i think there were many ways to prove this but the main message is that it was incorrect anyway if i only lost most my marks on the last question then i should still get an A, any grade predictions?
I predicted that 64 would be an A, but it seems that others agree on 60.
Btw did you mean to say the claim was incorrect or was that a typo?
How did everyone do the question where the patients were selected byt the wernt in the origional 500 sample
I used the table to see the values for the appropriate parts e.g between 41-66 (or whatever it was) then put this as the numerator and 500 as the denominator.
After that there were conditional probabilities, meaning that you need to take into account the "given that".
For example, if you are looking at an age range given that the band was 0, you need to use the total for band 0 as the denominator and the appropriate values on the table as the numerator.
The final part of that question was a random pick style question, meaning that the denominator reduces by 1 for each selection and you had to adjust your values for the numerator appropriately. Then you needed to times the probability you acquired by the number of ways that the value could be obtained, which many on here are saying was 6 so 3!, and then you have your answer.
yes but 166.67 times by 1.2 = 200.004 which is greater than 200 and so should not be counted
But if you round anywhere it makes you count 166.67: if you round 166.666... you get 166.67, or if you round 200.004 to the nearest cent you get €200 :/
What about question 3b? I got 0.007.... or something less than 0.01.
And the last question about the claims? The first claim is valid or not?? I just directly multiplied the c.i by 1.20 and said it is valid because 400 is within the new c.i. But some students calculate the new c.i using the mean+-standard error. The second claim said at most 25% and i got 17.5%, so I said it is invalid because it can not get to 25% as the highest is 17.5%.
I used the table to see the values for the appropriate parts e.g between 41-66 (or whatever it was) then put this as the numerator and 500 as the denominator.
After that there were conditional probabilities, meaning that you need to take into account the "given that".
For example, if you are looking at an age range given that the band was 0, you need to use the total for band 0 as the denominator and the appropriate values on the table as the numerator.
The final part of that question was a random pick style question, meaning that the denominator reduces by 1 for each selection and you had to adjust your values for the numerator appropriately. Then you needed to times the probability you acquired by the number of ways that the value could be obtained, which many on here are saying was 6 so 3!, and then you have your answer.
Apologies if that wasn't clearly explained.
I think you had to divide by 125 so that the denominator was 4 because only 4 of them picked, not 500....i think it was n/4 x n/3 x n/2 x n/1
But if you round anywhere it makes you count 166.67: if you round 166.666... you get 166.67, or if you round 200.004 to the nearest cent you get €200 :/
nah if u round 166.666 to 166.67 then this is shoudlnt be counted as it said less than but u could agrue that 200.01 rounds to 200 and so i dont know what the ms would be
Both were valid, for part i you had to convert the confidence interval (which was in £) into euros, after doing this 400euros lay within the new CI.
yeah i know , i did this but then changed my answer for the CI will i get method marks as i did convert to euros an stuff, my CI was just wrong, also will i get ECF marks for my claim? if not how mnay marks will i loose
For the one with 166.67 they should accept both if they aren't being *******s about it. I'm sure I got the first one wrong though since I just converted the mean into euros which was 381 and then said it was therefore wrong. ;(
I can't remember the exact question, so I will try to explain as best as I can.
There were 4 outcomes, two pairs of which were the same. As you were estimating from the sample, you divide each outcome from the total in the sample (500). Multiply them together, then multiply by 6 as this was the number of ways of arranging them.
This came down to the following calculation:
{[(176-28)/500]^2} x {[(140-6-3)/500]^2} x 6 = 0.030786 = 0.0308
Sorry this is a general answer and not specific, however I have forgotton most of the wording of the question