Hello Can anyone help with these? Q&A Bank part 4 Question 4.12 I'm not clear why 5 cards have to be drawn to demonstrate the existence of ESP at the 5% level. As the probability of 4 or less cards is 0.9672 wouldn’t 4 cards be sufficient for a one-sided test? Q&A Bank part 4 Question 4.15 I calculate the chi square test statistic to be 23.6 rather than 32.0 as given in the answers. Where am I going wrong? Thanks Simon
sorry my ct3 notes are at home so I can't try the problems now (currently at work) I looked at the corrections: http://www.acted.co.uk/html/paper_corrections.htm but couldn't find a reference to your questions....
i)I think you need 5 cards correct. As the probability of four cards or less is 0.9672, the probability of at least 5 cards is less then 0.05%, therefore we reject H0. The probability of at least 4 cards correct is greater then 0.05%, so 4 cards isn't sufficient. ii)I get 23.6 too... are we missing something?
Thanks for that. Looks like there is a misprint in the solutions to Q4.15. I think I see what you are saying re Q4.12. Realising that we should be comparing probabilties of the type "X cards or more" against 5% was the difficult bit for me. If I have understood correctly: P(four cards or less correct) = 0.9672. Therefore P(five cards or more correct) = 0.0328. As this is less than 5% we can reject H0. However P(three cards or less correct) = 0.8791. Therefore P(four cards or more correct) = 0.1209. As this is more than 5% we do not reject H0. Therefore 5 cards is the minimum number that will enable us to reject H0 as it is the lowest number of cards that takes us within the 5% region.
