10. Does not make sense. It could be that a decrease in D causes an increase in C rather than the other way around. Or both variables could be responding to a common underlying cause. Or the correlation could be a coincidence, if the sample is small. (To take an extreme example of this last possibility: if n is just 2, then the two data-points will necessarily lie on a straight line, and if that line happens to have negative slope, then there will be perfect negative correlation between the two variables --- but this will prove nothing, since n is so small as to make the experiment worthless.)
16.
(a) The variables are GPA and hours of TV watched per week.
(b) A line through the data would fit quite well.
The correlation is negative and strong.
(c) Most people who get good grades spend fewer
than 5 hours per week watching television.
28. (a) See the ASCII file I created
(view it in fixed-width mode).
(b) There is a strong negative correlation.
(c) The negative correlation could be explained by asserting
that in households with lower incomes, watching television
is an affordable and hence popular form of entertainment.
34. There is a positive correlation between iced tea sales and ticket sales. This is most likely due to a common underlying cause: warmer weather.
36. There is a positive correlation between the number of priests and attendance at movies. This is most likely due to a common underlying cause: a population increase over the period in question.
Section 6A:
14. The mean is 98.44, obtained by dividing the sum of the six numbers 10. The median is 98.4, obtained by averaging the middle two numbers when arranged in ascending order (the two numbers being averaged happen to both be equal to 98.4!). The mode is also 98.4 because it occurs most frequently.
Section 6B:
18. (a) Beethoven: The mean is 38.78 minutes, or 38.8 minutes,
or 39 minutes (rounding to the nearest minute makes the most sense
to me, since the durations of the individual symphonies were rounded
to the nearest minute); the median is 36 minutes; and the range is
68 – 26 = 42 minutes. Mahler: the mean is 75 minutes (no need
to round); the median is 80 minutes; and the range is 94 – 50 -
44 minutes.
(b) For Beethoven, the low value is 26, the lower quartile is
29 (the average of 28 and 30), the median is 36, the upper quartile
is 45 (the average of 40 and 50), and the high value is 68.
For Mahler, the low value is 50, the lower quartile is 62 (the
average of 52 and 72), the median is 80, the upper quartile is 87.5
(the average of 85 and 90), and the high value is 94.
(Boxplots omitted; too hard in ASCII.)
(c) For Beethoven, the standard deviation is
sqrt(((28-38.78)2+...+(68-38.78)2)/8)
= 13.13 minutes (13.1 and 13 are also acceptable answers).
For Mahler, the standard deviation is
sqrt(((52-75)2+...+(80-75)2)/8)
= 15.44 minutes (15.4 and 15 are also acceptable answers).
(d) The range rule of thumb suggests that the standard
deviation is about one quarter of the range. For the Beethoven
data this is 10.5. For the Mahler data it is 11. So, if you
use the rule of thumb, the ranges will lead you to underestimate
the actual standard deviations. Or, if you use the range rule
of thumb in the other direction, the standard deviations will
lead you to overestimate the ranges of the two data sets.
(e) The length of a typical Beethoven symphony is about
half the length of a typical Mahler symphony, even though they
had about the same amount of variation.
20. (a) Histograms omitted (most of you seem to understand
histograms on the earlier assignments; if you had trouble with
this part, send me email or come see me).
(b)
Set 1: the five numbers are 6, 6, 6, 6, 6.
Set 2: the five numbers are 5, 5, 6, 7, 7.
Set 3: the five numbers are 5, 5, 6, 7, 7.
Set 4: the five numbers are 3, 3, 6, 9, 9.
(c)
For each number in the set, the deviation is the difference
between it and the mean; for example 6 – 6 = 0. To find
the standard deviation for Set 1, sum the squares of the
deviations to get 0, divide by 6 (the size of the data set
minus 1), and then take the square root, obtaining 0.
The standard deviation of Set 2 is about sqrt(2/3), or about
0.8165. The standard deviation for Set 3 is exactly 1. The
standard deviation for Set 4 is exactly 3.
(d)
The standard deviation is a measure of the deviation from the mean.
There is no variation in Set 1, some variation in Sets 2 and 3,
and a large variation in Set 4.
22. Since the store probably charges a fixed price per one-pint serving (I don't know of any ice cream stores where the prices vary randomly!), what is really happening when the price per pint fluctuates is that the serving size fluctuates. A one-pint serving costs $1.75 regardless of who is working in the store that night, but Sam scoops out almost exactly one pint for each one-pint serving (small standard deviation) whereas Kevin's scooping is much less precise (the standard deviation is seven times as great). Kevin's sloppiness will lead to both gratitude (from those who pay for a pint and get more than a pint) and complaints (from those who pay for a pint and get less than a pint). This analysis assumes for simplicity that all customers order exactly one pint; in real life serving-sizes would vary, but the idea is the same.
24. Students convinced me on Wednesday that this is not a fair question. Most investors seeking long-term gains would favor a more a conservative approach (with lower risk), and go with Fund A (which has smaller standard deviation). Likewise, most investors seeking short-term gains would be more tolerant of risk and go with Fund B (which has higher expected growth). But this is not explained in the chapter.
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