Sample Solution

1. (a) In the natural history of a disease define: total preclinical phase, detectable pre-clinical phase and clinical phase. Suppose on 10 am on 25.01.2010 a disease A onset you biologically. Suppose test of the disease A can detect it exactly after completion of 1000 days of biologically onset. Suppose signs and symptoms develop exactly after completion of 1010 days of biologically onset. Suppose you consult doctor exactly after 1015 days of biologically onset of the disease. Suppose outcome the treatment is cure. What is duration of (i) total preclinical phase (ii) detectable pre-clinical phase (iii) clinical phase.

1. Total Preclinical Phase: This is the period between the biological onset of the disease and the appearance of clinical signs and symptoms. During this phase, the disease is present but not yet clinically apparent.
2. Detectable Preclinical Phase: This is the portion of the total preclinical phase during which the disease can be detected by screening tests, but before clinical symptoms appear. This phase starts when the disease becomes detectable by a test and ends when clinical signs and symptoms develop.
3. Clinical Phase: This begins when clinical signs and symptoms first appear and continues through diagnosis and treatment until the outcome (e.g., cure, chronic illness, or death).

• The biological onset of disease A: 10 am on 25.01.2010
• The disease can be detected by a test exactly 1000 days after the biological onset.
• Signs and symptoms develop exactly 1010 days after the biological onset.
• Consultation with a doctor happens exactly 1015 days after the biological onset.
• The outcome of the treatment is a cure.

1. Total Preclinical Phase:
   • Start: 10 am on 25.01.2010
   • End: Signs and symptoms appear 1010 days after the onset.
   • Duration: 1010 days
2. Detectable Preclinical Phase:
   • Start: The disease becomes detectable by a test 1000 days after the onset.
   • End: Signs and symptoms appear 1010 days after the onset.
   • Duration: 1010 days – 1000 days = 10 days
3. Clinical Phase:
   • Start: Signs and symptoms appear 1010 days after the onset.
   • End: The outcome of treatment (cure) is not specified to be at a particular time, but typically, the clinical phase would be considered from the appearance of symptoms until the consultation, which is 1015 days after the onset. This assumes the clinical phase includes the period where the patient actively seeks medical attention.
   • Duration: 1015 days – 1010 days = 5 days (until consulting a doctor)

• Total Preclinical Phase: 1010 days
• Detectable Preclinical Phase: 10 days
• Clinical Phase: 5 days

Proving the Relationship $q=\frac{p}{1-p}$$q=\frac{p}{1-p}$q=(p)/(1-p)q = \frac{p}{1-p}$q=\frac{p}{1-p}$

1. Definition of Proportion:
   • Proportion ($p$$p$pp$p$) is the fraction of a population that exhibits a certain characteristic. If $p$$p$pp$p$ is the proportion of smokers in a study, then $1-p$$1-p$1-p1-p$1-p$ is the proportion of non-smokers.
2. Definition of Odds:
   • Odds ($q$$q$qq$q$) is the ratio of the probability of an event occurring to the probability of it not occurring. If the odds of being a smoker are $q$$q$qq$q$, then it is the ratio of the proportion of smokers to the proportion of non-smokers.
3. Expressing Odds in Terms of Proportion:
   • By definition, odds can be written as:$$q=\frac{\text{Proportion of smokers}}{\text{Proportion of non-smokers}}$$$$q=\frac{\text{Proportion of smokers}}{\text{Proportion of non-smokers}}$$q=(“Proportion of smokers”)/(“Proportion of non-smokers”)q = \frac{\text{Proportion of smokers}}{\text{Proportion of non-smokers}}$$q=\frac{\text{Proportion of smokers}}{\text{Proportion of non-smokers}}$$
   • In terms of $p$$p$pp$p$, this becomes:$$q=\frac{p}{1-p}$$$$q=\frac{p}{1-p}$$q=(p)/(1-p)q = \frac{p}{1-p}$$q=\frac{p}{1-p}$$

Finding the Range of $q$$q$qq$q$

1. Range of $p$$p$pp$p$:
   • The proportion $p$$p$pp$p$ ranges between 0 and 1 (inclusive):$$0\le p\le 1$$$$0\le p\le 1$$0 <= p <= 10 \leq p \leq 1$$0\le p\le 1$$
2. Behavior of $q$$q$qq$q$ as $p$$p$pp$p$ Approaches Extremes:
   • When $p=0$$p=0$p=0p = 0$p=0$:$$q=\frac{0}{1-0}=0$$$$q=\frac{0}{1-0}=0$$q=(0)/(1-0)=0q = \frac{0}{1-0} = 0$$q=\frac{0}{1-0}=0$$
   • When $p$$p$pp$p$ approaches 1:$$q=\frac{p}{1-p}\to \mathrm{\infty }\text{as}p\to 1$$$$q=\frac{p}{1-p}\to \mathrm{\infty }\text{as}p\to 1$$q=(p)/(1-p)rarr oo” as “p rarr1q = \frac{p}{1-p} \to \infty \text{ as } p \to 1$$q=\frac{p}{1-p}\to \mathrm{\infty }\text{as}p\to 1$$
3. Exclusion of $p=1$$p=1$p=1p = 1$p=1$:
   • $p=1$$p=1$p=1p = 1$p=1$ would imply $q=\frac{1}{0}$$q=\frac{1}{0}$q=(1)/(0)q = \frac{1}{0}$q=\frac{1}{0}$, which is undefined.

Finding the Proportion of Smokers Given the Odds

1. Using the Relationship:
   $$q=\frac{p}{1-p}$$$$q=\frac{p}{1-p}$$q=(p)/(1-p)q = \frac{p}{1-p}$$q=\frac{p}{1-p}$$
   Given $q=0.25$$q=0.25$q=0.25q = 0.25$q=0.25$:
   $$0.25=\frac{p}{1-p}$$$$0.25=\frac{p}{1-p}$$0.25=(p)/(1-p)0.25 = \frac{p}{1-p}$$0.25=\frac{p}{1-p}$$
2. Solving for $p$$p$pp$p$:
   $$0.25(1-p)=p$$$$0.25(1-p)=p$$0.25(1-p)=p0.25 (1-p) = p$$0.25(1-p)=p$$
   $$0.25-0.25p=p$$$$0.25-0.25p=p$$0.25-0.25 p=p0.25 – 0.25p = p$$0.25-0.25p=p$$
   $$0.25=p+0.25p$$$$0.25=p+0.25p$$0.25=p+0.25 p0.25 = p + 0.25p$$0.25=p+0.25p$$
   $$0.25=1.25p$$$$0.25=1.25p$$0.25=1.25 p0.25 = 1.25p$$0.25=1.25p$$
   $$p=\frac{0.25}{1.25}=\frac{1}{5}=0.2$$$$p=\frac{0.25}{1.25}=\frac{1}{5}=0.2$$p=(0.25)/(1.25)=(1)/(5)=0.2p = \frac{0.25}{1.25} = \frac{1}{5} = 0.2$$p=\frac{0.25}{1.25}=\frac{1}{5}=0.2$$

Summary

• The relationship between proportion and odds is $q=\frac{p}{1-p}$$q=\frac{p}{1-p}$q=(p)/(1-p)q = \frac{p}{1-p}$q=\frac{p}{1-p}$.
• The range of $q$$q$qq$q$ is $0\le q<\mathrm{\infty }$$0\le q<\mathrm{\infty }$0 <= q < oo0 \leq q < \infty$0\le q<\mathrm{\infty }$.
• Given the odds of 0.25, the proportion of smokers in the study is 0.2 or 20%.

i) Regimens for Group I and Group II

• Group I:
  • Advised to do a 30-minute morning walk each day.
  • Advised to take light food each day.
• Group II:
  • Given one injection once a day.
  • Given one $100\mathrm{m}\mathrm{g}$$100\mathrm{m}\mathrm{g}$100mg100 \mathrm{mg}$100\mathrm{m}\mathrm{g}$ tablet once a day to control disease $X$$X$XX$X$.

ii) Efficacies in Group I and Group II

• Group I:
  • $90\mathrm{\%}$$90\mathrm{\%}$90%90\%$90\mathrm{\%}$ of the participants recovered from disease $X$$X$XX$X$.
• Group II:
  • $80\mathrm{\%}$$80\mathrm{\%}$80%80\%$80\mathrm{\%}$ of the participants recovered from disease $X$$X$XX$X$.

iii) Safety Issues in Group I and Group II

• Group I:
  • No specific safety issues reported in the regimen involving a morning walk and light food. Generally, this regimen is considered safe and has no known side effects.
• Group II:
  • The injection can cause loose motion in some cases, which is a safety issue.
  • The $100\mathrm{m}\mathrm{g}$$100\mathrm{m}\mathrm{g}$100mg100 \mathrm{mg}$100\mathrm{m}\mathrm{g}$ tablet has no side effects, so it is considered safe.

Summary

• Regimens:
  • Group I: Morning walk and light food.
  • Group II: Daily injection and $100\mathrm{m}\mathrm{g}$$100\mathrm{m}\mathrm{g}$100mg100 \mathrm{mg}$100\mathrm{m}\mathrm{g}$ tablet.
• Efficacies:
  • Group I: $90\mathrm{\%}$$90\mathrm{\%}$90%90\%$90\mathrm{\%}$ recovery.
  • Group II: $80\mathrm{\%}$$80\mathrm{\%}$80%80\%$80\mathrm{\%}$ recovery.
• Safety Issues:
  • Group I: No reported safety issues.
  • Group II: Injection may cause loose motion; the tablet is safe.