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    Hi guys.

    Can anyone take a look at question 2b.

    The wording is throwing me a bit. According to the mark scheme the answer to part b is just the actual significance of the test, which i correctly calculated as 0.0432 or 4.32%.

    However the significance of the test is 5% (2.5 at each tail). So if the actual significance is less than 5% surely you cannot incorrectly reject H(0). I would’ve thought that you could only incorrectly accept H(0).

    I would appreciate it if someone could clarify the question for me here.

    Paper-[link]http://pmt.physicsandmathstutor.com/download/Maths/A-level/S2/Papers-Edexcel/June 2012 QP - S2 Edexcel.pdf[/link]

    Markscheme-[link]http://pmt.physicsandmathstutor.com/download/Maths/A-level/S2/Papers-Edexcel/June 2012 MS - S2 Edexcel.pdf[/link]

    Thanks in advance.
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    notes for 2b

    M1 – adding their two critical regions’ probabilities together or may be awarded for awrt 0.0432 If they add their critical regions’ probabilities and then go on and get a different probability as their answer then it is M0A0 e.g. 0.0216 + 0.0216 = 0.0432 then 0.05 – 0.0432 = 0.0068 gets M0 A0 e.g. 0.0216 + 0.0216 = 0.0432 < 0.05 reject H0 gets M1 A1 e.g. 0.0216 + 0.0216 = 0.0432 so probability of rejecting H0 is 1 – 0.0432 = 0.9568 gets M0 A0
 
 
 
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