Download Friday, September 19 th

Solutions: Homework #1
Spring, 2003
1) The independent variable in the study was exercise, which was operationally defined as
the number of hours one spent in the gym. The dependent variable was depression, which
was operationally defined as self-reported ratings on a 10-point scale. The experiment
was observational in that the researchers merely measured on-going behavior. They did
not attempt to control either exercise or depression, nor were the subject randomly
assigned to conditions (exercise vs. no exercise). Therefore, the researchers cannot make
cause-and-effect conclusions regarding their data because other explanations are possible.
For example, depressed people may have less energy, which makes it less likely that they
will go to the gym. Or, it could be that people who are disposed towards regular exercise
are less disposed to depression.
2)
3)
Day
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sum
Hrs.
Sleep
x
2
6
7
7
6
7
4.5
8
36
49
49
36
49
20.25
64
45.5 303.25
x
6.5
6.5
6.5
6.5
6.5
6.5
6.5
x  x  x  x 2
-0.5
0.5
0.5
-0.5
0.5
-2
1.5
0.25
0.25
0.25
0.25
0.25
4
2.25
7.5
Mean =
=
=
(x) / 7
45.5 / 7
6.5
Median
=
4.5
6
6
7
7
7
8
There are seven observations, so we select the [(7/2) + ½]= 4th observation.
The 4th observation is 7(highlighted in red).
Mode =
Also equals 7; it occurs three times.
Short cut
s
2
=
=
=
=
=
=
=
s
((x ) – [(x) /n]) / n-1
[303.25 – (45.52 / 7)] / 6
[303.25 – (2070.25 / 7)] / 6
(303.25 – 295.75) / 6
7.5 / 6
1.25 (variance)
1.12 (standard deviation)
2
2
If you did it the long way, you would divide the sum of the squared deviations (7.5) by 6
to get the variance, and then take the square root to get the standard deviation.
4)
Statistics
N
Bed_WD
Bed_WE
Valid
124
124
Missing
0
0
Mean
12.7040
14.3944
Median
12.5000
14.0000
Mode
13.00
14.00
Std. Deviation
1.36894
1.09233
Variance
1.874
1.193
I assumed that most of you would use half-hour classes, so I decided to use one-hour
classes. Not just to be unique, but so that you could see how the graphs look different
using different class width. I daresay that the one-hour bin graph looks prettier than the
half-hour bin graph (although I recognize that ‘pretty’ probably is not the right word
here).
As for the skewness, I think you would have to conclude that the data were positivelyskewed (outliers toward the later end of the scale). This judgment is based on two
factors: the means for both data sets are greater than the medians (see Table above).
Because the mean is more sensitive to outliers than the median, it is pulled in the
direction of the skew (in this case ‘up’). As well, looking at the graphs, especially the
graph for the weekend bedtimes, there is a certain positive skewishness quality to them. I
wouldn’t describe either data set as highly skewed (the weekday bedtime is marginally
skewed at best), but given the relative values of the mean and median, let’s go with
positively skewed.