Introduction

• Increased number of patients admitted annually has been a major concern in Australia’s Healthcare.

• According to the Australian Institute of Health and Welfare, 11.3 million patients were admitted in public and private hospitals in 2017-18.

• Out of those 61% were in public hospitals with 42% cases being emergency admissions followed by childbirth and medical hospitalisations, whereas in private hospitals, more than 50% cases were surgical and mental health care.

• The length of stay (LOS) is a significant indicator of the performance in hospital management that influences the capacity of the health system, both in terms of bed availability and cost.

• Different hospitals should be able to treat a similar patient in the same amount of time. However, it might not be the case every time.

Problem Statement

Are patients discharged soon if they choose to stay in a medium over large hospitals?

• This investigation aims to identify whether there is any statistically significant difference in the average length of stay between large and medium hospitals.

• Assuming there is no difference in the length of stay on an average, we try to find evidence using hospitals peer-group particularly: large hospitals and medium hospitals over the average length of stay (ALOS).

Data

• The data was collected by the Australian Institute of Health and Welfare that represents records of admitted patients in the year 2017 - 2018 and is publicly available.

• Variable of interest
  • Hospital’s peer group(large and medium hospitals)
  • The average length of stay (ALOS)

• The average length of stay (ALOS) is calculated as the number of bed days for overnight stays divided by the number of overnight stays and is reported for selected.

• After loading data, Peer groups are converted to factors and NP values in Average length of stay (ALOS) are converted to NA values and are ignored for this analysis.

• For Peer group, large and medium hospitals are selected.

• Data Source: https://www.aihw.gov.au/reports-data/myhospitals/sectors/admitted-patients

Descriptive Statistics

Looking at the summary statistics, the average length of stay in a large hospital is 3.99 days with a standard deviation of 1.98 whereas in medium hospital, the average length of stay is 3.7 days with a standard deviation of 1.85.

hosp_summary <- data %>% filter(`Peer group`=="Large hospitals"|`Peer group`=="Medium hospitals")%>% 
          group_by(`Peer group`) %>% 
          summarise(
                  Mean = mean(`Average length of stay (days)`,na.rm=T),
                  S.D = sd(`Average length of stay (days)`,na.rm=T),
                  First_quartile = quantile(`Average length of stay (days)`,0.25,na.rm=T),
                  Third_quartile = quantile(`Average length of stay (days)`,0.75,na.rm=T),
                  Min = min(`Average length of stay (days)`,na.rm=T),
                  Max = max(`Average length of stay (days)`,na.rm=T),
                  Missing = sum(is.na(`Average length of stay (days)`)))

knitr::kable(hosp_summary,caption="Summary Statistics")

Data Visualisation I

The histogram shows that several patients were discharged within 0 to 5 days. However, there are longer durations of stay recorded which indicates the distribution is skewed to the right.

ggplot(data = data, aes(`Average length of stay (days)`)) +
  geom_histogram(bins=22) + 
  ylab('Frequency') + 
  ggtitle('Histogram of Average length of Stay (in days)')

Data Visualisation II

• From the visualisation, it is clear that large hospitals have a higher average length of stay (ALOS) compared to medium hospitals. An independent sample t-test is performed to identify the statistical significance and validate the assumption.

medium_large <- data  %>% filter(data$`Peer group`=='Large hospitals'|data$`Peer group`=='Medium hospitals')

ggplot(medium_large,aes(`Peer group`,`Average length of stay (days)`))+ 
  geom_boxplot(aes(fill= `Peer group`)) +
  ggtitle('Average Length of Stay between Large and Medium Hospitals')

Hypothesis Testing

Independent Sample t-test

Hypothesis Generation

\(H_0\): There is no difference in the mean of average length of stay between medium and large hospitals

\(H_a\): There is significant difference between the average length of stay between medium and large hospitals

Mathematically,

\(H_0: \mu_1 - \mu_2 = 0\)

\(H_a: \mu_1 - \mu_2 \, \star \, 0\)

Testing the Assumption for Normality

Viewing the normality plot, it is evident that the average length of stay for both large and medium hospitals are skewed to the right. However, due to the large sample size, normality is ignored.

Homogenity of Variance

Assuming equal variance, the Levene Test is performed to examine the homogeneity of Variance. The results with \(p\)-value < 0.01 provides evidence that Levene Test is statistically significant. This implies the variance between two groups; large and medium hospitals are not equal.

leveneTest(`Average length of stay (days)`~ `Peer group`,data=data)

## Levene's Test for Homogeneity of Variance (center = median)
##          Df F value    Pr(>F)    
## group     5  86.543 < 2.2e-16 ***
##       10032                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Independent Sample t-test (Assuming Unequal Variance)

t.test(
  `Average length of stay (days)`~ `Peer group`,
  data = medium_large,
  var.equal = FALSE,
  alternative = "two.sided"
  )

## 
##  Welch Two Sample t-test
## 
## data:  Average length of stay (days) by Peer group
## t = 5.2855, df = 4500.1, p-value = 1.313e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1668947 0.3637059
## sample estimates:
##  mean in group Large hospitals mean in group Medium hospitals 
##                       3.983052                       3.717752

Discussion

Major Findings

• A two-sample t-test was used to test for the significant difference in the average length of stay in a medium and large hospital.

• While the length of stay showed evidence of non-normality upon examination, the central limit theorem assured that the t-test could be applied due to the large sample size in each group.

• The Levene’s Test of Homogeneity of variance showed that equal variance could not be assumed. Hence, Welch Two Sample t-test is carried out.

• The results of Welch two-sample t-test assuming unequal variance found a statistically significant difference between the mean of length of stay (ALOS) of medium and large hospitals,t(df=4500.1) = 5.29, p < 0.01, 95% confidence interval for the difference in means [0.167,0.364].

• The outcome of the investigation infers that patients admitted to medium-size hospitals have a significantly lower length of stay than patients admitted in large hospitals.

References

• James Baglin 2 April, 2019, Viewed 5 May 2020 [https://astral-theory-157510.appspot.com/secured/MATH1324_Module_07.html#confidence_interval_approach]

• Australian Health and Welfare, Viewed May 4 2020 [https://www.aihw.gov.au/reports-data/myhospitals/sectors/admitted-patients]

• Karlijn Willems, 12 March 2019, Viewed 9 May 2020 [https://www.datacamp.com/community/tutorials/make-histogram-ggplot2]

• Chester Ismay,21 Jan 2019, Viewed 8 May 2020 [https://github.com/ismayc/moderndiver-book/blob/master/10-hypothesis-testing.Rmd]

• Martin Schmelzer, Viewed 9 May 2020 [https://stackoverflow.com/questions/37115276/control-alignment-of-two-side-by-side-plots-in-knitr]