Q&A Bank 4 question number 4.15

Discussion in 'CT3' started by Hemant Rupani, Sep 18, 2014.

  1. Hemant Rupani

    Hemant Rupani Senior Member

    Can anybody explain this question in layman language?
    I'll try to find solution but now need to understand question :(
     
  2. C2H6O

    C2H6O Member

    In part (i) we are testing

    Ho : p = 0.2 (Interviewee doesn't have ESP, she is like any normal person) VS
    H1 : p \(\neq\) 0.2 (Interviewee has ESP and can receive answers from audience through eye contact or whatever!)

    Now she failed to identify any of the 10 patterns and claimed that audience transmitted wrong answers to her.

    But under Ho,
    P( getting 0 correct answers out of 10) = 0.8 ^10 = 0.107
    which is greater than 5% (significance level)

    This is by no means convincing enough to support her claim (Alternative hypothesis)
    Any normal person can except to see results like this 10.7% of the times.

    Final wording: Fail to reject Ho at 5% level of significance. There is not sufficient evidence to support the claim that she has ESP powers.

    NOTE: P-value would be "2 * 0.107" because this is a two-tailed test.

    Part (ii) will be easy now. If you still have problems, feel free to ask.
     
  3. Hemant Rupani

    Hemant Rupani Senior Member

    Ohhh! that means ESP of audience is also considered
    Thank you C2H6O it makes sense now!
    But I doubt on your NOTE: (as this is two-tailed test is it necessary that p value is distributed half-half both sides like in normal distribution? I mean probability distribution is right tailed so.)

    And
    "State precisely the hypotheses that the interviewer could have specified before the experiment to prevent the interviewee from “cheating” in this way"
    I didn't get how interviewee can cheat.?
     
  4. C2H6O

    C2H6O Member

    Sorry my mistake. P-value will not be 2*0.107 because distribution is not symmetric. Binomial distribution is perfectly symmetric when p = 0.5 which is not in this case.

    I just performed this test in a software and it's showing me a p-value of 0.14
    Not sure how it calculated! But you need not worry about this because it's beyond the course.

    Cheating is what she did in this question! She put the blame on audience that they transmitted wrong answers to her.

    So to prevent this, interviewer could've specified the alternative hypothesis as
    "p > 0.2" before the experiment so that there's no rejection region on the left side of mean. Now in order to justify her claim, she has to give at least 5 correct answers.
     
    Last edited by a moderator: Sep 19, 2014
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  5. Hemant Rupani

    Hemant Rupani Senior Member

    Hi C2H6O, Suggest me a software to study myself ?
    Thank you! All doubts cleared. :)
     
  6. John Lee

    John Lee ActEd Tutor Staff Member

    You know I'm going to say "CT3 online classroom". Sample units on YouTube!

    So guess there's no point saying it.... ;)
     
  7. salj67

    salj67 Member

    Why is p=0.2?
     
  8. John Lee

    John Lee ActEd Tutor Staff Member

    There are five patterns, so the chance of choosing the correct pattern is 1/5 = 0.2.
     

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