Question

In: Statistics and Probability

The service time in minutes from admit to discharge for ten patients seeking care in a...

The service time in minutes from admit to discharge for ten patients seeking care in a hospital emergency department are 21, 136, 185, 156, 3, 16, 48, 28, 100, and 12. Construct a normal probability plot, an exponential probability plot, and a Weibull probability plot for these data in R. Based on the plots, which distribution is the best one to describe the outpatient service time?

Solutions

Expert Solution

Ran the below code to construct a normal probability plot

service.time = c(21, 136, 185, 156, 3, 16, 48, 28, 100,12)
# GET SIZE OF SAMPLE
n <- length(service.time)
# GET FRACTIONS (PERCENTILES) OF THE OBSERVED SAMPLE.
Fi <- (1:n-0.5)/n
# GET THE CORRESPONDING z - scores BY SOLVING Fi = Phi(Zi).
Zi <- qnorm(Fi,0,1)
# PUT THE OBSERVATIONS INTO INCREASING ORDER.
SP <- sort(service.time)
# PLOT SP VS. Zi. DRAW THE BEST FIT LINE THIS TIME.
plot(Zi,SP,ylab="Observed Service time",xlab="z - Percentile",main="Normal P-P Plot")
abline(lm(SP~Zi),col=2)

Ran the below code to construct a exponential probability plot

service.time = c(21, 136, 185, 156, 3, 16, 48, 28, 100,12)
# GET SIZE OF SAMPLE
n <- length(service.time)
# GET FRACTIONS (PERCENTILES) OF THE OBSERVED SAMPLE.
Fi <- (1:n-0.5)/n
# lambda Qi IS THE QUANTILE. SOLVE EQUATION Fi = 1 - exp(-lambda Qi)
Qi <- -log(1-Fi)
# PUT THE OBSERVATIONS INTO INCREASING ORDER.
SP <- sort(service.time)
# PLOT SP VS. Zi. DRAW THE BEST FIT LINE THIS TIME.
plot(SP, Qi,ylab="Service time Quantiles",xlab="Theoretical Exponential Quantiles",main="Exponential P-P Plot")
abline(lm(Qi ~SP),col=2)

Ran the below code to construct a Weibull probability plot

service.time = c(21, 136, 185, 156, 3, 16, 48, 28, 100,12)
# GET SIZE OF SAMPLE
n <- length(service.time)
# GET FRACTIONS (PERCENTILES) OF THE OBSERVED SAMPLE.
Fi <- (1:n-0.5)/n
# Qi/beta IS THE QUANTILE. SOLVE EQUATION Fi = 1 - exp(-(-Qi/beta )^alpha)
Eta <- log(-log(1-Fi))
# PUT THE OBSERVATIONS INTO INCREASING ORDER.
SP <- sort(service.time)
# PLOT SP VS. Zi. DRAW THE BEST FIT LINE THIS TIME.
plot(Eta,log(SP),ylab="Log Service time Quantiles", xlab="Theoretical Weibull Quantiles",main="Weibull P-P Plot")
abline(lm(log(SP) ~Eta),col=2)

Weibull distribution is the best one to describe the outpatient service time, because the estimated line in Weibull probability plot, approximately lie on the line y = x and the data points are closer to the estimated line.


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