HSA 523 Homework #3
Answer Key
Dr. Robert Jantzen
Economics Department
1. (generate the statistics w/ SPSS by using <Analyze><DescriptiveStats><Frequencies>
and clicking on the <Statistics> tab to select the statistics you want.)
Statistics
N
Valid
Missing
average cost
per licensed
bed (1000s)
15
full-time staff
per bed
15
0
0
Mean
238.47
3.533
Median
238.00
3.600
Std. Deviation
33.768
.4030
Minimum
165
2.9
Maximum
293
4.3
25
228.00
3.200
50
238.00
3.600
75
264.00
3.900
Percentiles
a. Mean measures the middle # w/ the average.
b. Median is the # exactly in the middle of an ordered series of #s.
c. The first quartile divides the bottom 25% of the #s from the top 75%.
The third quartile divides the bottom 75% from the top 25%.
d. The five # summaries show the minimum, first quartile, median, third quartile &
maximum. They also show how wide the intervals are that capture each quarter of all the
#s. If the middle two intervals are equally wide and the outer two are also equally wide,
the #s are symmetrically distributed.
e. The interquartile range (third quartile – first quartile) shows the interval that captures
the middle 50% of the #s.
(IQRs are 36 & .7)
f. The sample standard deviation shows how much the #s vary from the mean #.
g. The coefficient of variation shows how large the standard deviation is, as a % of the
mean. (CVs are .142 = 14.2% & .114 = 11.4%)
h. The pearson skewness coefficient shows whether the # series is skewed (absolute
value for pearson > .1 indicates skewed #s). (pearsons are .014 & -.166)
2. Box plot shows that although chicken dogs generally have fewer calories, the top 25%
of chicken dogs have more calories than the bottom 25% of either meat or beef dogs.
200.00
calories per hot dog
180.00
160.00
140.00
120.00
100.00
80.00
beef
meat
chicken
type of hot dog
3. (Generate the boxplots & histograms by using <Graphs><Interactive> & then the type
of graph you want to generate). Both chart types indicate that the number of beds and the
average cost of care are skewed to the high numbers.
30%
Percent
Percent
12%
8%
4%
20%
10%
0%
0%
250.00
500.00
750.00
1000.00
1000.00



1000.00

















750.00
2000.00
3000.00
av gcost- av g cost/da y routine care (ahd)
avgcost- avg cost/day routine care (ahd)
bedsahd - total beds (ahd)
bedsahd - total beds (ahd)
500.00
250.00
0.00
3000.00

2000.00




1000.00











(Generate the statistics w/ SPSS by using <Analyze><DescriptiveStats><Frequencies>
and clicking on the <Statistics> tab to select the statistics you want. Also turn off
<Display frequency tables> otherwise you’ll generate a lot of “output.”)
bedsahd total beds
(ahd)
N
Valid
Missing
Mean
Median
717
avgcost- avg
cost/day
routine care
(ahd)
717
0
0
199.1339
454.2999
154.0000
423.0000
165.20221
191.22003
Minimum
12.00
166.00
Maximum
1068.00
3006.00
Std. Deviation
Percentiles
25
78.5000
357.0000
50
154.0000
423.0000
75
276.0000
497.5000
a. 5 # summaries can be found above.
b. mean & median are above. Since the means are bigger than the medians, the #s could
be skewed to the high numbers.
c. the pearson skewness coefficients are .27 for beds size & .164 for average cost (mean
– median divided by std.dev.)
d. the IQRs are 197.5 & 140.5 (3rd Quartile – 1st Quartile). Shows the interval that
contains the middle 50% of the #s.
e. std. deviations are above. Use if #s are not skewed.
f. the coefficients of variation are .83 or 83% for beds and .421 or 42.1% for average cost
(CV = Std.Dev./Mean). Use if #s are not skewed.
4. If the mean cholesterol count is 280 w/ a standard deviation of 25:
a. for any kind of # distribution we can use Chebyshev’s Rule which says that at least
75% of the #s are within 2 SDs of the mean (between 230 and 330) and that at least 89%
are within 3 SDs of the mean (between 205 and 355).
b. for a symmetrical bell-shaped distribution we can use the Empirical Rule which says
that about 68% of the #s are within 1 SD of the mean (between 255 and 305), about 95%
are within 2 SDs (between 230 and 330) and nearly all (99.7%) are within 3 SDs of the
mean (between 205 and 355).
5. Given mean CEO pay of 300K w/ a SD of 40K and mean LPN pay of 40K w/ a SD of
2K:
a. a CEO earning 260K has a z score of -1 [=(260-300)/40] while an LPN earning 36K
has a z score of -2 [=(36-40)/2]. The LPN is more underpaid because (s)he is 2 standard
deviations below average while the CEO is only 1 SD below average.
b. in order to use the mean and SD in calculations, we must assume that the earnings
distributions for CEOs and LPNs are both bell-shaped and symmetrical.