(a) If c(x, t) denote observed proportion of females in the age group (x, x+t) and f(x, t) is the observed proportion of females giving birth to female children in the age group (x, x+t) at time t. Let us assume that X is uniformly distributed in (α,

Question

(a) If c(x, t) denote observed proportion of females in the age group (x, x+t) and f(x, t) is the observed proportion of females giving birth to female children in the age group (x, x+t) at time t. Let us assume that X is uniformly distributed in (α, β). Then show that

B̂_f(t)=[r̂_c,f|t σ̂_c σ̂_f (β-α) + ([T̂_f(t)]²)/((β-α)) 1/(Ĝ_f(t))],

where T̂_f(t) is the estimated total fertility.

B̂_f(t) is the estimated female birthrate at time t.

Ĝ_f(t) is the estimated General Fertility rate.

r̂_c,f|t represents product moment correlation coefficient between c and f given t.

σ̂_c, σ̂_f are observed standard deviations of c and f respectively.

(15 marks)

(b) What do you mean by Intelligence Quotient (I.Q.) ? Describe the procedure and test of measuring I.Q. How does an aptitude test differ from an Intelligence Test ?

The reliability coefficient of a test of 60 items is 0·65. How much the test should be lengthened to raise the self correlation to 0·95 ? What effect will the doubling and tripling the test's length have upon the reliability coefficients ? What is the reliability of a test having 135 comparable items ?

(15 marks)

(c) Define instantaneous force of mortality (μₓ).

Show that qₓ = (1/lₓ) ∫₀¹ μₓ₊ₜ lₓ₊ₜ dx

where qₓ is the probability of dying within one year following the attainment of age x.

Also prove that μₓ = (1/eₓ⁰) [1 + (deₓ⁰/dx)]

where eₓ⁰ is the complete expectation of life.

(20 marks)

UPSC Answer Check · Accepted Answer

(a) Let X, the age of a female, be uniform on (α, β), so its density is 1/(β−α). For fixed t write c=c(X,t), f=f(X,t). By definition of the product moment correlation,

Cov(c,f)=r̂(c,f|t) σ̂_c σ̂_f.

Also E(cf)=Cov(c,f)+E(c)E(f). Hence

(β−α)E(cf)=r̂(c,f|t) σ̂_c σ̂_f(β−α)+(β−α)E(c)E(f). (1)

Now the estimated female birth rate is

B̂_f(t)=∫_α^β c(x,t) f(x,t) dx=(β−α)E(cf).

The estimated total fertility is

T̂_f(t)=∫_α^β f(x,t) dx=(β−α)E(f), so E(f)=T̂_f(t)/(β−α).

The estimated general fertility rate is

Ĝ_f(t)=T̂_f(t)/∫_α^β c(x,t) dx, so ∫_α^β c dx=T̂_f(t)/Ĝ_f(t),

and therefore

E(c)=T̂_f(t)/[(β−α)Ĝ_f(t)].

Substituting in (1),

B̂_f(t)=r̂(c,f|t) σ̂_c σ̂_f(β−α)+(β−α)·T̂_f(t)/[(β−α)Ĝ_f(t)]·T̂_f(t)/(β−α)

= r̂(c,f|t) σ̂_c σ̂_f(β−α)+[T̂_f(t)]²/[(β−α)Ĝ_f(t)].

This proves the required result.

(b) I.Q. means Intelligence Quotient. Historically,

I.Q.=(Mental Age/Chronological Age)×100.

In modern tests, deviation I.Q. is used: scores are standardised to mean 100 and standard deviation 15. It measures general cognitive ability, reasoning, verbal comprehension, numerical ability, memory and problem solving. Procedure: a standardised intelligence test such as Stanford–Binet, Wechsler, Raven’s Progressive Matrices is administered under fixed instructions and time limits; raw scores are converted to mental-age or standard scores using norms for age, sex, culture and education; the quotient or deviation score is then interpreted. Reliability and validity are checked by test–retest, split-half and criterion methods.

An aptitude test differs from an intelligence test as follows. Intelligence tests measure present general mental ability and broad intellectual functioning. Aptitude tests measure a specific potential or capacity to acquire a particular skill, such as mechanical, clerical, musical, teaching or engineering aptitude. Intelligence tests are broad and general; aptitude tests are more specific and predictive of future training or job success. Aptitude tests are used in educational and vocational guidance, whereas intelligence tests are used to assess general cognitive level.

Using the Spearman–Brown prophecy formula,

r_n=n r/[1+(n−1)r],

where r=0.65 and n is the factor by which test length is multiplied. To raise reliability to 0.95:

0.95=0.65n/[1+0.65(n−1)]=0.65n/[0.35+0.65n].

Thus

n=0.95×0.35/[0.65×0.05]=0.3325/0.0325=10.23077.

Original length=60 items. New length=60×10.23077=613.85≈614 items. So the test should be lengthened by about 554 items.

For doubling, n=2:

r_2=2×0.65/[1+0.65]=1.30/1.65=0.7879.

For tripling, n=3:

r_3=3×0.65/[1+2×0.65]=1.95/2.30=0.8478.

For 135 comparable items, n=135/60=2.25:

r=2.25×0.65/[1+1.25×0.65]=1.4625/1.8125=117/145=0.8069.

(c) The instantaneous force of mortality μₓ is the limiting annual death rate at exact age x:

μₓ=lim_h→0+ [P(T≤x+h | T>x)/h]=−(1/lₓ) dlₓ/dx,

where T denotes age at death and lₓ is the number surviving to age x.

Number dying between ages x and x+1 is

∫₀¹ lₓ₊ₜ μₓ₊ₜ dt.

Hence the probability qₓ of dying within one year after age x is

qₓ=(1/lₓ)∫₀¹ μₓ₊ₜ lₓ₊ₜ dt.

This is the required identity.

For the second result, let eₓ⁰ be complete expectation of life:

eₓ⁰=∫₀∞ lₓ₊ₜ/lₓ dt=∫₀∞ pₓ(t) dt.

Differentiating with respect to x,

deₓ⁰/dx=∫₀∞ ∂/∂x(lₓ₊ₜ/lₓ) dt.

Since ∂lₓ₊ₜ/∂x=−μₓ₊ₜ lₓ₊ₜ and ∂(1/lₓ)/∂x=μₓ/lₓ,

deₓ⁰/dx=∫₀∞ [−μₓ₊ₜ lₓ₊ₜ/lₓ+μₓ lₓ₊ₜ/lₓ] dt

=μₓ eₓ⁰−∫₀∞ (lₓ₊ₜ/lₓ) μₓ₊ₜ dt.

But

d/dt(lₓ₊ₜ/lₓ)=−(lₓ₊ₜ/lₓ) μₓ₊ₜ,

so

∫₀∞ (lₓ₊ₜ/lₓ) μₓ₊ₜ dt=−[lₓ₊ₜ/lₓ]₀∞=1.

Therefore

deₓ⁰/dx=μₓ eₓ⁰−1.

Rearranging,

μₓ=(1/eₓ⁰)[1+deₓ⁰/dx].

Thus proved.

(a) If c(x, t) denote observed proportion of females in the age group (x, x+t) and f(x, t) is the observed proportion of females…

Model answer

What "Derive" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Statistics 2021 Paper II