OpenIntro Statistics 3rd Edition
5.5 Working backwards, Part I. A 95% confidence interval for a population mean, \(\mu\), is given as (18.985, 21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t-distribution in any calculations.
\(CI:\ \bar{x} \pm t^{\star}_{df} SE\)
\(df = n - 1 = 36 - 1 = 35\)
\(95\%\ CI:\ \bar{x} \pm t^{\star}_{35} SE = (18.985, 21.015)\)
\(\bar{x} + t^{\star}_{35} SE = 21.015\)
\(\bar{x} - t^{\star}_{35} SE = 18.985\)
\(2\bar{x} = 21.015 + 18.985\)
samp_mean <- (21.015 + 18.985)/2\(\bar{x} = 40 / 2 = 20\)
\(t^{\star}_{35} SE = 21.015 - 20 = 20 - 18.985 = 1.015\)
\(SE = 1.015 / t^{\star}_{35} = s / \sqrt{n} = s / \sqrt{36} = s/6\)
\(s = 6 (1.015) / t^{\star}_{35}\)
t_score <- qt(0.975, df = 35)
samp_sd = sqrt(36) * 1.015 / t_score\(s = 6 (1.015) / 2.0301079 = 2.9998405 \approx 3\)