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Size sample required to construct a 95% C.I. with margin of error

  1. Apr 11, 2013 #1
    N = 8
    Mean = 2.10
    stDev = 0.537

    What size sample is required to construct a 95% confidence interval with a margin of error of 0.3?
    try n = 12, t(.975) with 11 deg. of freedom = 2.2001
    so n = ( (.537* 2.2001) / (.3))^2 = 15.5

    try n = 14, i get n = 14.95
    n = 15, n = 14.74
    Why is this a trial and error?
    What I don't get is what are we trying to get n to equal?

    the answer: n must be between 14 and 15 so, use n = 15.
     
  2. jcsd
  3. May 28, 2013 #2

    statdad

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    Homework Helper

    First question: what is [itex] N = 8 [/itex] in your question? Second question: are you trying to look for the sample size to use to generate a CI for a mean? If so, have you not encountered the formula

    [tex]
    n = \left(\frac{z \times \sigma}{E}\right)^2
    [/tex]

    (E is the margin of error)? The general rule is to use this and round the result to the next highest integer. If you use the method you suggest (which I haven't seen in any text)
    * you are correct, you get a game of guessing
    * it isn't obvious to me that it would ever converge - that is, if you repeated the same step with 15, then with
    the result of that calculation, and so on, I am not sure you would ever zero in on a single value.

    The benefit of the formula I presented is that it does give an idea of an appropriate sample size, IF, your guess at the size of the standard deviation is in the correct ballpark.
     
    Last edited: May 28, 2013
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