Question

a key hospital outcome measure of clinical performance is length of stay (los), that is, the number of days a patient is hospitalized. for patients at one hospital with acute myocardial infarction (heart attack), the length of stay over the past four years has consistently decreased. the hospital also has data for various treatment options such as the percentage of patients who received aspirin upon arrival and cardiac medication for left ventricular systolic dysfunction (lvsd). the data are as follows:
year average los aspirin on arrival lvsd medication
2007 4.28 days 92% 90%
2008 4.16 days 97% 94%
2009 4.34 days 99% 97%
2010 4.17 days 100% 98%
calculate descriptive statistics for the data. do not round intermediate calculations. round your answers to four decimal places, when needed.
descriptive statistics: los, aspirin, lvsd
variable mean standard error standard deviation minimum median maximum
los (days)
aspirin (%)
lvsd (%)
calculate correlations for the data. do not round intermediate calculations. use a minus sign for negative values, if any. round your answers to three decimal places.
correlations: los, aspirin, lvsd
los aspirin
aspirin
lvsd

Explanation:

Step1: Calculate the mean of LOS

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For LOS values $x = [4.28,4.16,4.34,4.17]$, $n = 4$, $\sum_{i=1}^{4}x_{i}=4.28 + 4.16+4.34 + 4.17=16.95$, so $\bar{x}_{LOS}=\frac{16.95}{4}=4.2375$.

Step2: Calculate the mean of Aspirin

For Aspirin values $x = [92,97,99,100]$, $\sum_{i = 1}^{4}x_{i}=92 + 97+99+100 = 388$, so $\bar{x}_{Aspirin}=\frac{388}{4}=97$.

Step3: Calculate the mean of LVSD

For LVSD values $x=[90,94,97,98]$, $\sum_{i = 1}^{4}x_{i}=90 + 94+97+98 = 379$, so $\bar{x}_{LVSD}=\frac{379}{4}=94.75$.

Step4: Calculate the standard - deviation of LOS

The formula for the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
$(4.28-4.2375)^{2}=0.00180625$, $(4.16 - 4.2375)^{2}=0.00600625$, $(4.34-4.2375)^{2}=0.01050625$, $(4.17-4.2375)^{2}=0.00455625$.
$\sum_{i = 1}^{4}(x_{i}-4.2375)^{2}=0.00180625+0.00600625 + 0.01050625+0.00455625=0.022875$, $s_{LOS}=\sqrt{\frac{0.022875}{3}}\approx0.0871$.

Step5: Calculate the standard - deviation of Aspirin

$(92 - 97)^{2}=25$, $(97-97)^{2}=0$, $(99 - 97)^{2}=4$, $(100 - 97)^{2}=9$.
$\sum_{i = 1}^{4}(x_{i}-97)^{2}=25 + 0+4+9 = 38$, $s_{Aspirin}=\sqrt{\frac{38}{3}}\approx3.5590$.

Step6: Calculate the standard - deviation of LVSD

$(90 - 94.75)^{2}=22.5625$, $(94 - 94.75)^{2}=0.5625$, $(97 - 94.75)^{2}=5.0625$, $(98 - 94.75)^{2}=10.5625$.
$\sum_{i = 1}^{4}(x_{i}-94.75)^{2}=22.5625+0.5625 + 5.0625+10.5625=38.75$, $s_{LVSD}=\sqrt{\frac{38.75}{3}}\approx3.5972$.

Step7: Calculate the standard error

The standard error $SE=\frac{s}{\sqrt{n}}$. For LOS, $SE_{LOS}=\frac{0.0871}{\sqrt{4}}\approx0.0436$. For Aspirin, $SE_{Aspirin}=\frac{3.5590}{\sqrt{4}}\approx1.7795$. For LVSD, $SE_{LVSD}=\frac{3.5972}{\sqrt{4}}\approx1.7986$.

Step8: Find the minimum, median and maximum of LOS

The minimum of $[4.28,4.16,4.34,4.17]$ is $4.16$, the median of $[4.16,4.17,4.28,4.34]$ is $\frac{4.17 + 4.28}{2}=4.225$, the maximum is $4.34$.

Step9: Find the minimum, median and maximum of Aspirin

The minimum of $[92,97,99,100]$ is $92$, the median of $[92,97,99,100]$ is $\frac{97+99}{2}=98$, the maximum is $100$.

Step10: Find the minimum, median and maximum of LVSD

The minimum of $[90,94,97,98]$ is $90$, the median of $[90,94,97,98]$ is $\frac{94 + 97}{2}=95.5$, the maximum is $98$.

Step11: Calculate the correlation between LOS and Aspirin

The formula for the correlation coefficient $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x}_{x})(y_{i}-\bar{y}_{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x}_{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y}_{y})^{2}}}$.
Let $x$ be LOS and $y$ be Aspirin.
$(4.28-4.2375)(92 - 97)=-0.2125$, $(4.16-4.2375)(97 - 97)=0$, $(4.34-4.2375)(99 - 97)=0.205$, $(4.17-4.2375)(100 - 97)=-0.2025$.
$\sum_{i = 1}^{4}(x_{i}-4.2375)(y_{i}-97)=-0.2125+0 + 0.205-0.2025=-0.21$.
$\sum_{i = 1}^{4}(x_{i}-4.2375)^{2}=0.022875$, $\sum_{i = 1}^{4}(y_{i}-97)^{2}=38$.
$r_{LOS - Aspirin}=\frac{-0.21}{\sqrt{0.022875\times38}}\approx - 0.220$.

Step12: Calculate the correlation between LOS and LVSD

Let $x$ be LOS and $y$ be LVSD.
$(4.28-4.2375)(90 - 94.75)=-0.2025$, $(4.16-4.2375)(94 - 94.75)=0.058125$, $(4.34-4.2375)(97 - 94.75)=0.230625$, $(4.17-4.2375)(98 - 94.75)=-0.219375$.
$\sum_{i = 1}^{4}(x_{i}-4.2375)(y_{i}-94.75)=-0.2025+0.058125+0.230625 - 0.219375=-0.133125$.
$\sum_{i = 1}^{4}(x_{i}-4.2375)^{2}=0.022875$, $\sum_{i = 1}^{4}(y_{i}-94.75)^{2}=38.75$.
$r_{LOS - LVSD}=\frac{-0.133125}{\sqrt{0.022875\times38.75}}\approx - 0.143$.

Step13: Calculate the correlation between Aspirin and LVSD

Let $x$ be Aspirin and $y$ be LVSD.
$(92 - 97)(90 - 94.75)=23.75$, $(97 - 97)(94 - 94.75)=0$, $(99 - 97)(97 - 94.75)=4.5$, $(100 - 97)(98 - 94.75)=9.75$.
$\sum_{i = 1}^{4}(x_{i}-97)(y_{i}-94.75)=23.75+0+4.5 + 9.75=38$.
$\sum_{i = 1}^{4}(x_{i}-97)^{2}=38$, $\sum_{i = 1}^{4}(y_{i}-94.75)^{2}=38.75$.
$r_{Aspirin - LVSD}=\frac{38}{\sqrt{38\times38.75}}\approx0.989$.

Answer:

Variable	Mean	Standard Error	Standard Deviation	Minimum	Median	Maximum
Aspirin (%)	97	1.7795	3.5590	92	98	100
LVSD (%)	94.75	1.7986	3.5972	90	95.5	98
Correlations: LOS, Aspirin, LVSD	LOS	Aspirin
Aspirin	-0.220	1
LVSD	-0.143	0.989	1