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 99% confidence interval estimate of the mean skull width. 127.9 138.1 126.3 131.8 142.7 134.7 139.1 128.8 mm < μ < mm (round to two decimal places as needed)

Explanation:

Step1: Calculate sample mean ($\bar{x}$)

$$\bar{x}=\frac{127.9 + 138.1+126.3 + 131.8+142.7+134.7+139.1+128.8}{8}$$
$$=\frac{1069.4}{8}=133.675$$

Step2: Calculate sample standard deviation ($s$)

First, calculate deviations from the mean:
\((127.9 - 133.675), (138.1 - 133.675),(126.3 - 133.675),(131.8 - 133.675),(142.7 - 133.675),(134.7 - 133.675),(139.1 - 133.675),(128.8 - 133.675)\)
\(=- 5.775,4.425,-7.375,-1.875,9.025,1.025,5.425,-4.875\)
Then, square these deviations:
\((-5.775)^{2}=33.350625,(4.425)^{2}=19.580625,(-7.375)^{2}=54.380625,(-1.875)^{2}=3.515625,(9.025)^{2}=81.450625,(1.025)^{2}=1.050625,(5.425)^{2}=29.430625,(-4.875)^{2}=23.765625\)
Sum of squared deviations: \(33.350625 + 19.580625+54.380625+3.515625+81.450625+1.050625+29.430625+23.765625 = 246.525\)
Sample variance \(s^{2}=\frac{246.525}{8 - 1}=\frac{246.525}{7}\approx35.2179\)
Sample standard deviation \(s=\sqrt{35.2179}\approx5.93\)

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

Degrees of freedom \(n-1=8 - 1 = 7\), \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\)
From \(t -\)distribution table, \(t_{0.005,7}=3.499\)

Step4: Calculate margin of error ($E$)

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.499\times\frac{5.93}{\sqrt{8}}\)
\(=3.499\times\frac{5.93}{2.828}\approx3.499\times2.097\approx7.34\)

Step5: Calculate confidence interval

Lower limit: \(\bar{x}-E=133.675 - 7.34 = 126.34\)
Upper limit: \(\bar{x}+E=133.675+7.34 = 141.02\)

Answer:

\(126.34\) mm \(<\mu<141.02\) mm