Q 3.79. - Elementary Statistics

Question

 Medieval Cremation Burials. In the article "Material Culture as Memory: Combs and Cremations in Early Medieval Britain" Early  Medieval, Vol. 12. Issue 2, pp. 89-128, H. Williams discussed the frequency of cremation burials found in 17 archaeological sites in eastern England. Here are the data.

 $$\begin{gathered} 23, 64, 46, 48, 523, 35, 34, 265, 2484, \\ 46, 385, 21, 86, 429, 51, 258, 119 \end{gathered}$$

a) Obtain the sample standard deviation of these data.

b) Do you think that in this case, the sample standard deviation provides a good measure of variation? Explain your answer. 

Step-by-Step Solution

Answer

a)Standard deviation of the data $=586.32$ cremation burials

b) No, the sample standard deviation does not provides a good measure of variation in this case because of it's lacks in resistance.  

Step 1. Given information.

We have given data: $$\begin{gathered} 23, 64, 46, 48, 523, 35, 34, 265, 2484, \\ 46, 385, 21, 86, 429, 51, 258, 119 \end{gathered}$$

Part a) Step 2. To find standard deviation of the data.

$$\begin{gathered} Mean =\frac{Sum of observation}{Number of observation} \\ =\frac{23+ 64+4+48+ 523+ 35+34+ 265+ 2484+46+385+21+86+429+51+258+119}{17} \\ =\frac{4977}{17} \\ =292.76 \end{gathered}$$

$x$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$$\begin{gathered} 23 \\ 64 \\ 46 \\ 48 \\ 523 \\ 35 \\ 34 \\ 265 \\ 2484 \\ 46 \\ 385 \\ 21 \\ 86 \\ 429 \\ 51 \\ 258 \\ 119 \end{gathered}$$	$$\begin{gathered} -209.8 \\ -228.8 \\ -246.8 \\ -244.8 \\ 230.2 \\ -257.8 \\ -258.8 \\ -27.8 \\ 2191.2 \\ -246.8 \\ 192.2 \\ -271.8 \\ -206.8 \\ 136.2 \\ -241.8 \\ -34.8 \\ -173.8 \end{gathered}$$	$$\begin{gathered} 44016.04 \\ 52349.44 \\ 60910.24 \\ 59927.04 \\ 52992.04 \\ 66460.84 \\ 66977.44 \\ 772.84 \\ 4801357.44 \\ 60910.24 \\ 8522.84 \\ 73875.24 \\ 42766.24 \\ 18550.44 \\ 58467.24 \\ 1211.04 \\ 30206.44 \end{gathered}$$

$$\begin{gathered} S\tan dard deviation=\sqrt{\frac{{\displaystyle \sum _{}}{\left(x-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{5500251.08}{16}} \\ =586.31535 \\ 586.32 \end{gathered}$$

Part b) Step 3. Explain the sample standard deviation is a good measure of variance.

Because of its lack of resistance, the sample standard deviation does not provide a meaningful measure of variation in this circumstance.