Question

which is more affected by extreme observations, the mean or median? and how about the standard deviation or iqr?

mean, standard deviation

mean, iqr

median, standard deviation

median, iqr

Explanation:

• Mean vs Median: The mean is calculated by summing all values and dividing by the number of values. Extreme values (very large or very small) will pull the mean in their direction. The median is the middle - value when data is ordered. Extreme values on either end do not affect the position of the middle - value (as long as the number of data points is fixed). For example, in the data set \(1,2,3\), the mean is \(\frac{1 + 2+3}{3}=2\) and the median is \(2\). If we add an extreme value like \(100\) to get the data set \(1,2,3,100\), the mean is \(\frac{1+2 + 3+100}{4}=26.5\) and the median is \(\frac{2 + 3}{2}=2.5\).
• Standard Deviation vs IQR: The standard deviation formula is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\). Extreme values will increase the sum \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\), thus increasing the standard deviation. The inter - quartile range (IQR) is \(Q_{3}-Q_{1}\), where \(Q_{1}\) is the first quartile (25th percentile) and \(Q_{3}\) is the third quartile (75th percentile). Extreme values in the tails of the data distribution do not affect the calculation of the quartiles (since quartiles are based on the position of data points, not their actual magnitudes). For example, in the data set \(1,2,3,100\), \(Q_{1}=1.5\), \(Q_{3}=99.5\) (using one method of calculating quartiles for a small data set), \(IQR=99.5 - 1.5=98\). If we remove the \(100\) (data set \(1,2,3\)), \(Q_{1}=1.5\), \(Q_{3}=2.5\), \(IQR = 1\). But if we change the extreme value (say to \(200\) in the data set \(1,2,3,200\)), \(Q_{1}=1.5\), \(Q_{3}=199.5\), \(IQR=198\). The change in the extreme value has a much larger impact on the standard deviation (which is sensitive to the distance of each data point from the mean) than on the IQR.

Answer:

mean, standard deviation