Answers for Unit 39

To calculate the mean for this data set we simply:
- Sum up the numbers
  Sum = 222 + 120 + 222 + 176 + 473 + 538 + 456 = 2207 ppb lead
- And divide by the number of numbers
  Avg = 2315 / 7 = 315
To calculate the median of these numbers, we can:
- Sort the numbers from smallest to largest
  120
  176
  222
  222
  456
  473
  538
- And then select the middle number:
  120
  176
  222
  222 <--- the median
  456
  473
  538
  
  So the median for this data set is 222 ppb lead
To calculate the mode of this data, we can:
- Count the number of occurences for each number
  #     Occurences
  ------------------
  120     1
  176     1
  222     2
  456     1
  473     1
  538     1
- And then select the number that occurs most frequently:
  #     Occurences
  ------------------
  120     1
  176     1
  222     2 <------ the mode
  456     1
  473     1
  538     1
  So the mode for this data set is the same as the median: 222 ppb lead
Which measure seems the most appropriate for evaluating a public safety risk? Each measure has its advantages. The mode tells which value is most frequent, which is certainly important for accessing risk, since this is the value most commonly encountered. The median is useful to, since it gives an estimate for the 'middlemost' value that is not overly-influenced by extreme values. The median and the mode for this data set are the same (222 parts per billion of lead) and are quite lower than the mean (315 ppb lead). We might be tempted to believe that some extreme high values contributed to the difference between the median and mean values. Look again at the time-series of values:
```
        Year     Lead (in parts per billion)
        ----     ---------------------------
        1995                 222
        1996                 120
        1997                 222
        1998                 176
        1999                 473
        2000                 538
        2001                 456
```
Notice how all the higher values come in the later years? This does seem to suggest (not prove!) that there has been a shift in the levels of lead since 1999. This apparent shift could be just that - mere appearance - and the diffence between the pre and post plant values could simply be a random occurence. One way of thinking quantitatively about these differences is to use the variance and the standard deviation to determine how large a typical difference might be. We'll do this in the next section:
To calculate the variance of this data, we can:
- Calculate the difference between each value and the mean
  222 - 315 = -93
  120 - 315 = -195
  222 - 315 = -93
  176 - 315 = -139
  473 - 315 = 158
  538 - 315 = 223
  456 - 315 = 141
- And then square each of these values:
  -93 * -93 = 8649
  -195 * -195 = 38025
  -93 * -93 = 8649
  -139 * -139 = 19321
  158 * 158 = 24964
  223 * 223 = 49729
  141 * 141 = 19881
- And then sum these squared values:
  Sum = 8649 + 38025 + 8649 + 19321 + 24964 + 49729 + 19881 = 169218 ppb lead
- And then divide this sum by the number of numbers minus one:
  Variance = 169218 / (7-1) = 28203 ppb lead
The variance can be difficult to interpret, especially since is units will be the square of the original variable's units. A more easily understood statistic is the standard deviation which we can calculate by:
Hmmmmmmm. The mean lead level for the first four years was (222+120+222+176)/4 = 185, and the mean lead level for the last 3 years (after the plant was in operation) was (473+538+456)/3 = 489. So the amount of lead in the years following the building of the plant was more than 2 and a half times it's pre-1999 levels. Furthermore, the size of the jump between pre and post plant means was 489 - 185 = 304 parts per billion of lead, or almost 2 whole standard deviations. This is unlikely to occur totally by chance, and we suggest that further reasearch into the Romullus and Remus lead pipe factory is in order.