Official Thread: OCR MEI S2/M1

When using the normal tables with a z value to 4+ significant figures do you guys take an average from the 'add' column since my values always seem to be 1 out in the 4th decimal place but we were never taught to take an average.

you dont have to do that!!! the mark scheme always says correct answer to 3sf!!

Do probabilities to 4.d.p and other values to 3 or 4 significant figures depending on the value.

For example 0.487548375384 should be 0.4875
125.3258 should be 125.3
83.2142352 should be 83.2 or 83.21 depending on which you prefer, both get the marks.

Just never go beyond 4 s.f for solutions. However remember to never use rounded values in your calculations.

hey can anyone give me an explanation on how to interpret contributions to the x^2 statistic? confused...

Pls help. WHat definitions and worded answers do we need to know of for s2

THe x^2 contribution is a way of quanitvativly diplaying the difference between the result observed and the result expected. Thus the larger the discrepency between the two the larger the contribution. If the sum of the contributions are larger than the x^2 critical value for your test then the result is significant.

Furmulae is (O-E)^2 / E where O is observed value and E is expected.

Do that for each cell and sum the total to find the x^2 contrivution value!

Hope that helps mate and just ask if need any help with anything else

When we calculate contributions do we use the rounded expected value e.g 10.46 or the exact value 136/13

S2 is really easy if you do past papers.

If you don't, you're screwed.

Its such an anal unit. If you dont hit that buzzword you lose marks, adn the boundires are usually v high. Its v v annoyinh although not very hard to get your head around

There are many approaches, but here's how I do these:

Since we need to find a range (a<x<b) such that 90% (p=0.9) of the data points fit within that range, I like to go for a symmetrical range.

Therefore the range from p=0.05 to p=0.95
In this range, 90% of the data is here.

Therefore, we have two things to solve:
1) P( x < a) = 0.05 (we want to find the value of a for which only 5% of the data is under it)
Solving this is pretty simple, simply being:
P( z < [a-mean]/s.d) = 0.05
And therefore [a-mean]/s.d = inverse-phi(0.05)
= -inverse-phi(0.95)
= -1.645
Then you can just solve a, since you would've been given the mean and the standard deviation in the question

2) P(x > b) = 0.05 (we want to find the value of b for which only 5% of the data is over it)
Solve the same as normal.

This will give two values (a and b) such that they collectively contain 90% (0.9) of the data. So say for example we get a = 135 and b=165
It would be true that P(135 < x < 165) = 0.9
Therefore 90% of the data is between 135 and 165

Hope this helped!