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Homework Help: Determining a sample size

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data

    An investigator, interested in estimating a population mean, wants to be sure that the length of the 95% confidence interval does not exceed 5. What sample size should she use if σ = 18?

    3. The attempt at a solution
    the formula I found in my book is n = [(z_(α/2) σ)/E]^2

    z_(α/2) = z.025 = 1.96

    I am fairly certain if the length of the interval can't exceed 5, then 5 will be the max error so
    E = 5
    n = [1.96(18)/5]^2 = 49.8

    Am I doing this correctly?
  2. jcsd
  3. Nov 21, 2015 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Avoid using canned formulas; rather, work things out from first principles. So ask yourself: if ##X_1, X_2, \ldots, X_n## are iid random variables from the distribution ##N(\mu,\sigma^2)##, what is the distribution of the sample mean
    [tex] \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \: ? [/tex]
    Now you need to know how large to make ##n## in order to have
    [tex] P(-2.5 \leq \bar{X} - \mu \leq 2.5 ) = 0.95, [/tex]
    assuming that you know ##\sigma = 18##. At that point you are ready to state with absolute confidence the appropriate test to use. (And no, I will not tell you if you are correct or not!)
  4. Nov 21, 2015 #3

    P(-z_(α/2) ≤ (X-μ)/(σ/sqrt(n)) ≤ z_(α/2)) = .95

    P(-z.025 ≤ (X-μ)/(18/sqrt(n)) ≤ z.025) = .95

    -1.96 ≤ sqrt(n)(X-μ)/18 ≤ 1.96

    -1.96(18)/(X-μ) ≤ sqrt(n) ≤ 1.96(18)/(X-μ)

    [-1.96(18)/(X-μ)]^2 ≤ n ≤ [1.96(18)/(X-μ)]^2

    n = [1.96(18)/(X-μ)]^2

    which is what I got before if E = (X-μ)
    I think it does, but I am now not sure if it is equal to 5
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