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 90% confidence interval estimate of the mean skull width.
127.6 137.8 125.9 131.7 143.2 135.1 138.9 128.9
mm < μ < mm
(round to two decimal places as needed.)

Explanation:

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

$\bar{x}=\frac{127.6 + 137.8+125.9+131.7+143.2+135.1+138.9+128.9}{8}$
$=\frac{1069.1}{8}=133.64$

Step2: Calculate sample standard deviation $s$

First, calculate deviations from the mean:
$(127.6 - 133.64)=-6.04$, $(137.8 - 133.64)=4.16$, $(125.9 - 133.64)=-7.74$, $(131.7 - 133.64)=-1.94$, $(143.2 - 133.64)=9.56$, $(135.1 - 133.64)=1.46$, $(138.9 - 133.64)=5.26$, $(128.9 - 133.64)=-4.74$

Squares of deviations:
$(-6.04)^2 = 36.4816$, $(4.16)^2=17.3056$, $(-7.74)^2 = 60.0276$, $(-1.94)^2=3.7636$, $(9.56)^2 = 91.3936$, $(1.46)^2=2.1316$, $(5.26)^2=27.6676$, $(-4.74)^2=22.4676$

Sum of squares: $36.4816+17.3056 + 60.0276+3.7636+91.3936+2.1316+27.6676+22.4676=261.242$

$s=\sqrt{\frac{261.242}{8 - 1}}=\sqrt{\frac{261.242}{7}}\approx6.10$

Step3: Determine $t$-value

For $n = 8$, degrees of freedom $df=n - 1=7$, and $90\%$ confidence level, $\alpha=1 - 0.90 = 0.10$, $\frac{\alpha}{2}=0.05$
From $t$-distribution table, $t_{\frac{\alpha}{2},df}=t_{0.05,7}=1.895$

Step4: Calculate margin of error $E$

$E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}=1.895\times\frac{6.10}{\sqrt{8}}\approx1.895\times2.16\approx4.09$

Step5: Construct confidence interval

Lower bound: $\bar{x}-E=133.64 - 4.09 = 129.55$
Upper bound: $\bar{x}+E=133.64+4.09 = 137.73$

Answer:

$129.55\ \text{mm}<\mu<137.73\ \text{mm}$