Chapter 3.2, Problem 18P

Archaeology: Ireland The Hill of Tara in Ireland is a place of great archaeological importance. This region has been occupied by people for more than 4000 years. Geomagnetic surveys detect subsurface anomalies in the earth's magnetic field. These surveys have led to many significant archaeological discoveries. After collecting data, the next step is to begin a statistical study. The following data measure magnetic susceptibility (centimeter-gram-second x 10^-6) on two of the main grids of the Hill of Tara (Reference: Tara: An ArchaeologicalSurvey by Conor Newman. Royal Irish Academy. Dublin).

Grid E: x variable
13.20	5.60	19.80	15.05	21.40	17.25	27.45
16.95	23.90	32.40	40.75	5.10	17.75	28.35
Grid H: y variable
11.85	15.25	21.30	17.30	27.50	10.35	14.90
48.70	25.40	25.95	57.60	34.35	38.80	41.00
31.25

(a) Compute $\sum x,\sum x^2,\sum y,\text{ and }\sum y^2$.

(b) Use the results of part (a) to compute the sample mean, variance, and standard deviation for x and for y.

(c) Compute a 75% Chebyshev interval around the mean for x values and also for y values. Use the intervals to compare the magnetic susceptibility on the two grids. Higher numbers indicate higher magnetic susceptibility. However, extreme values, high or low. could mean an anomaly and possible archaeological treasure.

(d)Interpretation Compute the sample coefficient of variation for each grid. Use the CVs to compare the two grids. If s represents variabilityin the signal (magnetic susceptibility) and $\bar{x}$ represents the expected level of the signal, then $s/\bar{x}$ can he thought of as a measure of the variability per unit of expected signal. Remember, a considerable variability in the signal (above or below average) might indicate buried artifacts, Why, in this case, would a large CV be better or at least more exciting? Explain.

To determine

To find: ${\displaystyle \sum x},{{\displaystyle \sum x}}^2,{\displaystyle \sum y},and{{\displaystyle \sum y}}^2$.

Answer to Problem 18P

Solution: ${\displaystyle \sum x}=284.95,{{\displaystyle \sum x}}^2=7046.8,{\displaystyle \sum y}=421.5,and{{\displaystyle \sum y}}^2=14,562.29$.

Explanation of Solution

Calculation: The values of ${\displaystyle \sum x},{{\displaystyle \sum x}}^2,{\displaystyle \sum y},and{{\displaystyle \sum y}}^2$ can be found in the following way:

x	y	x2	y2
13.20	11.85	174.24	140.4225
5.60	15.25	31.36	232.5625
19.80	21.30	392.04	453.6900
15.05	17.30	226.50	299.2900
21.40	27.50	457.96	756.2500
17.25	10.35	297.56	107.1225
27.45	14.90	753.50	222.0100
16.95	48.70	287.30	2371.6900
23.90	25.40	571.21	645.1600
32.40	25.95	1049.76	673.4025
40.75	57.60	1660.56	3317.7600
5.10	34.35	26.01	1179.9225
17.75	38.80	315.06	1505.4400
28.35	41.00	803.72	1681.0000
	31.25		976.5625
284.95	421.50	7046.80	14562.29

The above table shows the variable x, y, and their corresponding squares.

Therefore, from above table ${\displaystyle \sum x}=284.95,{{\displaystyle \sum x}}^2=7046.8,{\displaystyle \sum y}=421.5,and{{\displaystyle \sum y}}^2=14,562.29$.

Interpretation: ${\displaystyle \sum x}=284.95,{{\displaystyle \sum x}}^2=7046.8,{\displaystyle \sum y}=421.5,and{{\displaystyle \sum y}}^2=14,562.29$.

To determine

To find: The sample mean, variance, and standard deviation for x and for y.

Answer to Problem 18P

Solution: For x: $\overline{x}=20.35,{\text{ s}}^2=96\text{ and s=9}\text{.79}$;

For y: $\overline{y}=28.1,{\text{ s}}^2=194\text{ and s=13}\text{.93}$.

Explanation of Solution

Calculation: The results obtained in the part (a) as: ${\displaystyle \sum x}=284.95,{{\displaystyle \sum x}}^2=7046.8,{\displaystyle \sum y}=421.5,and{{\displaystyle \sum y}}^2=14,562.29$.

For (Grid E) x, calculation of $\overline{x}$, $s^2$ and s as follows:

Here, $n=14$.

The sample mean $\overline{x}$ is defined as:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{284.95}{14}\\ =20.35\end{array}$

The sample variance $s^2$ is defined as:

$\begin{array}{c}{s}^2=\frac{{{\displaystyle \sum x}}^2-{({\displaystyle \sum x})}^2/n}{n-1}\\ =\frac{7046.8-{(284.95)}^2/14}{14-1}\\ =\frac{7046.8-5799.75}{13}\\ =95.93\\ \approx 96\end{array}$

The sample standard deviation s is calculated as follows:

$\begin{array}{c}s=\sqrt{\text{sample variance}}\\ =\sqrt{96}\\ =9.79\end{array}$ $$

For y (Grid H) calculation of $\overline{x}$, $s^2$ and s as follows:

Here, $n=15$.

The sample mean $\overline{x}$ is defined as:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{421.50}{15}\\ =28.1\end{array}$

The sample variance $s^2$ is defined as:

$\begin{array}{c}{s}^2=\frac{{{\displaystyle \sum y}}^2-{({\displaystyle \sum y})}^2/n}{n-1}\\ =\frac{14562.29-{(421.5)}^2/15}{15-1}\\ =\frac{14562.29-11844.15}{14}\\ =194.15\\ \approx 194\end{array}$

The sample standard deviation s is calculated,

$\begin{array}{c}s=\sqrt{\text{sample variance}}\\ =\sqrt{194}\\ =13.93\end{array}$ $$

Interpretation: For x (Grid E): $\overline{x}=20.35,{\text{ s}}^2=96\text{ and s=9}\text{.79}$;

For y (Grid H): $\overline{y}=28.1,{\text{ s}}^2=194\text{ and s=13}\text{.93}$.

To determine

To find: A 75% Chebyshev's interval around the mean for x values and for y values, and also compare these intervals.

Answer to Problem 18P

Solution: A 75% Chebyshev's interval around the mean for x values is 0.77 to 39.93 and for y values is 0.24 to 55.96. Grid H (y) shows a wider 75% range of values.

Explanation of Solution

Calculation: According to results of Chebyshev's theorem, at least 75% of the data must fall within 2 standard deviations of the mean.

For a 75% Chebyshev's interval around the mean for x values: the mean is $\overline{x}$ = 20.35 and the standard deviation is s = 9.79, the interval is

$\begin{array}{c}\overline{x}-2s\text{ to }\overline{x}+2s\\ 20.35-2(9.79)\text{ to }20.35+2(9.79)\\ 0.77\text{ to 39}\text{.93}\end{array}$

For a 75% Chebyshev's interval around the mean for y values: the mean is $\overline{y}$ = 28.1 and the standard deviation is s = 13.93, the interval is

$\begin{array}{c}\overline{y}-2s\text{ to }\overline{y}+2s\\ 28.1-2(13.93)\text{ to 28}\text{.1+2(13}\text{.93)}\\ \text{ 0}\text{.24 to 55}\text{.96}\end{array}$

A 75% Chebyshev's interval around the mean for x values is 0.77 to 39.93 and for y values is 0.24 to 55.96.

It can be seen from 75% Chebyshev's interval around the mean for x and y, 75% of magnetic susceptibility on the Grid H falls within a wider range than those of the Grid E. Hence, In particular, the Grid H indicates higher magnetic susceptibility.

Interpretation: A 75% Chebyshev's interval around the mean for x values is 0.77 to 39.93 and for y values is 0.24 to 55.96.

To determine

To find: The coefficient of variation for each fund.

Answer to Problem 18P

Solution: For Grid E x, coefficient of variation is 48%; for the Grid H y, coefficient of variation is 50%.

Explanation of Solution

Calculation: For x, coefficient of variation is calculated as:

The sample coefficient of variation CV is defined to be

$CV=\frac{s}{\overline{x}}\times 100$ $$

In this data set, the $\overline{x}=20.35\text{ and }s=9.79$

Hence, the coefficient of variation is calculated as:

$\begin{array}{c}CV=\frac{9.79}{20.35}\times 100\\ =48.11\\ \approx 48\end{array}$

For y, coefficient of variation is calculated as:

$CV=\frac{s}{\overline{x}}\times 100$ $$

In this data set, the $\overline{y}=28.1\text{ and }s=13.93$

Hence, the coefficient of variation is calculated as:

$\begin{array}{c}CV=\frac{13.93}{28.1}\times 100\\ =49.57\\ \approx 50\end{array}$

The coefficient of variation (CV), for x is 49% and for y is 50%. CV for x, is small than y. since, the lower CV, the smaller the level of dispersion around the mean. Grid H demonstrates slightly greater variability per expected signal. The CV, together with the confidence interval, indicates that Grid H might have more buried artifacts.

Interpretation: Hence, the coefficient of variation for x is 49% and the coefficient of variation for y is 50%.