Baanee’s Version

Problem 9

We were given the sensitivity, specificity, and prevalence.

Table	\(D+\)	\(D-\)	Total
\(T+\)	\(0.00085\)	\(0.04995\)	\(0.0508\)
\(T-\)	\(0.00015\)	\(0.94905\)	\(0.9492\)
Total	\(0.001\)	\(0.999\)	\(1\)

| Table | $D+$ | $D-$ | Total |
| ----- | -- | -- | ----- |
| $T+$ | $0.00085$ | $0.04995$ | $0.0508$ |
| $T-$ | $0.00015$ | $0.94905$ | $0.9492$ |
| Total | $0.001$ | $0.999$ | $1$

\[P(D- \| T-) = \frac{P(T- \cap D-)}{P(T-)} = 0.99984\] $$P(D- \| T-) = \frac{P(T- \cap D-)}{P(T-)} = 0.99984$$

This means that if the test yields a negative result, there is a high probability that the true status is also negative.

Problem 10

The positive predictive value with the default prevalence of \(P(D+) = 0.001\) is \(\frac{0.00085}{0.05080} = 0.0167\). This means there are high chances of false positives. This is not because the test is necessarily a bad one, but because it is such a rare condition that there are more likely to be false positives than true positives.

We can study the behavior of positive predictive value if we make it into a function of an increasing variable \(P(D+)\). After reviewing the code done in class, I recreated this plot under this context.

# constant for sensitivity and specificity
sensitivity <- 0.85
specificity <- 0.95

# predictive positive value function
ppv <- function(x, sen, spec){ 
  sen*x/(sen*x+(1-spec)*(1-x))
}

# plotting
xxx <- seq(0.001,1,by = 0.01)
yyy <- ppv(xxx, sensitivity, specificity)
plot(xxx,yyy, type = "l", lwd = 2, col = "red", xlab = "Prevalence", ylab = "Positive Predictive Value")

This code yielded this plot: http://127.0.0.1:26887/graphics/e16ee78b-1027-4ab1-85ae-65bd907580c2.png

Mrunal’s Version

Problem 9

We were given the sensitivity, specificity, and prevalence.

Table	\(D+\)	\(D-\)	Total
\(T+\)	\(0.00085\)	\(0.04995\)	\(0.0508\)
\(T-\)	\(0.00015\)	\(0.94905\)	\(0.9492\)
Total	\(0.001\)	\(0.999\)	\(1\)

| Table | $D+$ | $D-$ | Total |
| ----- | -- | -- | ----- |
| $T+$ | $0.00085$ | $0.04995$ | $0.0508$ |
| $T-$ | $0.00015$ | $0.94905$ | $0.9492$ |
| Total | $0.001$ | $0.999$ | $1$ |

\[P(D- \| T-) = \frac{P(T- \cap D-)}{P(T-)} = 0.99984\] $$P(D- \| T-) = \frac{P(T- \cap D-)}{P(T-)} = 0.99984$$

This means that if the test yields a negative result, there is a high probability that the true status is also negative.

Problem 10

The positive predictive value with the default prevalence of \(P(D+) = 0.001\) is \(\frac{0.00085}{0.05080} = 0.0167\). This means there are high chances of false positives. This is not because the test is necessarily a bad one, but because it is such a rare condition that there are more likely to be false positives than true positives.

We can study the behavior of positive predictive value if we make it into a function of an increasing variable \(P(D+)\). After reviewing the code done in class, I recreated this plot under this context.

# constant for sensitivity and specificity
sensitivity <- 0.85
specificity <- 0.95

# predictive positive value function
ppv <- function(x, sen, spec){ 
  sen*x/(sen*x+(1-spec)*(1-x))
}

# plotting
xxx <- seq(0.001,1,by = 0.01)
yyy <- ppv(xxx, sensitivity, specificity)
plot(xxx,yyy, type = "l", lwd = 5, col = "navy", xlab = "Prevalence", ylab = "Positive Predictive Value")

This code yielded this plot: http://127.0.0.1:26887/graphics/ba64323d-aa40-4c9d-a58f-8f95c43acda5.png