I am a bit confused when it comes to 68-95-99.7 rule of confidence interval of normal distribution. Normally I could see confidence interval of 95% for a sample statistic (s) with margin of error lets say e. So the confidence interval is s +- e. So 95% means that if we do the sampling 100 times and calculate the sample statistic s 100 times then for each we will have different s+-e. However 95 times the true population parameter p lies within s+-e.
This is what I have understood about confidence interval. The confidence interval is s+-e and the confidence level is 95%.
Now when it comes to normal distribution, how the above idea fits in. I mean here confidence interval is mui+-sigma and the confidence level is 68%?. According to wikipedia it is something like this
About 68% of values drawn from a normal distribution are within one standard deviation σ away from the mean; about 95% of the values lie within two standard deviations; and about 99.7% are within three standard deviations
I am confused where is the sample statistic here, it says the values drawn(which is directly the random variable).
Best Answer
It sounds like the statement you are quoting is not about confidence intervals but rather about the probability content within k standard deviations of the mean for k=1, 2, 3. This result plays a role in constructing confidence intervals for the mean of a normal sample though, because if Xi iid N(μ,σ) the sample mean Xb is N(μ,σ/√n) So for known σ [Xb-2σ/√n,Xb+2σ/√n] is approximately a 95% confidence interval for μ (actually 95.4%). When σ is unknown replacing σ with the sample standard deviation s will still give an approximate 95% confidence interval for μ when n is large. For small n the exact distribution to construct the confidence interval is student t with n-1 degrees of freedom. So 2 should be replaced by the (larger) appropriate percentile from the t distribution to get the 95% confidence.