The minimum injection pressure (psi) for injection molding specimens of high armylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty),

(a) The sample mean can be found by adding up all the values and dividing by the sample size: {eq}bar x = \frac{14.7+12.9+17.9+14.1+12.3+16.4+9.1+8.0}{8} = 13.575 {/eq}. The sample median is the middle value when the data is arranged in order: {eq}bar x = 13.5 {/eq}, as the fourth and fifth values (12.3 and 14.1) average to 13.2 and are in the middle. To find the {eq}12.5 % {/eq} trimmed mean, we first remove the lowest and highest 12.5% of the data (i.e. one value from each end). That leaves us with the following remaining six values: 14.7, 12.9, 17.9, 14.1, 16.4, and 9.1. The trimmed mean is then the average of these values: {eq}bar x_{tr} = \frac{14.7+12.9+17.9+14.1+16.4+9.1}{6} = 14.183 {/eq}. (b) To compare these values, we can consider how far apart they are from each other. The mean is slightly larger than the trimmed mean, but not by much. However, the median is noticeably larger than both the mean and the trimmed mean. This suggests that there is some negative skewness to the data set (i.e. some values are dragging the median higher than the "typical" value). (c) To determine the maximum increase that would not affect the median, we need to first figure out which value(s) in the data set are at the midpoint. Since we have an even number of observations, there are two values at the middle: 12.9 and 14.1. If we increase the smallest observation (8.0), it will not affect the median as long as it stays below 12.9. Thus, the smallest observation could be increased by up to {eq}12.9-8.0 = 4.9 {/eq} without affecting the sample median.