Solutions Homework #1

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) Hrs. Day Sleep x2 Sunday 6 36 Monday 7 49 Tuesday 7 49 Wednesday 6 36 Thursday 7 49 Friday 4.5 20.25 Saturday 8 64 Sum 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 2 2 ((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) 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.