Question

Mercury is a persistent and dispersive environmental contaminantfound in many ecosystems around the world.When released as an industrial by-product, it often finds itsway into aquatic systems where it can have deleteriouseffects on various avian and aquatic species. The accompanyingdata on blood mercury concentration (mg/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).

.20 .22 .25 .30 .34 .41 .55 .56

1.42 1.70 1.83 2.20 2.25 3.07 3.25

a. Determine the values of the sample mean and sample median and explain why they are different. (Hint:\(\sum {{{\bf{x}}_{\bf{1}}}{\bf{ = 18}}{\bf{.55}}} \))

b. Determine the value of the 10% trimmed mean and compare to the mean and median.

c. By how much could the observation .20 be increased without impacting the value of the sample median?

Short answer

1. The sample mean for x is 1.237 mg/g.The median value of x is 0.56 mg/g. The two values are different due to positive skewness in the distribution of values.
2. The 10% trimmed mean is 1.118mg/g. The value lies between median and mean value of 0.56 and 1.237 respectively.
3. The observation .20 can be increased to 0.36 without impacting the value of the sample median.

Step-by-step solution

Given information

The data on the blood mercury concentration (mg/g) for adult females near contaminated rivers in Virginia is provided.

Compute the sample mean and sample median

1. Let x represents the blood mercury concentration (mg/g) for adult females near contaminated rivers in Virginia.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.20 + 0.22 + 0.25 + ... + 3.25}}{{15}}\\ &=& \frac{{18.55}}{{15}}\\ &=& 1.237\end{array}\)

Thus, the sample mean for x is 1.237 mg/g.

For the odd number of observations, the median value is computed as,

\(\begin{array}{c}\tilde x &=& {\left( {\frac{{n + 1}}{2}} \right)^{th}}{\rm{ordered}}\;{\rm{value}}\\ &=& {\left( {\frac{{15 + 1}}{2}} \right)^{th}}{\rm{ordered}}\;{\rm{value}}\\ &=& {\left( 8 \right)^{th}}{\rm{ordered}}\;{\rm{value}}\end{array}\)

Thus, the median value of x is 0.56.

The sample mean represents the average of values

The sample median separates the data into equal halves.

The two values are different in magnitude because the distribution of values is not symmetric but positively skewed.

Compute the 10% trimmed mean

b.

10% trimmed mean is the mean value obtained after removing 10% of the extreme (largest and smallest value each) from the data.

In this case, 10% of 15 is 1.5. Thus, the following procedure is carried out to compute trimmed mean.

A 10% trimmed mean is computed by removing the smallest and the largest observation from the data.

Remove one extreme value from each end to obtain the data as follows,

0.22

0.25

0.3

0.34

0.41

0.55

0.56

1.42

1.7

1.83

2.2

2.25

3.07

The sample mean is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr1} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.22 + 0.25 + ... + 3.07}}{{13}}\\ &=& 1.162\end{array}\)

Now, again remove the smallest and largest value from the remaining dataset.

The data after removing is as follows,

0.25

0.3

0.34

0.41

0.55

0.56

1.42

1.7

1.83

2.2

2.25

The sample mean is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr2} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.25 + 0.3 + ... + 2.25}}{{11}}\\ &=& 1.0736\end{array}\)

Now, the 10% trimmed mean for the data is computed as,

\(\begin{array}{c}{{\bar x}_{tr}} &=& \frac{{{{\bar x}_{\left( {tr1} \right)}} + {{\bar x}_{\left( {tr2} \right)}}}}{2}\\ &=& \frac{{1.162 + 1.0736}}{2}\\ &=& 1.118\end{array}\)

Thus, the 10% trimmed mean is 1.118.

Therefore, it can be observed that the trimmed mean (1.118) falls between the median value (0.56) and the mean value (1.237).

Obtain the maximum value to which 0.20 can be increased to keep sample median constant

The median value of x is 0.56.

The value that the observation .20 be increased without impacting the value of the sample median is computed as,

\(\begin{array}{c}\frac{{0.20}}{{0.56}} = 0.357\\ \approx 0.36\end{array}\)

Thus, the observation .20 can be increased to 0.36 without impacting the value of the sample median.