Question

archaeologists 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. 128.1 137.9 125.7 132.1 143.1 136.1 128.9 128.7 \\(\square\\) mm < \\(\mu\\) < \\(\square\\) mm (round to two decimal places as needed.)

Explanation:

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

First, sum up all the data points:
$128.1 + 137.9 + 125.7 + 132.1 + 143.1 + 135.1 + 128.9 + 128.7$
$= (128.1 + 137.9) + (125.7 + 132.1) + (143.1 + 135.1) + (128.9 + 128.7)$
$= 266 + 257.8 + 278.2 + 257.6$
$= 266 + 257.8 = 523.8$; $523.8 + 278.2 = 802$; $802 + 257.6 = 1059.6$

There are $n = 8$ data points. So, $\bar{x} = \frac{1059.6}{8} = 132.45$

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

First, find the squared differences from the mean:

• $(128.1 - 132.45)^2 = (-4.35)^2 = 18.9225$
• $(137.9 - 132.45)^2 = (5.45)^2 = 29.7025$
• $(125.7 - 132.45)^2 = (-6.75)^2 = 45.5625$
• $(132.1 - 132.45)^2 = (-0.35)^2 = 0.1225$
• $(143.1 - 132.45)^2 = (10.65)^2 = 113.4225$
• $(135.1 - 132.45)^2 = (2.65)^2 = 7.0225$
• $(128.9 - 132.45)^2 = (-3.55)^2 = 12.6025$
• $(128.7 - 132.45)^2 = (-3.75)^2 = 14.0625$

Sum these squared differences:
$18.9225 + 29.7025 + 45.5625 + 0.1225 + 113.4225 + 7.0225 + 12.6025 + 14.0625$
$= 18.9225 + 29.7025 = 48.625$; $48.625 + 45.5625 = 94.1875$; $94.1875 + 0.1225 = 94.31$; $94.31 + 113.4225 = 207.7325$; $207.7325 + 7.0225 = 214.755$; $214.755 + 12.6025 = 227.3575$; $227.3575 + 14.0625 = 241.42$

The sample variance ($s^2$) is $\frac{\sum (x_i - \bar{x})^2}{n - 1} = \frac{241.42}{7} \approx 34.4886$

Then, $s = \sqrt{34.4886} \approx 5.8727$

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

For a 95% confidence interval and $n - 1 = 7$ degrees of freedom, $\alpha = 0.05$, so $\alpha/2 = 0.025$. From the t-distribution table, $t_{0.025, 7} \approx 2.3646$

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

$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 2.3646 \times \frac{5.8727}{\sqrt{8}} \approx 2.3646 \times \frac{5.8727}{2.8284} \approx 2.3646 \times 2.076 \approx 4.91$

Step5: Construct the confidence interval

Lower bound: $\bar{x} - E = 132.45 - 4.91 = 127.54$
Upper bound: $\bar{x} + E = 132.45 + 4.91 = 137.36$

Answer:

$127.54$ mm $< \mu < 137.36$ mm

(Note: Depending on rounding precision during intermediate steps, slight variations may occur. For example, if more precise $s$ or $t$-value is used, the interval might adjust slightly, but the above is accurate with standard calculations.)