Edexcel Mathematics S3 23rd May 2018 - Unofficial Mark Scheme & Discussion

Thanks for finding that - so it is find average rank and then PMCC? And don't use the Spearman formula at all?

1.96=1.25/(4.8/root(n)) gets you 57. The z value is 1.96 because you're looking at half of p=0.05.

Ok thanks sorry so its because the questions specifically says this confidence interval and the probability of the mean being outside is 10%?

Pretty much, yeah. The tool you've been given instead of a statistic to carry out this test is the confidence interval, and its properties determine your significance level.

can someone explain how to do last part???

https://i.imgur.com/OtdurXS.jpg

Thank you!!
Sorryyy last q this june 15 papers really messing with me aha do you think you could explain why 5d is done like that, I follow that you cannot use the variance formula like usual because they are not independent (like you do for part ii)a )but I dont really understand what they are doing and why, I would have just thought it meant you couldn't work out the variance?

Because they're all based on U, you have to make your U₁ into 5U₁/5 and combine the fractions into one term. This is not something you could do when your samples were taken from 2 different populations.

Is this anywhere in the textbook? I dont remember learning this
So if they are from the same population you find the variance a different way by combining them and then use the variance formula despite them not being independent?

He's combined the male and female data into just one row, so "people who reported their weight as being..." one of the three columns. He wants to know whether the three body types are chose equally - by this he means whether they follow a discrete uniform distribution, where the expected value is 50 respondents for each category. Calculate x-squared, then check in the tables in the formula booklet for the smallest sig level % given your 2 degrees of freedom for which the chi-squared value is smaller than x-squared, such that H1 is accepted.

It's kind of the stuff in the textbook taken to a higher conceptual level, it is really tricky stuff.

Well the reason you combine them to one fraction is so you have the correct fraction to square when finding variance. Once you find that one fraction, you combine them to find E(whatever) and Var(whatever) normally - it is an independent sample, it's just got a different probability to the other independent sample since that one was from 2 different populations, which is why you need to calculate almost the same thing again.

i did that before but isnt the H0 that a uniform distribution is suitable for this model. and H1 says it isn't suitable?
how is this an association hypothesis? wouldn't we use r-1 c-1 but then that would give 0 which isn't possible. sigh...........

Ohhh that makes sense your making it into one statistic so that its from one population and solve normally from there? Is it just me or is that paper on such a different level to the others

Yeah, essentially. I agree, it's a nasty paper. Lower grade boundaries though!

https://i.imgur.com/sM58oAU.jpg
for my confidence interval is
(-1.06,9.06)
how do I convert this into the time?
part b(I)

For a hypothesis test where we're testing whether a binomial distribution with p= (e.g.) 0.3 is a suitable model, should you give the p value in the hypothesis or not? In some mark schemes you lose a mark for giving the p value whilst in other papers you lose a mark if you do not (e.g. June 2016 question 6). Thank you!

Depends on what we're testing.

If we're testing whether a binomial distribution with p = 0.3 is a suitable model, then yes, of course you should give the p value in the hypothesis because this is what we're testing! (example, June 2016 6a: penalise omission of p = 0.3)

However if we're testing whether a binomial distribution is a suitable model, then no, because we're not testing a specific p, just Binomial - after we have assumed H0 we use the data to calculate p and then use this to find our expected values. (example, June 2014 5d: penalise p = 0.6)

Hope this helps