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Hospital Statistics - Math










































The following table shows the LOS for a sample of eleven discharged patients. Using
the data from this table, calculate the mean, range, variance, and standard
deviation. Then answer as well questions a and f
Patient Patient Length of Stays(days)
1 1
2 3
3 5
4 3
5 2
6 29
7 3
8 4
9 2
10 1
11 2
a) Mean =
b) Range =
c) Variance =
d) Standard Deviation =
e) What value is affecting the mean and standard deviation of this sample?
f) Does this mean adequately represent this distribution? If not, what would be a
better measure of central tendency for this data set?


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Apr 24, 2012
Hospital Statistics
by: Staff

Part I

Question:

The following table shows the LOS for a sample of eleven discharged patients. Using
the data from this table, calculate the mean, range, variance, and standard
deviation. Then answer as well questions a and f
Patient Length of Stays (days)

1 1
2 3
3 5
4 3
5 2
6 29
7 3
8 4
9 2
10 1
11 2
a) Mean =
b) Range =
c) Variance =
d) Standard Deviation =
e) What value is affecting the mean and standard deviation of this sample?
f) Does this mean adequately represent this distribution? If not, what would be a
better measure of central tendency for this data set?



Answer:

a) Calculation of the Mean


Population Mean (population average) = µ

µ = the simple average of the number of data points for the entire population (all LOS patients, for all time, for the hospital). This is equal to the sum of all LOS data points for the entire population divided by the number of data points for the entire population, N.

the POPULATION MEAN, µ, DOES NOT APPLY to this problem. The problem statement lists the 11 data points for LOS as a “sample”.


SAMPLE MEAN (sample average) = x̄

x̄ = the sum of all eleven LOS data points in the sample divided by ten (N-1 = 11 - 1 = 10, the number of LOS data points in the sample minus one)

>> the SAMPLE MEAN, x̄, APPLIES to this problem






Data points (sorted, lowest to highest):

1
1
2
2
2
3
3
3
4
5
29

N = 11 sample data points

N - 1 = 10

x̄ (sample average) = (1 + 1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 + 5 + 29) / 10

x̄ (sample average) = (55) / 10

x̄ (sample average) = 5.5 days


b) Calculation of the Range

Range = Highest LOS - Lowest LOS

Range = 29 - 1

Range = 28 days


c) Calculation of the Variance

We have already calculated x̄ (the sample average) to be 5.5 days.


The “variance” is another kind of average. It is the “average of the squared differences from the mean” to the LOS data points.



Population Variance = σ²

σ² = (sum of all the squared differences from the mean, µ, for the entire population/N)

the POPULATION VARIANCE, σ², DOES NOT APPLY to this problem. The problem statement lists the 11 data points for LOS as a “sample”.


Sample Variance = s²

N = sample size = 11

s² = (sum of all the squared differences from the sample mean, x̄)/(N-1)

>> the SAMPLE VARIANCE, s², APPLIES to this problem



1 - 5 = -4

1 - 5 = -4

2 - 5 = -3

2 - 5 = -3

2 - 5 = -3

3 - 5 = -2

3 - 5 = -2

3 - 5 = -2

4 - 5 = -1

5 - 5 = 0

29 - 5 = 24



(-4)² = 16

(-4)² = 16

(-3)² = 9

(-3)² = 9

(-3)² = 9

(-2)² = 4

(-2)² = 4

(-2)² = 4

(-1)² = 1

(0)² = 0

(24)² = 576


s² = (16 + 16 + 9 + 9 + 9 + 4 + 4 + 4 + 1 + 0 + 576)/10

s² = (648)/10

s² = 64.8

--------------------------------------------

Apr 24, 2012
Hospital Statistics
by: Staff


--------------------------------------------

Part II


d) Calculation of the Standard Deviation


Population Standard Deviation = σ

N = the number of data points for the entire population (all LOS patients, for all time, for the hospital)

σ = √(sum of all the squared differences from the mean for the entire population/N)

the POPULATION STANDARD DEVIATION, σ, DOES NOT APPLY to this problem because the problem statement lists the data points as a “sample”



Sample Standard Deviation = s

N-1 = the number of data points for the sample population minus one

s = √[(sum of all the squared differences from the mean for the sample)/(N-1)]

s = √(sample variance)

s = √(s²)


>> the SAMPLE Standard Deviation, s, APPLIES to this problem



s = √(s²)

s = √(64.8)

s = 8.04984


e) What value is affecting the mean and standard deviation of this sample?


LOS = 29, the highest value of LOS, is distorting the mean and standard deviation of this sample.


f) Does this mean adequately represent this distribution? If not, what would be a
better measure of central tendency for this data set?

The mean does not adequately represent this distribution because it is skewed upwards by LOS = 29.

The median is probably a better measure of the central tendency of the data.

1
1
2
2
2
3>>median
3
3
4
5
29



>>> final answers:

a) Sample Mean, x̄ = 5.5 days

b) Range = 28 days

c) Sample Variance, s² = 64.8

d) Sample Standard Deviation, s = 8.04984

e) LOS = 29, the highest value of LOS, is distorting the mean and standard deviation of this sample.

f) The mean, x̄, does not adequately represent this distribution because it is skewed upwards by LOS = 29.

The sample median, M = 3, is probably a better measure of the central tendency of the data.



Thanks for writing.

Staff
www.solving-math-problems.com


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