I have one sample of only 86 values, which is not normally distributed according to the Shapiro-Wilk normality test. Can I still use this formula/code (sorry R code) to estimate the 95% confidence interval?
me <- qt(.975, length(sample) - 1) * sd(sample)/sqrt(length(sample)) lower <- mean(sample) - me upper <- mean(sample) + me
I would think that it is OK, as the sample mean should be normally distributed according to the central limit theorem? As far as I understand the central limit theorem, the means of several samples of sufficient size (30?) should be normally distributed, even if the distribution is, for example, multi modal. Now what confused me about this scenario is that I only have 1 sample. The function qt is the t-distribution's quantile - in this case for the 95% percentile. This should deal with the smallish sample size and the non normality?
I appreciate that this is basic stuff for you experts, but any feedback would be appreciated. Thanks!
I am aware of bootstrapping but would like to use "classical statistics" to estimate the CI.
This is related but there is no accepted answer. Here Ben's answer suggests that I am "right". Tony Ladson's answer looks also interesting. Should I use this approach and if so do I have to test for log normality?