Question

archeologists have studied sizes of egyptian skulls in an attempt to determine whether breeding occurred between different cultures. listed below are the widths (mm) of skulls from 150 a d. construct a 95% confidence interval estimate of the mean skull width. 127.8 137.6 125.6 131.8 143.1 134.8 138.9 128.9 mm < \mu < mm (round to two decimal places as needed)

Explanation:

Step1: Calculate the sample mean $\bar{x}$

The sample data is \(x = [127.8, 137.6, 125.6, 131.8, 143.1, 134.8, 138.9, 128.9]\).
The formula for the sample mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
\(\sum_{i=1}^{8}x_{i}=127.8 + 137.6+125.6+131.8+143.1+134.8+138.9+128.9 = 1068.5\)
\(n = 8\), so \(\bar{x}=\frac{1068.5}{8}=133.5625\)

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(127.8 - 133.5625)^{2}=(- 5.7625)^{2}=33.20640625\)
\((x_{2}-\bar{x})^{2}=(137.6-133.5625)^{2}=(4.0375)^{2}=16.30140625\)
\((x_{3}-\bar{x})^{2}=(125.6 - 133.5625)^{2}=(-7.9625)^{2}=63.40140625\)
\((x_{4}-\bar{x})^{2}=(131.8-133.5625)^{2}=(-1.7625)^{2}=3.10640625\)
\((x_{5}-\bar{x})^{2}=(143.1-133.5625)^{2}=(9.5375)^{2}=90.96440625\)
\((x_{6}-\bar{x})^{2}=(134.8-133.5625)^{2}=(1.2375)^{2}=1.53140625\)
\((x_{7}-\bar{x})^{2}=(138.9-133.5625)^{2}=(5.3375)^{2}=28.49940625\)
\((x_{8}-\bar{x})^{2}=(128.9-133.5625)^{2}=(-4.6625)^{2}=21.74840625\)
\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=33.20640625+16.30140625 + 63.40140625+3.10640625+90.96440625+1.53140625+28.49940625+21.74840625=258.75\)
\(s=\sqrt{\frac{258.75}{8 - 1}}=\sqrt{\frac{258.75}{7}}\approx6.06\)

Step3: Determine the critical value \(t_{\alpha/2}\)

The confidence level is \(95\%\), so \(\alpha=1 - 0.95 = 0.05\) and \(\alpha/2=0.025\).
The degrees of freedom \(df=n - 1=8 - 1 = 7\)
From the \(t\)-distribution table, \(t_{0.025,7}=2.365\)

Step4: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval for the population mean (when \(\sigma\) is unknown) is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E=2.365\times\frac{6.06}{\sqrt{8}}\)
\(\sqrt{8}\approx2.828\), \(\frac{6.06}{2.828}\approx2.14\)
\(E=2.365\times2.14\approx5.06\)

Step5: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
\(\bar{x}-E=133.5625-5.06 = 128.50\)
\(\bar{x}+E=133.5625 + 5.06=138.62\)

Answer:

\(128.50\) mm \(<\mu<138.62\) mm