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Finding or estimating confidence interval for populaion mean

  1. Aug 25, 2011 #1
    From Koosis, I pieced together the following algorithm.

    Is sigma known?

    Yes? Then calculate the exact confidence interval using a normal distribution to estimate that of the sample means, with mean = the mean of sample means = the mean of the population, [itex]\mu_{\overline{x}}=\mu[/itex], and standard deviation [itex]\sigma_{\overline{x}}=\sigma/\sqrt{n}[/itex].

    No? Then is the poplation normal?

    Yes? Then (a) estimate the confidence interval with a Student's t distribution for the sample means, using degrees of freedom dof = n - 1, and standard deviation [itex]s_{\overline{x}}=s\sqrt{n}[/itex], or (b) for a slightly inferior result, and only if [itex]n\geq 30[/itex], estimate the confidence interval using the normal distribution with mean [itex]\mu_{\overline{x}}=\mu[/itex], and standard deviation [itex]s_{\overline{x}}=s\sqrt{n}[/itex].

    No or don't know? Then is [itex]n\geq 30[/itex]?

    Yes? Then estimate the confidence interval, using a normal distribution to estimate that of the sample means, with mean [itex]\mu_{\overline{x}}=\mu[/itex], and standard deviation [itex]s_{\overline{x}}=s\sqrt{n}[/itex].

    No? Then can't do.


    But Sanders has the following, somewhat different algorithm.

    Is [itex]n\geq 30[/itex]?

    Yes? Then use z values in computations.

    No? Then are population values known to be normally distributed?

    Yes? If the population standard deviation of the population is known, use z values in computations. Otherwise, use t values in computations.

    No? Cannot use z or t values in computations.


    Any comments on which is the best procedure? Actually Koosis presented the z test first, as if he, like Sanders, assumed that one would choose this over the t test wherever possible, even though he said it wasn't as good when both choices were possible. I wonder why z beats t in that case? Is it because the difference in accuracy is negligible then and the computations for t potentially more time consuming than those for z? (And if so, is this still the case with current software; both books are a few years old.)
  2. jcsd
  3. Aug 27, 2011 #2

    I like Serena

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    Hi Rasalhague! :smile:

    The more information you use the more accurate the result.

    If you know sigma beforehand, you have to use it to get the most accurate results.
    However, in practice, sigma is often not known, so an estimate has to be made using a sample.
    This is the reason the t-distribution has been thought up in the first place.

    Whenever you use a sample to estimate the population mean or the variance, there is a significant risk on a selective bias, so whenever possible this should be avoided.

    If your n is large enough the difference between the normal distribution and the t-distribution becomes negligible, so you can choose.
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