74?

When I do it on my calculator, I put in: 200*(8C5)*(0.5)^8 which gives me the 43.75 the mark scheme says.

I'm not 100% sure how you calculate it using the tables (I only know the calculator method), but I'm assuming you do the cumulative probability for x=5 minus the probability for x=4 and times this by 200, and this gives me 200(0.8555 -0.6367)=43.76

However I see what you mean about them not adding up to 200. It's only like 0.01 difference so it won't make a massive difference to the test.
(edited 9 years ago)
Reply 4
Avatar for MrB24
MrB24
OP
3
Original post by rayquaza17
How did you get 43.74?

When I do it on my calculator, I put in: 200*(8C5)*(0.5)^8 which gives me the 43.75 the mark scheme says.

I'm not 100% sure how you calculate it using the tables (I only know the calculator method), but I'm assuming you do the cumulative probability for x=5 minus the probability for x=4 and times this by 200, and this gives me 200(0.8555 -0.6367)=43.76


i did 200-(sum of other expected frequencies)
i got R and S by use of the tables, and in the mark scheme they said my answers for those are right (43.76 and 54.68)
Original post by MrB24
i did 200-(sum of other expected frequencies)
i got R and S by use of the tables, and in the mark scheme they said my answers for those are right (43.76 and 54.68)


Using 200-(sum of other expected frequencies) is fine to do, but tbh I think it is quicker just to work out out normally in this case!

I think you should get the mark for doing that still. The mark scheme you are looking at says "provisional mark scheme", so maybe they allowed your method in the real mark scheme? I don't know, I didn't do S3, but in my university exam on this, you would have gotten full marks for doing it your way!!
Reply 6
Avatar for MrB24
MrB24
OP
3
Original post by rayquaza17
Using 200-(sum of other expected frequencies) is fine to do, but tbh I think it is quicker just to work out out normally in this case!

I think you should get the mark for doing that still. The mark scheme you are looking at says "provisional mark scheme", so maybe they allowed your method in the real mark scheme? I don't know, I didn't do S3, but in my university exam on this, you would have gotten full marks for doing it your way!!


ohh cool thanks for ur help

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