Chapter 1.3, Problem 35E

Mercury is a persistent and dispersive environmental contaminant found in many ecosystems around the world. When released as an industrial by-product, it often finds its way into aquatic systems where it can have deleterious effects on various avian and aquatic species. The accompanying data on blood mercury- concentration (μg/g) for adult females near contaminated rivers in Virginia was read from a graph in the article “Mercury Exposure Effects the Reproductive Success of a Free-Living Terrestrial Songbird, the Carolina Wren” (The Auk. 2011:759-769: this is a publication of the American Ornithologists’ Union).

1. a. Determine the values of the sample mean and sample median and explain why they are different. [Hint: Σx₁ = 18.55.]
2. b. Determine the value of the 10% trimmed mean and compare to the mean and median.
3. c. By how much could the observation .20 be increased without impacting the value of the sample median?

To determine

Find the mean blood mercury concentration level for the female adults.

Find the median blood mercury concentration level for the female adults.

Explain the reason for the difference in mean and median of blood mercury concentration level.

Answer to Problem 35E

The mean blood mercury concentration level for the female adults is 1.24$\left(\frac{\mu g}{g}\right)$.

The median blood mercury concentration level for the female adults is 0.556$\left(\frac{\mu g}{g}\right)$.

The mean and median of blood mercury concentration level are due to the skewness in the distribution.

Explanation of Solution

Given info:

The data represents the blood mercury concentration levels $\left(\frac{\mu g}{g}\right)$ for a random sample of female adults from Virginia near contaminated rivers.

Calculation:

Mean blood mercury concentration level:

The sum of the ${\displaystyle \sum x_i}$ is obtained in the following table,

Blood mercury concentration x
0.20
0.22
0.25
0.30
0.34
0.41
0.55
0.56
1.42
1.70
1.83
2.20
2.25
3.07
3.25
${\displaystyle \sum x_i}=18.55$

Thus, the total blood mercury concentration level is 18.55.

The mean blood mercury concentration level is:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{18.55}{15}\\ =1.24\end{array}$

Thus, the mean blood mercury concentration level is 1.24$\left(\frac{\mu g}{g}\right)$.

Median blood mercury concentration level:

Median:

The median is the middle value in an ordered sequence of data. If there are no ties, half of the observations will be smaller than the median, and half of the observations will be larger than the median.

In addition, the median is unaffected by extreme values in a set of data. Thus, whenever an extreme observation is present, it may be more appropriate to use the median rather than the mean to describe a set of data.

In case of discrete data, if the sample size is an odd number then the middle observation of the ordered data is the median and if the sample size is even number then the average of middle two observations of the ordered data is the median.

To obtain median, the data has to be arranged in ascending order.

The data set is arranged in ascending order.

Blood mercury concentration x
0.20
0.22
0.25
0.30
0.34
0.41
0.55
0.56
1.42
1.70
1.83
2.20
2.25
3.07
3.25

There are 15 observations in the data. Since 15 is odd, median of the data is the middle value.

That is, 8^th observation.

Here, the 8^th observation is 0.556.

Thus, the median blood mercury concentration level is 0.556$\left(\frac{\mu g}{g}\right)$.

Reason:

The mean blood mercury concentration level for the female adults is 1.24$\left(\frac{\mu g}{g}\right)$.

The median blood mercury concentration level for the female adults is 0.556$\left(\frac{\mu g}{g}\right)$.

For symmetric distribution the values of mean, median and mode will be same.

For the skewed distribution, the values of mean and median will not be same.

The mean and median blood mercury concentration level are different from each other.

The median is unaffected by extreme values in a set of data. Thus, whenever an extreme observation is present, it may be more appropriate to use the median rather than the mean to describe a set of data.

The distribution of blood mercury concentration level is examined through histogram.

Histogram:

Software procedure:

Step by step procedure to draw histogram using MINITAB software is given below:

• Choose Graph>Histogram.
• Enter the column blood mercury concentration in Graph variables.
• In options, select mean.
• Click OK

Output obtained from MINITAB is given below:

From the histogram, the distribution of blood mercury concentration levels is positive skewed.

Hence, there is a difference between the sample mean and sample median of blood mercury concentration level.

To determine

Find the 10% trimmed mean blood mercury concentration level.

Compare the trimmed mean with usual mean and median.

Answer to Problem 35E

The 10% trimmed mean blood mercury concentration level is 1.1201$\left(\frac{\mu g}{g}\right)$.

Explanation of Solution

Calculation:

The data set contains 15 observations. That is, the sample size is 15 and the trimmed percentage is 10.

The trimmed data is calculated as,

$\begin{array}{c}10\%\text{trimmed}\text{data}=10\%\times \text{sample}\text{size}\\ \text{=10\%}\times 15\\ =1.5\end{array}$

Therefore, the ten percentage of the sample size 15 is 1.5.

In general, the smallest and largest observations have to be trimmed.

Here, 1.5 is not an integer it is a decimal number.

Therefore, it is not appropriate to calculate the 10% trimmed mean directly as it is impossible to remove 1.5 numbers of observations from the data set.

It is known that, average of 6.7% and 13.3% is 10%.

Hence, the 10% trimmed mean is obtained as the average of 6.7% trimmed mean and 13.3% trimmed mean.

6.7% trimmed mean blood mercury concentration level:

The data set contains 15 observations. That is, the sample size is 15 and the trimmed percentage is 6.7.

The trimmed data is calculated as,

$\begin{array}{c}6.7\%\text{trimmed}\text{data}=6.7\%\times \text{sample}\text{size}\\ \text{=6}\text{.7\%}\times 15\\ \simeq 1\end{array}$

Therefore, the 6.7 percentage of the sample size 15 is approximately 1.

In general, the smallest and largest observations have to be trimmed.

Thus, the 6.7% percentage trimmed mean is the mean of the data set after removing the one smallest and one largest observation.

That is, the mean of the data set after removing 0.20 and 3.25 from the data set.

The trimmed mean is obtained as below:

Home sales x
0.22
0.25
0.30
0.34
0.41
0.55
0.56
1.42
1.70
1.83
2.20
2.25
3.07
${\displaystyle \sum x_{x_i{}_{tr\left(6.7\right)}}}=15.1$

Thus, the total blood mercury concentration level is 15.1.

The mean of the data set is:

$\begin{array}{c}{\overline{x}}_{tr\left(6.7\right)}=\frac{{\displaystyle \sum x_i{}_{tr\left(6.7\right)}}}{n}\\ =\frac{15.1}{13}\\ =1.162\end{array}$

Thus, 6.7% trimmed mean is 1.1662.

13.3% trimmed mean blood mercury concentration level:

The data set contains 15 observations. That is, the sample size is 15 and the trimmed percentage is 13.3.

The trimmed data is calculated as,

$\begin{array}{c}13.3\%\text{trimmed}\text{data}=13.3\%\times \text{sample}\text{size}\\ \text{=13}\text{.3\%}\times 15\\ \simeq 2\end{array}$

Therefore, the 13.3 percentage of the sample size 15 is approximately 2.

In general, the smallest and largest observations have to be trimmed.

Thus, the 13.3% percentage trimmed mean is the mean of the data set after removing the two smallest and two largest observations.

That is, the mean of the data set after removing 0.20, 0.22, 3.07 and 3.25 from the data set.

The trimmed mean is obtained as below:

Home sales x
0.25
0.30
0.34
0.41
0.55
0.56
1.42
1.70
1.83
2.20
2.25
${\displaystyle \sum x_{x_i{}_{tr\left(6.7\right)}}}=11.81$

Thus, the total blood mercury concentration level is 11.81.

The mean of the data set is:

$\begin{array}{c}{\overline{x}}_{tr\left(13.3\right)}=\frac{{\displaystyle \sum x_i{}_{tr\left(13.3\right)}}}{n}\\ =\frac{11.81}{11}\\ =1.074\end{array}$

Thus, 13.3% trimmed mean is 1.074.

10% trimmed mean blood mercury concentration level:

The 10% trimmed mean is the average of 6.7% and 13.3% trimmed mean blood mercury concentration level.

6.7% trimmed mean blood mercury concentration level is 1.1662.

13.3% trimmed mean blood mercury concentration level is 1.074.

10% trimmed mean blood mercury concentration level is,

$\begin{array}{c}{\overline{x}}_{tr\left(10\right)}=\frac{{\overline{x}}_{tr\left(6.7\right)}+{\overline{x}}_{tr\left(13.3\right)}}{2}\\ =\frac{1.1662+1.074}{2}\\ =1.1201\end{array}$

Thus, 10% trimmed mean blood mercury concentration level is 1.1201$\left(\frac{\mu g}{g}\right)$.

Comparison:

The mean blood mercury concentration level for the female adults is 1.24$\left(\frac{\mu g}{g}\right)$.

The 10% trimmed mean blood mercury concentration level is 1.1201$\left(\frac{\mu g}{g}\right)$.

Even though there is not much difference between the usual mean and the trimmed mean, the trimmed mean will give good result, because it is not affected by the outliers.

To determine

Explain the level of increment of the observation 0.20 without affecting the sample median.

Answer to Problem 35E

The observation 0.20 can be increased up to 0.36 without affecting the sample median.

Explanation of Solution

Median:

The median is the middle value in an ordered sequence of data. If there are no ties, half of the observations will be smaller than the median, and half of the observations will be larger than the median.

In addition, the median is unaffected by extreme values in a set of data. Thus, whenever an extreme observation is present, it may be more appropriate to use the median rather than the mean to describe a set of data.

In case of discrete data, if the sample size is an odd number then the middle observation of the ordered data is the median and if the sample size is even number then the average of middle two observations of the ordered data is the median.

The median blood mercury concentration level is 0.556$\left(\frac{\mu g}{g}\right)$.

Here, the sample median is the 8the observation.

The sample median will not change if the number of observations in the data set does not change.

The observation 0.20 can be raised up to 0.56, as the sample median value is 0.56.

The increment should be in such a way that the number of observations below the median value should not change.

Hence, the observation 0.20 can be increased up to 0.36 without affecting the sample median.