Paper 2 Section B Practice: Sampling and Estimation — Answers

Questions: Sampling and Estimation Drills.

Question 1

(a) Year groups may differ systematically in study time. Stratification guarantees representation from every year and can improve precision when the strata differ. [2]

(b) The total population is $3000$. Proportional allocations are

$$100\left(\frac{1200}{3000},\frac{900}{3000},\frac{600}{3000},\frac{300}{3000}\right)=(40,30,20,10).$$

[2]

(c) Construct a sampling frame for each year group and use independent random numbers, or an equivalent random mechanism, to select the required number without replacement within each stratum. [2]

Question 2

(a) By the central limit theorem,

$$\bar{X}\approx N\left(72,\frac{15^2}{36}\right)=N(72,6.25).$$

The sample must be random and sufficiently large for the approximation to be appropriate. [3]

(b)

$$P(69<\bar{X}<76)=P\left(-1.2<Z<1.6\right)\approx \mathrm{0.830.}$$

[3]

(c) For sample size $n$,

$$P(|\bar{X}-72|<3)=P\left(|Z|<\frac{3\sqrt{n}}{15}\right).$$

The central probability exceeds $0.95$ when

$$\frac{3\sqrt{n}}{15}>{\Phi }^{-1}(0.975)\approx \mathrm{1.95996.}$$

Thus

$$n>{\left(5\times 1.95996\right)}^2\approx 96.04,$$

so the smallest integer is $n=97$. [4]

Question 3

(a)

$$\bar{x}=\frac{4+7+9+10+15}{5}=9.$$

[2]

(b) The squared deviations from $9$ sum to

$$25+4+0+1+36=66.$$

Therefore

$$s^2=\frac{66}{5-1}=\mathrm{16.5.}$$

The denominator is $n-1$ because an unbiased variance estimate is required. [3]

(c) Over repeated random samples of the same size,

$$E(\bar{X})=\mu .$$

Thus the estimator has no systematic upward or downward bias. [1]