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T-Distribution and confidence interval)

  1. Dec 10, 2015 #1
    I have a question I've been working on and I've noticed that my solution differs from the textbook solution. I would have considered it a mistake, but the next 4 questions are similar with the same solution "mistake"?

    Aside from not converting the % to 0.885, I am wondering about part b. It says to cut the confidence interval to one third; in the solution they cut it to 2, why is that?

    (oddly enough, in the next question, they ask to cut the confidence interval to 1/2 but use 1/3 in the solution)

    is there something I am not understanding?

    question.png solution.png
     
  2. jcsd
  3. Dec 10, 2015 #2

    RUber

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    That all seems good. It appears that there is a problem in the question itself. Either it is asking to shrink your interval to 1/3 of its size or 1/2 of its size.
    The process they outlined seems reasonable.
    If you were not using a t test, it would be simpler...
    ## E = \pm z_{\alpha/2}\frac{\sigma}{\sqrt{N}}##
    ## E/3 = \pm z_{\alpha/2}\frac{\sigma}{\sqrt{N_2}}##
    So N_2 would be 9*N, or an increase of 8*N.
    With the t-test, you should need slightly fewer, since you reduce the size of your ##t_{\alpha/2, n-1} ## term as n increases.

    Not changing the percent to decimal does not affect the solution, since the units for the standard deviation and the mean were the same (%).
     
  4. Dec 10, 2015 #3
    Yah the process seemed similar to mine, but I wasnt sure why in the solution the cut the interval in half instead of 1/3

    So its safe to say that its wrong, and if i was asked to cut by 1/3, it should be 0.42/3 and not 0.42/2, correct?
     
  5. Dec 10, 2015 #4

    RUber

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    Right. The question does not seem to match te posted solution.
     
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