Question

Effects of an Outlier Listed below are platelet counts (1000 cells>mL) from subjects included in Data Set 1 “Body Data.” Identify the outlier and then comment on the effect it has on the mean and standard deviation by finding the values of those statistics with the outlier included and then with the outlier excluded. 263 206 185 246 188 191 308 262 198 253 646

Short answer

The summarized results from the data are shown below.

	Sample mean	Sample standard deviation
Including outlier	267.8	131.6
Excluding outlier	230.0	42.0

Both measures have changed by a large margin.

Step-by-step solution

Given information

The platelet count of different subjects is in terms of $\left(1000\text{cells}/\mu L\right)$ .

263 206 185 246 188 191 308 262 198 253 646

Formula for mean and sample standard deviation

For n observations denoted by x, the mean value is computed using the following formula.

$\overline{x}=\frac{\sum x}{n}$

The sample standard deviation is computed using the following formula.

$s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}}$

Compute the values of mean and sample standard deviation

Substitute the value for mean as follows.

$$\begin{gathered} \overline{x}=\frac{263+206+185+...+646}{11} \\ =\frac{2946}{11} \\ =267.8182 \end{gathered}$$

Thus, the sample mean value is 267.8$\left(1000\text{cells}/\mu L\right)$.

Substitute the values for sample standard deviation as follows.

$$\begin{gathered} s=\sqrt{\frac{{\left(263-267.8\right)}^2+{\left(206-267.8\right)}^2+...+{\left(646-267.8\right)}^2}{11-1}} \\ =\sqrt{\frac{23.21+3821.49+...+143021.5}{10}} \\ =\sqrt{\frac{173215.64}{10}} \\ =131.6114 \end{gathered}$$

Thus, the sample standard deviation is 131.6 $\left(1000\text{cells}/\mu L\right)$.

Identify the outlier

Outliers are one or more observations that are unusual from the set of observations. It can be termed as extreme among other numerical values.

In the given set of observations, 646 is the outlier as it is extreme compared to the range of the dataset from 185 to 308.

Compute the mean and the sample standard deviation value, excluding the outlier

Substitute the values for mean excluding the outlier 646 as follows.

$$\begin{gathered} \overline{x}=\frac{263+206+185+...+253}{11} \\ =\frac{2300}{10} \\ =230 \end{gathered}$$

Thus, the sample mean value is 230.0$\left(1000\text{cells}/\mu L\right)$.

Substitute the values for sample standard deviation as follows.

$$\begin{gathered} s=\sqrt{\frac{{\left(263-230\right)}^2+{\left(206-230\right)}^2+...+{\left(253-230\right)}^2}{10-1}} \\ =\sqrt{\frac{1089+576+...+529}{9}} \\ =\sqrt{\frac{15892}{9}} \\ =42.02 \end{gathered}$$

Thus, the sample standard deviation is 42.0 $\left(1000\text{cells}/\mu L\right)$.

Compare the mean and sample standard deviation values 

The summary result is shown below.

	Sample mean	Sample standard deviation
Including outlier	267.8	131.6
Excluding outlier	230.0	42.0

Thus, both measures have changed significantly.

Specifically, the sample standard deviation has reduced by a vast measure.