Given set m: 0, 1, 3, 4, 7 and set n: 1, 2, 3, 4, 5, 9. calculate the mean and mean absolute deviation for both sets. then, answer the following questions. how many data points in set m are closer than one mean absolute deviation from the mean of set m? how many data points in set n are further than two mean absolute deviations from the mean of set n? find the difference between the two means. how does this difference relate to the mean absolute deviation of each set? for example, is it the same as the mean absolute deviation, twice as much, half as much, etc?

For set m, the average is (0 + 1 + 3 + 4 + 7)/5 = 15/5 = 3. the mean absolute deviation can be solved by taking the absolute values of deviations: |0-3| + |1-3| + |3-3| + |4-3| + |7-3| = 3 + 2 + 0 + 1 + 4 = 10, and dividing by 5 to get mean absolute deviation = 2.for set n, the average is (1 + 2 + 3 + 4 + 5 + 9)/6 = 24/6 = 4. the mean absolute deviation can be solved in a similar way as for set m: |1-4| + |2-4| + |3-4| + |4-4| + |5-4| + |9-4| = (3 + 2 + 1 + 0 + 1 + 5)/6 = 2.in set m, the range of 1 mean absolute deviation from the mean is 3 +/- 2, which is from 1 to 5. three of the five points (1, 3 and 4) are within this range.in set n, the range of 2 mean absolute deviations from the mean is 4 +/- 2*2 = 4 +/- 4, which is from 0 to 8. five of the six data points (1, 2, 3, 4, and 5) are within this range.the difference between the means of set n minus set m is 4 - 3 = 1. in this particular case, the difference between the means is half the mean absolute deviation of each set (both of which have a mean absolute deviation of 2).the comparison of the difference between the means versus each group's mean absolute deviation is a way to see whether there is a statistical difference between the two groups. the larger the difference between the means is compared to each group's mean absolute deviation, the larger the probability that the difference between the groups is statistically significant.