Solution to Practice Quiz 2
The full quiz is here. The
answers appear below. Comments, which are not part of the answers, are italicized.
For the batch {1, 2} compute the
- Mean: (1+2)/2 = 1.5.
- Five-letter summary: N=2, so the order of the median is 1h and the depths
of the hinges are (1h + 1)/2 = 1. Whence median = 1.5, upper hinge =
maximum = 2, lower hinge = minimum = 1, and h-spread = range = 1. (Arrange
these in the usual way to exhibit the 5-letter summary.)
- Standard deviation: sqrt(variance) = sqrt(0.5) = 0.71, approx. (see
#4).
- Variance: [(1-1.5)2 + (2-1.5)2]/(2-1) = 0.5.
- Skewness (use the method of moments estimator): The mm estimator of
variance is 0.25 (see #4 and use "2" in the denominator instead of
"2-1"), so the mm sd is sqrt(0.25) = 0.5. The standardized
values are therefore {(1-1.5)/0.5, (2-1.5)/0.5} = {-1, 1}. Their mean
cube is 0.
- Kurtosis (use the method of moments estimator): the fourth powers of the
standardized values are 1, so their mean is 1.
- MAD: the absolute deviations from the median are {0.5, 0.5}; the median of
this set is 0.5.
- Geometric mean: sqrt(1*2) = sqrt(2) = 1.41, approx.
- 25% trimmed mean: trimming does not remove anything from this batch
because N=2 and 25% of 2 is 0.5, which is less than 1. Therefore the
trimmed mean is the mean, which is 1.5.
Consider the batches A = {0, 1, -1}, B = {0, 1,
3}, C = {0, 1, 10}, D = {100, 200, 300}, E = {100,
200, 400}, F = {0, 9, 10}, and G = {1000, 1000, 1000}.
- Order the batches from least to greatest variance.
The variance is unchanged when a constant is added to all values, so the
variance of A is the variance of {0, 1, 2}. Fixing the
first two values and increasing the maximum must increase the variance, so
A < B < C. Similar
reasoning gives D < E. Negating the
values will not change the variance, either, so negating all values in F
and adding 10 demonstrates F = C. The
variance of D must be 100*100 times the variance of {1, 2,
3}, which has the same variance as A (just subtract 2 from
each value). Clearly this is much larger than the variance of A,
B, C, or F. Finally, the
variance of G is zero since all values are the same, but all
other variances are greater than zero. Therefore:
G < A < B < C =
F < D < E.
- Order the batches from least to greatest skewness.
Evidently F has negative skewness and the skewness of G
is not defined. All the other skewnesses are positive. Use the
same reasoning as in #10, but note that skewness depends only on standardized
values in a batch: that is, it is unchanged when a all values are rescaled
or shifted by a constant. We can rescale and shift D
to A (multiply by 0.01 and add -2) and E to B
(multiply by 0.01 and add -1). Therefore
F < A = D <
B = E < C.
- Order the batches from least to greatest kurtosis.
The reasoning is the same as in #11, except to note that F
cannot have negative kurtosis; instead, it must have the same kurtosis as C.
Therefore the correct order is
A = D < B =
E < C = F.
