Answer to Question #246784 in Statistics and Probability for s12

Question

Maximum breadth of samples of male Egyptian skulls from 4000 B.C. and 150 A.D.

(based on data from Ancient Races of the Thebaid by Thomson and Randall-Maciver):

 

           4000 B.C. :     131 119 138 125 129 126 131 132 126 128 128 131 

           150 A.D. :       136 130 126 126 139 141 137 138 133 131 134 129 

Changes in head sizes over time suggest interbreeding with people from other regions.  Find the sample standard deviation and use a 90% confidence interval to determine whether the head sizes appear to have changed from 4000 B.C. to 150 A.D. Explain your result. 

Answer 1

At 4000 B.C.:

Mean:

$\bar{x_1} = \frac{131+119+...+128+131}{12}= 128.67$

Standard deviation:

$s_1 = \sqrt{\frac{(131-128.67)^2 +(119-128.67)^2 +...+(128-128.67)^2+(131-128.67)^2}{12-1}} = 4.6384$

At 150 A.D.:

Mean:

$\bar{x_2} = \frac{136+130+...+134+129}{12}=133.33$

Standard deviation:

$s_2 = \sqrt{\frac{(136-133.33)^2 +(130-133.33)^2 +...+(134-133.33)^2 +(129-133.33)^2}{12-1}}=5.0151$

Confidence interval $(CI) = (\bar{x_1} -\bar{x_2}) ± t_c \sqrt{\frac{(n_1-1)s^2_1 +(n_2-1)s^2_2}{n_1+n_2-2} (\frac{1}{n_1} + \frac{1}{n_2})}$

$n_1=n_2=12 \\ df= n_1+n_2 -2 = 22$

Critical value $t_c = 1.72$ for 90% and df=22

$CI = (128.67 -133.33) ± 1.72 \sqrt{\frac{(11)(4.6384)^2 +(11)(5.0151)^2}{12+12-2} (\frac{1}{12} + \frac{1}{12})} \\ CI = -4.66 ± 1.72 \times 1.97 \\ CI = -4.66 ± 3.3862 \\ CI = (-8.0462, -1.2738)$

CI does not include 0. We conclude that head sizes appear to have changed from 4000 B.C. to 150 A.D.