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Sample size without standard Deviation

  1. Jun 3, 2009 #1
    Hello again,

    I have a question here that asks me to find how large a sample size is, but I have no Standard deviation. How would you tackle this>

    How large a sample size do we need to estimate the mean annual income of natives in New York, correct to within $1000 with probability 0.99? No information is available to us about the standard deviation of their annual income. We guess that nearly all of the incomes fall between $0 and $120,000 and that this distribution is approximately normal.

    Here's what I see:
    1 - alpha = 0.99 therefore alpha =0.01 /2 = 0.005
    This gives a Z* of 2.575 (because it states normal distribution)
    The 99% CI is (0, 120000).

    I know that n = [Z*s/m]squared, buty I have neither s, nor m....

    As always, the feedback and guidance is appreciated =)
  2. jcsd
  3. Jun 3, 2009 #2


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    Practically the endpoints of the distribution are 0 and 120K. Normal dist. is symmetric, so you can figure out the mean. As for standard dev., I would assume 99% of the people are within 0 to 120K, and find out how many standard deviations it would take to get 99% of people (within ___ standard deviations around the mean).
  4. Jun 3, 2009 #3
    I don't think this question can be answered without further information. Suppose that the income is distributed with a mean of $60,000, and a standard deviation of $1. After a small number of observations we would learn that the std. dev. is small, and realize we don't need to take many more samples.

    On the other hand suppose that the income is distributed with a mean of $60,000 and a standard deviation of $20,000. In that case we'd have to take a much larger number of samples to achieve the same confidence.
  5. Jun 4, 2009 #4


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    The statement "correct to within $1000 with probability 0.99" implies a standard deviation by Chebyshev's inequality: The probability an observation is with k standard deviations of the mean is less than [itex]1/k^2[/itex]. The largest k that has [itex]1/k^2< .99[/itex] is 2 so 1000 must be no more than 2 standard deviations. The smallest standard deviation that will work is $500.
  6. Jun 4, 2009 #5


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    You can also try use the (very crude) approximation that

    \sigma \approx \frac{\text{Range}}{4}

    presented in some texts. I suggest to students to use 6 rather than 4.
  7. Jun 4, 2009 #6
    Chebyshev's inequality says the probability an observation is _not_ within k std. dev. of the mean is <= 1/k^2.
  8. Jun 4, 2009 #7


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    Had the true std. dev. (σ) been known, you'd use N = (zσ/x)^2, where x is the margin of error = $1,000 (or x = 1 if you express everything in $1,000). When σ is unknown the process is more complicated and you may have to iterate. This page explains how.
    Last edited: Jun 4, 2009
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