My last couple of posts (here<\/a> and here<\/a>) I’ve talked about confidence intervals for the mean. Yesterday I mentioned finding one using Excel or LibreOffice Calc’s CONFIDENCE() function.<\/p>\n While the CONFIDENCE() function assumes the population standard deviation is known, I pointed out that, with sample size n≥31, the t-distribution approximates the normal closely enough that the sample standard deviation can be used. Today, I’ll make a direct comparison.<\/p>\n Yesterday’s post considered a sample mean of 67.3, known population standard deviation of 12.4, and sample size 42. The input<\/p>\n =confidence(0.05,12.4,42)<\/p>\n gave the result 3.75, meaning a confidence interval of 67.3\u00b13.75, or 63.55 to 71.05.<\/p>\n LibreOffice Calc’s CONFIDENCE.T() function has the following format:<\/p>\n =confidence(1-confidence_level, sample_standard_deviation, sample_size)<\/p>\n Since it uses the sample<\/em> standard deviation, CONFIDENCE.T() calculates the confidence interval from the t-distribution. By constrast, CONFIDENCE() takes the population<\/em> standard deviation, so uses the normal distribution to calculate the confidence interval.<\/p>\n The following input<\/p>\n =confidence.t(0.05, 12.4, 42)<\/p>\n gives the result 3.864, implying a confidence interval of 67.3\u00b13.864 or 63.44 to 71.16. Obviously this is not much different from the confidence interval 63.55 to 71.05 gotten using =confidence(0.05,12.4,42).<\/p>\n So, the CONFIDENCE.T() function seems to demonstrate that, for a sample size n≥31, the t-distribution approximates the normal distribution closely enough that the sample standard deviation can be used when the population standard deviation is unavailable.<\/p>\n HTH:)<\/p>\n