- #1
Rasalhague
- 1,383
- 2
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, \mu_{\overline{x}}=\mu, and standard deviation \sigma_{\overline{x}}=\sigma/\sqrt{n}.
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 s_{\overline{x}}=s\sqrt{n}, or (b) for a slightly inferior result, and only if n\geq 30, estimate the confidence interval using the normal distribution with mean \mu_{\overline{x}}=\mu, and standard deviation s_{\overline{x}}=s\sqrt{n}.
No or don't know? Then is n\geq 30?
Yes? Then estimate the confidence interval, using a normal distribution to estimate that of the sample means, with mean \mu_{\overline{x}}=\mu, and standard deviation s_{\overline{x}}=s\sqrt{n}.
No? Then can't do.
*
But Sanders has the following, somewhat different algorithm.
Is n\geq 30?
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.)
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, \mu_{\overline{x}}=\mu, and standard deviation \sigma_{\overline{x}}=\sigma/\sqrt{n}.
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 s_{\overline{x}}=s\sqrt{n}, or (b) for a slightly inferior result, and only if n\geq 30, estimate the confidence interval using the normal distribution with mean \mu_{\overline{x}}=\mu, and standard deviation s_{\overline{x}}=s\sqrt{n}.
No or don't know? Then is n\geq 30?
Yes? Then estimate the confidence interval, using a normal distribution to estimate that of the sample means, with mean \mu_{\overline{x}}=\mu, and standard deviation s_{\overline{x}}=s\sqrt{n}.
No? Then can't do.
*
But Sanders has the following, somewhat different algorithm.
Is n\geq 30?
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.)
