Consider a random variable, X, with a normal distribution and a sample of size 82 from this distribution with a sample mean of 49.75 and sample standard deviation 73.12. What is the upper limit of a 96.92% confidence interval for the population mean?
+33666 Form the standard error: s = 73.12/√82
Assuming a 2-sided confidence interval then the upper limit is at a cumulative probability value of 96.92+(100 - 96.92)/2 = 98.46%
Using the t distribution (because we only know the sample standard deviation, not the true population value) this cumulative probability occurs (with 81 degrees of freedom) when t = 2.198 standard errors.
Hence upper limit = 49.75+2.198*73.12/√82
$${\mathtt{ul}} = {\mathtt{49.75}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2.198}}{\mathtt{\,\times\,}}{\mathtt{73.12}}}{{\sqrt{{\mathtt{82}}}}}} \Rightarrow {\mathtt{ul}} = {\mathtt{67.498\: \!307\: \!504\: \!130\: \!928\: \!9}}$$
or at approximately 67.5
.
+33666 Form the standard error: s = 73.12/√82
Assuming a 2-sided confidence interval then the upper limit is at a cumulative probability value of 96.92+(100 - 96.92)/2 = 98.46%
Using the t distribution (because we only know the sample standard deviation, not the true population value) this cumulative probability occurs (with 81 degrees of freedom) when t = 2.198 standard errors.
Hence upper limit = 49.75+2.198*73.12/√82
$${\mathtt{ul}} = {\mathtt{49.75}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2.198}}{\mathtt{\,\times\,}}{\mathtt{73.12}}}{{\sqrt{{\mathtt{82}}}}}} \Rightarrow {\mathtt{ul}} = {\mathtt{67.498\: \!307\: \!504\: \!130\: \!928\: \!9}}$$
or at approximately 67.5
.
