I85PREP
⌘ K
IAI CS2 study materials
IAI CS2 study resources

IAI CS2 practice questions

One question at a time. Build your IAI CS2 understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Survival models and lifetime distributions

In survival analysis, which function gives the probability that a lifetime exceeds time t?

  1. The hazard function h(t)
  2. The cumulative hazard H(t)
  3. The probability density f(t)
  4. The survival function S(t) Correct answer.

Explanation

The survival function is defined as S(t) = P(T > t), so it directly returns the probability of surviving beyond time t. The hazard, cumulative hazard and density describe instantaneous risk or density, not the survival probability itself.

Question 2Survival models and lifetime distributions

Which expression correctly relates the survival function to the cumulative hazard H(t)?

  1. S(t) = H(t)/t
  2. S(t) = 1 - H(t)
  3. S(t) = log H(t)
  4. S(t) = exp(-H(t)) Correct answer.

Explanation

Integrating the hazard gives H(t) = -log S(t), and inverting this yields S(t) = exp(-H(t)). This identity holds for continuous lifetime distributions and underlies the Nelson-Aalen and Cox model formulas.

Question 3Survival models and lifetime distributions

A lifetime T follows an exponential distribution with rate lambda = 0.05 per year. What is the median survival time?

  1. About 20.0 years
  2. About 10.0 years
  3. About 6.9 years
  4. About 13.9 years Correct answer.

Explanation

The median solves S(t) = exp(-lambda t) = 0.5, giving t = log(2)/lambda. With lambda = 0.05 this is 0.6931/0.05, approximately 13.9 years.

Question 4Survival models and lifetime distributions

Which property is characteristic of the exponential lifetime distribution?

  1. A hazard that increases linearly with time
  2. A constant hazard rate over time Correct answer.
  3. A hazard that decreases as a power of time
  4. A hazard that is zero until a threshold age

Explanation

The exponential distribution has hazard h(t) = lambda, independent of t, which is the memoryless property in survival terms. Increasing, decreasing and threshold hazards correspond to Weibull, Gompertz or shifted models instead.

Question 5Survival models and lifetime distributions

A right-censored observation at time t contributes which term to the likelihood under a parametric survival model with density f and survival S?

  1. h(t)
  2. 1 - S(t)
  3. f(t)
  4. S(t) Correct answer.

Explanation

For right censoring we only know the subject survived past t, so the contribution is the probability of that event, S(t). An exact failure time instead contributes the density f(t).

Question 6Survival models and lifetime distributions

In the Kaplan-Meier estimator, how is the survival probability updated at an event time t(i) when there are d(i) events among n(i) subjects at risk?

  1. Divide the previous estimate by (1 - d(i)/n(i))
  2. Subtract d(i)/n(i) from the previous estimate
  3. Multiply the previous estimate by (1 - d(i)/n(i)) Correct answer.
  4. Add d(i)/n(i) to the previous estimate

Explanation

Kaplan-Meier is a product-limit estimator: S(t) is the running product of factors (1 - d(i)/n(i)) across event times. Each factor is the estimated conditional probability of surviving that event time given being at risk.

Question 7Survival models and lifetime distributions

Which function in R computes the Nelson-Aalen or Kaplan-Meier survival estimate directly from a survival object?

  1. survfit Correct answer.
  2. survdiff
  3. survreg
  4. coxph

Explanation

survfit() produces nonparametric estimates of the survival curve, including the Kaplan-Meier and Fleming-Harrington (Nelson-Aalen) estimators. The other functions fit regression models or compare groups.

Question 8Survival models and lifetime distributions

The Weibull survival model has hazard h(t) = lambda * p * t^(p-1). What does a shape parameter p greater than one imply?

  1. The hazard decreases with time
  2. The hazard is constant over time
  3. The hazard is zero after a fixed time
  4. The hazard increases with time Correct answer.

Explanation

With p > 1 the term t^(p-1) grows with t, so the Weibull hazard is monotonically increasing, describing wear-out or ageing. A constant hazard requires p = 1 and a decreasing hazard requires p < 1.

Question 9Survival models and lifetime distributions

A study observes 40 subjects at risk just before time t, and 4 deaths occur at time t with no censoring at that instant. Using Kaplan-Meier, what is the multiplicative factor applied to the survival estimate at t?

  1. 0.40
  2. 0.96
  3. 0.10
  4. 0.90 Correct answer.

Explanation

The factor is 1 - d/n = 1 - 4/40 = 0.90, and the survival estimate is multiplied by this value. Censoring at the same time would reduce the at-risk count and change the factor.

Question 10Survival models and lifetime distributions

Why is the Nelson-Aalen estimator of the cumulative hazard preferred over the negative log of Kaplan-Meier when survival probabilities are very small?

  1. It gives a smooth parametric estimate of the hazard
  2. It remains finite and better behaved where the Kaplan-Meier estimate reaches zero Correct answer.
  3. It requires no assumption about censoring
  4. It always has smaller variance at every time point

Explanation

When the largest observed time is an event, Kaplan-Meier can fall to zero, so -log S(t) becomes infinite. The Nelson-Aalen estimator sums d(i)/n(i), which stays finite and yields more stable estimates in the tail.

Question 11Survival models and lifetime distributions

In a Cox proportional hazards model, what is the role of the baseline hazard h0(t)?

  1. It is estimated jointly with the coefficients by maximum likelihood
  2. It must be assumed exponential for estimation
  3. It equals the hazard of the average covariate value
  4. It is left unspecified and cancels out of the partial likelihood Correct answer.

Explanation

Cox's partial likelihood conditions on the risk sets, so the common baseline hazard factor cancels and the coefficients can be estimated without specifying h0(t). The baseline hazard can be estimated afterwards, for example with Breslow's method.

Question 12Survival models and lifetime distributions

A Cox model gives a coefficient beta = 0.4 for a binary treatment covariate. What is the hazard ratio for treated versus untreated?

  1. About 0.67
  2. About 1.49 Correct answer.
  3. About 1.40
  4. About 0.40

Explanation

The hazard ratio is exp(beta) = exp(0.4), approximately 1.49, meaning treated subjects have about 49 percent higher hazard. The coefficient itself is on the log hazard scale.

Question 13Survival models and lifetime distributions

Which statement about the proportional hazards assumption in a Cox model is correct?

  1. The hazard ratio between any two covariate profiles is constant over time Correct answer.
  2. The survival curves of the groups must not cross the value 0.5
  3. The baseline hazard must be constant over time
  4. Covariate effects must be linear in time

Explanation

Proportional hazards means h(t|x1)/h(t|x2) does not depend on t, so hazard curves are proportional. The baseline hazard itself may vary freely with time.

Question 14Survival models and lifetime distributions

A log-rank test comparing two groups returns a p-value of 0.03. What is the most appropriate interpretation?

  1. The median survival times are exactly equal
  2. The Kaplan-Meier curves are identical
  3. The hazard ratio is exactly 1.0
  4. There is evidence the survival distributions differ, assuming the proportional hazards pattern holds Correct answer.

Explanation

The log-rank test assesses whether the hazard functions differ across groups, and small p-values indicate evidence against equality of survival. It is most powerful under proportional hazards but does not estimate the size of the effect.

Question 15Survival models and lifetime distributions

Which distribution is the standard parametric choice when the log hazard is assumed to increase linearly with time?

  1. Weibull
  2. Lognormal
  3. Log-logistic
  4. Gompertz Correct answer.

Explanation

The Gompertz model has h(t) = a * exp(b t), so log h(t) is linear in t with slope b. The Weibull has log hazard linear in log t, and the lognormal and log-logistic hazards are non-monotone.

Question 16Survival models and lifetime distributions

A parametric model is fitted to data with heavy right censoring and the log-likelihood is maximized. Which comparison of nested models is appropriate?

  1. A comparison of the raw log-likelihood values without regard to degrees of freedom
  2. A t-test on the estimated regression coefficients
  3. A log-rank test on the fitted residuals
  4. A likelihood ratio test using twice the log-likelihood difference against a chi-square distribution Correct answer.

Explanation

For nested parametric models, the likelihood ratio statistic 2(l1 - l0) is asymptotically chi-square with degrees of freedom equal to the number of extra parameters. This accounts for model complexity when comparing fits.

Question 17Survival models and lifetime distributions

In an accelerated failure time model, what does the acceleration factor exp(-beta x) represent?

  1. The factor by which the time scale of survival is multiplied for a unit change in covariates Correct answer.
  2. The probability of surviving past the median
  3. The cumulative hazard at the mean covariate
  4. The constant hazard ratio over all time

Explanation

AFT models write log T = mu + beta x + sigma W, so a covariate rescales the time axis by exp(-beta x). This differs from the Cox model, where covariates act multiplicatively on the hazard.

Question 18Survival models and lifetime distributions

Which R function fits a parametric survival regression model such as a Weibull AFT model?

  1. survfit
  2. coxph
  3. survreg Correct answer.
  4. survdiff

Explanation

survreg() fits parametric survival models by maximum likelihood, with the distribution chosen via the dist argument, for example Weibull or lognormal. coxph fits the semiparametric Cox model instead.

Question 19Survival models and lifetime distributions

A dataset has administrative censoring at the end of follow-up. What does this mean for the likelihood contribution of subjects still alive at that point?

  1. They are dropped from the analysis entirely
  2. They contribute the survival probability at their censoring time Correct answer.
  3. They contribute a factor of one half
  4. They contribute the density at their censoring time

Explanation

Administrative censoring means follow-up simply ends, so these subjects are known only to have survived to that time and contribute S(t) to the likelihood. Dropping them would bias the estimate by removing long survivors.

Question 20Survival models and lifetime distributions

Why can the Kaplan-Meier estimator be biased when censoring is informative?

  1. Because the estimate assumes censored subjects have the same future risk as those still at risk Correct answer.
  2. Because it ignores the at-risk set at event times
  3. Because the product-limit formula cannot handle ties
  4. Because it requires a parametric baseline hazard

Explanation

Kaplan-Meier relies on independent censoring, meaning censored subjects are representative of those remaining at risk. If censoring depends on prognosis, the risk sets are distorted and the estimate is biased.

Question 21Survival models and lifetime distributions

In a Cox model with a time-varying covariate, which construction is required in the survival package in R?

  1. A separate model fitted for each interval
  2. A single row per subject with the covariate fixed at baseline
  3. Recoding the covariate as a factor with one level
  4. A start-stop data format with one row per covariate interval Correct answer.

Explanation

Time-varying covariates are handled with the (start, stop, event) counting-process format, giving one row per interval during which the covariate is constant. This lets the partial likelihood use the covariate value in force at each event time.

Question 22Survival models and lifetime distributions

Cox-Snell residuals are computed as r_i = H0(t_i) * exp(beta x_i). What property should these residuals have if the fitted model is correct?

  1. They behave like a censored sample from a unit exponential distribution Correct answer.
  2. They are uniformly distributed between zero and one
  3. They equal the observed event times
  4. They are exactly normally distributed with mean zero

Explanation

Under a correctly specified Cox model, the Cox-Snell residuals follow a unit exponential distribution, and a plot of the cumulative hazard of the residuals against the residuals should be a straight line of slope one. This provides a graphical check of overall model fit.

Question 23Survival models and lifetime distributions

A Weibull model is fitted and the shape parameter is estimated as 0.6 with a confidence interval excluding one. What does this suggest about the hazard?

  1. The hazard increases over time, consistent with wear-out
  2. The hazard is non-monotone with a single peak
  3. The hazard decreases over time, consistent with early failure or infant mortality Correct answer.
  4. The hazard is constant over time

Explanation

For the Weibull model, a shape parameter below one implies a hazard proportional to t^(p-1) with negative exponent, so it decreases over time. This pattern is typical of infant mortality or early failures.

Question 24Survival models and lifetime distributions

A stratified Cox model is fitted with strata defined by centre. What is the effect of stratification on the baseline hazard and on the estimated coefficients?

  1. The baseline hazard is common and the strata are ignored in estimation
  2. Each stratum has its own coefficient vector and a common baseline hazard
  3. Stratification forces the baseline hazards to be proportional across strata
  4. Each stratum has its own baseline hazard while the covariate coefficients are shared Correct answer.

Explanation

Stratification allows the baseline hazard to differ arbitrarily between strata, removing the proportional hazards requirement across strata. The covariate effects are still assumed common and are estimated from a stratified partial likelihood.

Question 25Survival models and lifetime distributions

A frailty model is used in a multivariate survival setting. What does the frailty term represent?

  1. The baseline cumulative hazard at time zero
  2. An unobserved random effect shared within a cluster that induces dependence among lifetimes Correct answer.
  3. A fixed covariate measured without error
  4. A censoring indicator for the last observation in each cluster

Explanation

A frailty is an unobserved multiplicative random effect on the hazard, shared by members of a cluster, which creates association between their survival times. Ignoring it can bias estimates and understate dependence.

Question 26Survival models and lifetime distributions

Which quantity does the survival function S(t) represent in a lifetime model?

  1. The probability that the lifetime exceeds t Correct answer.
  2. The expected number of events occurring in the interval from 0 to t
  3. The probability that failure occurs exactly at time t
  4. The instantaneous rate of failure at time t conditional on survival to t

Explanation

The survival function is defined as S(t) = P(T > t), the probability that the random lifetime T exceeds t. It is a non-increasing function starting at 1 and tending to 0 as t grows. The hazard, cumulative hazard and density are distinct but related quantities.

Question 27Survival models and lifetime distributions

A random lifetime T follows an exponential distribution with constant hazard lambda = 0.05 per year. What is the median lifetime?

  1. About 13.86 years Correct answer.
  2. About 0.05 years
  3. About 20 years
  4. About 6.93 years

Explanation

For the exponential model S(t) = exp(-lambda t), setting S(t) = 0.5 gives t = ln(2)/lambda. With lambda = 0.05 this equals 0.6931/0.05, approximately 13.86 years. The mean would be 1/lambda = 20 years, so the median and mean differ.

Question 28Survival models and lifetime distributions

Which expression gives the hazard function h(t) in terms of the density f(t) and survival function S(t)?

  1. h(t) = S(t) / f(t)
  2. h(t) = f(t) * S(t)
  3. h(t) = -f(t) / S(t)
  4. h(t) = f(t) / S(t) Correct answer.

Explanation

The hazard is the instantaneous failure rate conditional on survival, defined as h(t) = f(t)/S(t). Equivalently h(t) = -d/dt log S(t), which links it to the cumulative hazard. The ratio form makes clear that the hazard is a conditional density, not a marginal one.

Question 29Survival models and lifetime distributions

A censored observation at time t contributes which term to the likelihood for a parametric survival model with right censoring?

  1. 1 - S(t), the probability of failure before the censoring time
  2. h(t), the hazard evaluated at the censoring time
  3. S(t), the probability of surviving beyond the censoring time Correct answer.
  4. f(t), the density evaluated at the censoring time

Explanation

A subject censored at t is known only to have survived past t, so the likelihood contribution is the survival probability S(t). Uncensored subjects contribute the density f(t). Combining these terms over all subjects gives the full likelihood used for maximum likelihood estimation.

Question 30Survival models and lifetime distributions

In a proportional hazards model, the survival function for a subject with covariate vector x can be written in terms of the baseline survival S0(t). Which form is correct?

  1. S(t|x) = S0(t) + exp(beta x)
  2. S(t|x) = S0(t)^exp(beta x) Correct answer.
  3. S(t|x) = exp(beta x) / S0(t)
  4. S(t|x) = S0(t) * exp(beta x)

Explanation

Under proportional hazards, H(t|x) = H0(t) exp(beta x), and since S = exp(-H) the covariate effect becomes a power on the baseline survival. This power relationship preserves the ordering of survival curves across time. It is the direct consequence of multiplying the baseline hazard by a constant factor.

Question 31Survival models and lifetime distributions

A Cox model is fitted and a covariate has estimated coefficient beta = -0.7 with standard error 0.25. What is the approximate 95% confidence interval for the hazard ratio?

  1. About (0.50, 0.99)
  2. About (0.70, 1.30)
  3. About (0.30, 0.81) Correct answer.
  4. About (0.21, 0.57)

Explanation

The hazard ratio is exp(-0.7) = 0.497, and the interval on the log scale is -0.7 plus or minus 1.96*0.25, giving -1.19 to -0.21. Exponentiating yields approximately 0.30 to 0.81. The interval lies entirely below one, indicating a significant protective effect.

Question 32Survival models and lifetime distributions

Which feature distinguishes the log-logistic from the log-normal accelerated failure time distribution in survival analysis?

  1. The log-logistic has a decreasing hazard, while the log-normal hazard is monotonically increasing
  2. The log-logistic has a monotonically increasing hazard, while the log-normal hazard is constant
  3. The log-logistic has a hazard that can be non-monotone, first rising then falling, while the log-normal hazard always eventually decreases after an early peak Correct answer.
  4. The log-logistic has a proportional hazards form, while the log-normal does not

Explanation

Both distributions are AFT models with heavy tails, but the log-logistic hazard can rise and then decline, giving a non-monotone shape. The log-normal hazard typically increases early and then decreases toward zero. Neither has a constant or purely monotone increasing hazard in general.

Question 33Survival models and lifetime distributions

In the Cox partial likelihood, which subjects appear in the risk set at an event time t?

  1. All subjects still under observation and event-free just before t Correct answer.
  2. Only the subjects who experience an event at t
  3. All subjects enrolled in the study regardless of their status at t
  4. Only the subjects censored before t

Explanation

The risk set at an event time consists of every subject who has not yet failed and has not been censored before t, including the subject who fails at t. The partial likelihood compares the failing subject against this risk set. Subjects censored before t leave the risk set and contribute no further terms.

Question 34Survival models and lifetime distributions

A survival dataset contains tied event times, and the analyst uses the Efron approximation in a Cox model. What is the purpose of this choice?

  1. To test the proportional hazards assumption
  2. To estimate the baseline hazard nonparametrically
  3. To correct the survival estimates for informative censoring
  4. To approximate the partial likelihood when several events occur at the same recorded time Correct answer.

Explanation

Tied event times make the exact partial likelihood computationally awkward, so approximations such as Breslow and Efron are used. The Efron method is generally more accurate than Breslow when ties are numerous. It modifies the denominator terms in the partial likelihood rather than addressing censoring or model checking.

Question 35Survival models and lifetime distributions

A survival curve is estimated for a group and the Greenwood formula is applied. What does the resulting quantity describe?

  1. The hazard ratio between two groups
  2. The expected number of future events in the group
  3. The variance of the estimated survival function at a given time Correct answer.
  4. The median survival time of the group

Explanation

The Greenwood formula provides an estimate of the variance of the Kaplan-Meier survival estimate, which supports pointwise confidence intervals. It accumulates terms over event times involving the number at risk and the number of events. It does not estimate hazard ratios or median survival directly.

Question 36Survival models and lifetime distributions

A parametric model is compared with a Cox model on the same data, and the analyst wants to check whether the parametric assumption is reasonable. Which approach is appropriate?

  1. Compare the fitted parametric survival curve with the Kaplan-Meier curve and examine whether the parametric hazard shape matches the observed data Correct answer.
  2. Compare the number of covariates retained by each model
  3. Report the Cox partial likelihood and the parametric log-likelihood directly as comparable values
  4. Use the Greenwood variance of the parametric fit as a goodness-of-fit statistic

Explanation

Graphical comparison of the parametric fitted curve against the nonparametric Kaplan-Meier estimate reveals systematic lack of fit. The shapes of the fitted hazard, such as monotonicity, should also be checked against the data. Likelihood values from Cox and parametric models are not directly comparable because the partial likelihood is not a full likelihood.

Question 37Survival models and lifetime distributions

A two-group study reports a hazard ratio of 1.0 with a wide confidence interval that includes 1, and the log-rank test gives p = 0.60. What is the most defensible conclusion?

  1. The hazard ratio should be reported as 1.0 to indicate a clinically important effect
  2. There is no evidence of a difference in survival between the groups in this sample Correct answer.
  3. The groups are proven to have identical survival distributions
  4. The study design was inadequate and should be discarded

Explanation

A non-significant log-rank test and a confidence interval spanning one indicate insufficient evidence of a difference, not proof of equality. Absence of evidence is not evidence of absence, especially with limited sample size. The estimate and interval should be reported together to convey uncertainty.

Question 38Survival models and lifetime distributions

In a Weibull proportional hazards model, the coefficient beta for a covariate enters the hazard as h(t|x) = lambda p t^(p-1) exp(beta x). How does a one-unit increase in x affect the hazard?

  1. It adds beta to the hazard at every time t
  2. It changes the shape parameter p by an amount beta
  3. It shifts the survival curve in time by an amount beta
  4. It multiplies the hazard by exp(beta) at every time t, leaving the shape unchanged Correct answer.

Explanation

In this parameterization the covariate term exp(beta x) multiplies the baseline hazard uniformly over time, so the hazard ratio is exp(beta). The shape parameter p is common to all subjects. A shift in time would correspond to the AFT parameterization, which is a different interpretation of the same model.

Question 39Survival models and lifetime distributions

A clinician argues that because a Cox model shows a significant hazard ratio, the model can be used to predict absolute survival probabilities for new patients. What additional quantity is required?

  1. An estimate of the baseline cumulative hazard, together with the covariate values Correct answer.
  2. The p-value of the log-rank test
  3. Only the hazard ratio and the covariate values
  4. The Greenwood variance of the Kaplan-Meier curve

Explanation

Predicted survival requires S(t|x) = exp(-H0(t) exp(beta x)), so the baseline cumulative hazard H0(t) must be estimated, for example with the Breslow estimator. The hazard ratio alone gives relative, not absolute, risk. Without the baseline, no absolute survival probability can be computed.

Question 40Survival models and lifetime distributions

A survival model assumes a piecewise exponential hazard that is constant within pre-specified intervals but may change between them. What is the main advantage of this specification over a single exponential model?

  1. It makes the hazard ratio independent of the interval boundaries
  2. It relaxes the constant-hazard assumption while retaining a closed-form likelihood that is easy to maximize Correct answer.
  3. It removes the need for any censoring assumptions
  4. It guarantees that the proportional hazards assumption holds for all covariates

Explanation

Piecewise exponential models allow the baseline hazard to vary across intervals, capturing non-constant risk while keeping a simple product-form likelihood. This flexibility comes at the cost of choosing interval cut points. The model still relies on assumptions about censoring and proportionality, which are not automatically satisfied.

Question 41Estimation and graduation of survival models

In a survival model, which quantity does the force of mortality μ(x) represent at exact age x?

  1. The proportion of the original cohort still alive at age x
  2. The expected remaining lifetime of a life aged x
  3. The probability that a life aged x dies before reaching age x+1
  4. The instantaneous rate of mortality at age x, conditional on survival to that age Correct answer.

Explanation

The force of mortality is the instantaneous hazard, defined as the limit of the conditional death probability over a short interval divided by that interval. It is a rate, not a probability, and it conditions on survival to exact age x. It is related to the survival function by μ(x) = -d/dx log S(x).

Question 42Estimation and graduation of survival models

A survival model is described as having a constant force of mortality μ over each year of age. Within such a year, how does the survival function behave?

  1. It declines linearly with age
  2. It declines along a curve of the Gompertz form
  3. It declines exponentially with age Correct answer.
  4. It stays constant throughout the year

Explanation

With a constant hazard μ, the survival function within the year is S(x+t) = S(x)·exp(-μt), which is exponential decay. This is the basis of the constant-force (exponential) assumption used in fractional-age calculations. The linear decline belongs to the uniform distribution of deaths assumption instead.

Question 43Estimation and graduation of survival models

Under the assumption of a uniform distribution of deaths between integer ages, what is the value of the one-year death probability q(x) in terms of the force of mortality at age x?

  1. q(x) = 1 - exp(-μ(x))
  2. q(x) = μ(x)
  3. q(x) = exp(-μ(x))
  4. q(x) = μ(x)/(1 + 0.5μ(x)) Correct answer.

Explanation

Uniform deaths imply that the force of mortality rises within the year according to μ(x+t) = μ(x)/(1 - t·μ(x)), with μ(x) equal to q(x)/(1 - 0.5q(x)). Inverting this relation gives q(x) = μ(x)/(1 + 0.5μ(x)). This is the standard link between the initial force and the annual death probability under UDD.

Question 44Estimation and graduation of survival models

Which statement correctly describes the actuarial estimate q̂(x) obtained by the census (exposure) method for a single year of age?

  1. It equals the central exposed to risk divided by the deaths during the year
  2. It equals deaths during the year divided by the number exposed to risk at the start of the year
  3. It equals deaths during the year divided by the central exposed to risk Correct answer.
  4. It equals the number of survivors at the end of the year divided by the number at the start

Explanation

The actuarial estimate divides observed deaths by the central (or average) exposed to risk, which accounts for lives entering and leaving during the year. Under a constant force of mortality this ratio is the maximum likelihood estimate of μ. It differs from the initial-exposure estimate, which uses lives at the start of the year.

Question 45Estimation and graduation of survival models

In a mortality investigation, a life aged exactly 60.4 is observed to die at age 60.7. Under the assumption of a uniform distribution of deaths, how many years of exposure does this life contribute to the age-60 to 61 interval?

  1. 0.4 years
  2. 0.7 years
  3. 0.3 years Correct answer.
  4. 1.0 years

Explanation

Under UDD the life is treated as exposed from entry at age 60.4 until death at 60.7, giving 0.3 years of exposure within the year of age. The exposure measures the time at risk before the death occurs. The remaining 0.6 years of the year of age are not counted because the life has died.

Question 46Estimation and graduation of survival models

A life is under observation from age 40.25 and is still alive at age 41. Under the assumption of a uniform distribution of deaths, what exposure does this life contribute to the year of age 40 to 41?

  1. 0.75 years Correct answer.
  2. 0.25 years
  3. 0.50 years
  4. 1.00 years

Explanation

The life enters observation at age 40.25 and remains alive to age 41, so the observed time within the year of age 40 is 41 minus 40.25, which equals 0.75 years. Under UDD no adjustment is needed for the part-year at entry because exposure is measured directly in time. The life contributes nothing to later years in this example.

Question 47Estimation and graduation of survival models

What is the main purpose of graduating a set of crude mortality rates?

  1. To remove random fluctuations and produce a smooth series of rates suitable for pricing and reserving Correct answer.
  2. To increase the number of deaths counted in each age group
  3. To test whether the crude rates are statistically significant
  4. To convert central death rates into probabilities of death

Explanation

Graduation fits a smooth curve or formula to the crude rates so that sampling noise is reduced while the underlying pattern is preserved. The graduated rates are then used for valuation, pricing and projection. Graduation is distinct from the arithmetic conversion of rates into probabilities.

Question 48Estimation and graduation of survival models

Which of the following is a standard test of smoothness applied to a set of graduated mortality rates?

  1. The likelihood ratio test for nested models
  2. The cumulative deviations (sign) test
  3. The chi-square goodness-of-fit test
  4. A test based on third differences of the graduated rates Correct answer.

Explanation

Smoothness tests examine the successive differences of the graduated rates, and a common approach looks at third differences to detect irregular curvature. Goodness-of-fit tests, by contrast, compare graduated rates with the crude data. The two families of tests address different requirements of a graduation.

Question 49Estimation and graduation of survival models

In a graduation, the chi-square test statistic is computed and found to be unusually small relative to its degrees of freedom. What does this most likely indicate?

  1. The graduated rates fit the data too closely, suggesting over-fitting or that the data were smoothed excessively Correct answer.
  2. The exposed to risk figures are too large
  3. The graduated rates are biased upwards at all ages
  4. The model has too few parameters to describe the data

Explanation

A chi-square statistic much smaller than its degrees of freedom suggests the graduated rates are closer to the crude rates than random variation would normally allow. This can arise from over-fitting or from the graduated curve having been influenced by the data too strongly. It signals a lack of credibility rather than a good fit.

Question 50Estimation and graduation of survival models

Which of the following is a parametric law of mortality that models the force of mortality as increasing exponentially with age?

  1. Gompertz's law Correct answer.
  2. de Moivre's law
  3. The Weibull law
  4. Makeham's law

Explanation

Gompertz's law sets μ(x) = B·c^x, so the force of mortality grows exponentially with age. Makeham extended this by adding a constant term representing age-independent causes. Both are widely used in graduating adult mortality.