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HSA 523 Homework #5
Answer Key
Dr. Robert Jantzen
Economics Department
1.
A. The occupancy rate that places a hospital in the top 10% (at the 90th
percentile) is 77.84% (see Excel calculation below).
B. If St.Elsewhere is at 50%, it is .215 standard deviations below average (see Z
score of -.215 from Excel calculation below). 41.5% of hospitals have rates below it.
Common Data
Mean
Standard
Deviation
54
18.6
Find X and Z Given Cum. Pctage.
Cumulative Percentage
90.00%
Z Value
1.281552
X Value
77.83687
Probability for X <=
X Value
50
Z Value
-0.215054
P(X<=50)
0.4148627
2.
Given mean of 5 and standard deviation of .05 mg., probability that the content
will be less than 4.9 is .02275, so 2.28% of the capsules will contain < 4.9 mgs. The
probability that the content will be > 5.2 is 0 so none of the capsules will be > 5.2 mgs.
Common Data
Mean
Standard
Deviation
5
0.05
Probability for X <=
X Value
4.9
Z Value
-2
P(X<=4.9)
0.0227501
Probability for X >
X Value
5.2
Z Value
4
P(X>5.2)
0.0000
3.
a. The probability that the duration is between 250 and 300 days is .8246, so
82.46% of the pregnancies will be there.
b. The top 5% duration (95th percentile) is 292.32 days and the top 1% duration
th
(99 percentile) is 303.22 days.
c. The probability that the duration will be > 275 days is .2869 so only 28.69% of
pregnancies will be covered.
Common Data
Mean
Standard Deviation
266
16
Probability for X >
X Value
275
Z Value
0.5625
P(X>275)
0.2869
Probability for a Range
From X Value
250
To X Value
300
Z Value for 250
-1
Z Value for 300
2.125
P(X<=250)
0.1587
P(X<=300)
0.9832
P(250<=X<=300)
0.8246
Find X and Z Given Cum. Pctage.
Cumulative Percentage
95.00%
Z Value
1.644853
X Value
292.3177
Find X and Z Given Cum. Pctage.
Cumulative Percentage
99.00%
Z Value
2.326347
X Value
303.2216
4.a. To test whether a set of numbers are normally distributed you can assess whether the
#s conform to the theoretical properties of a normal distribution which state that:
 Bell shaped symmetrical distribution
 Mean = median
 Interquartile Range = 1.33 times the Standard Deviation (SD)
 #s follow the 68 – 95 – 99.7 rule, i.e., 68% are within 1 SD of the mean, 95% are
within 2 SDs, and 99.7% are within 3 SDs.
An alternative test is to generate a normal probability plot. If the plot is a straight line,
the #s are normally distributed.
4.b. (generate the statistics w/ SPSS by using <Analyze><DescriptiveStats>
<Frequencies> or <Analyze><DescriptiveStats><<Explore>
Statistics
# of staffed services
N
Valid
Missing
Descriptives
Statistic
23.3400
50
0
Mean
23.3400
Median
23.0000
Std. Deviation
6.53596
# of staffed
services
Mean
95%
Confidence
Interval for
Mean
Lower
Bound
Upper
Bound
21.4825
25.1975
Minimum
11.00
5% Trimmed Mean
Maximum
23.3556
36.00
Median
23.0000
Percentiles
25
19.0000
Variance
50
23.0000
Std. Deviation
75
28.0000
Minimum
11.00
Maximum
36.00
Range
25.00
Interquartile Range





Std.
Error
.92432
42.719
6.53596
9.00
A boxplot shows the #s clustered around the middle and symmetrically
distributed.
The mean is about equal to the median
The pearson skewness measure is (23.34-23)/6.54 = .052 which is indicative of a
symmetrical distribution.
The IQR is 9 while 1.33 * the SD is 8.7
The mean +- 1 SD = 16.8 thru 29.9 captures 62% (31 of 50) of the #s (versus
68%). The mean +- 2 SDs = 10.3 thru 36.4 captures 94% (49 of 50) of the #s
(versus 95%). The mean +- 3 SDs = 3.75 thru 42.9 captures 100% of the #s
(versus 99.7%)
Hence it appears the #s have close to a normal distribution.
The normal probability plot plots the actual z values for each # against the expected z
values. It’s a pretty straight line indicating a normal distribution.
# of staffed services
35.00
30.00
25.00
20.00
15.00
Normal Probability Plot of # of staffed services
4
Expected Normal Value
2
0
-2
-4
-4
-2
0
Standardized Observed Value
2
4
5. Given a failure rate of .2 for the hip replacement parts and a sample size of 10.
Binomial Probabilities: This spreadsheet (adapted from PHSTAT2) makes some
common calculations for outcomes that can be modeled with the binomial distribution.
NOTES: Edit the values in BLUE to reflect the sample size and
probability of success in your sample. (Note: the sample size
must be <= 20). The spreadsheet will then calculate the
probability of observing exactly X successes in the sample,
as well as <= X successes, > X successes, etc.
Data
Sample size
Probability of
success
10
0.2
The probability that 0 of 10 fail is .107374
The probability that >2 fail is .3222
The probability that all 10 fail is 1.02 E-07 =
.000000102
Statistics
Mean
Variance
Standard deviation
2
1.6
1.264911
Binomial Probabilities Table
X
0
1
2
3
4
5
6
7
8
9
10
P(X)
0.107374
0.268435
0.30199
0.201327
0.08808
0.026424
0.005505
0.000786
7.37E-05
4.1E-06
1.02E-07
P(<=X)
0.107374
0.37581
0.6778
0.879126
0.967207
0.993631
0.999136
0.999922
0.999996
1
1
P(<X)
0
0.107374
0.37581
0.6778
0.879126
0.967207
0.993631
0.999136
0.999922
0.999996
1
P(>X)
0.892626
0.62419
0.3222
0.120874
0.032793
0.006369
0.000864
7.79E-05
4.2E-06
1.02E-07
0
P(>=X)
1
0.892626
0.62419
0.3222
0.120874
0.032793
0.006369
0.000864
7.79E-05
4.2E-06
1.02E-07
Note: binomial model can be used when:
 Drawing a random sample for a binary outcome (yes/no, works/doesn’t,
lives/dies, etc.)
 The probability of “success” remains the same across sampled observations. This
generally holds true if the sample size is <= .05 of the population size, or if
sampling w/ replacement.
The binomial model calculates the probability of observing any number (X) of
“successes” within the drawn sample.