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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.
Selected subtopics: Piecewise densities and CDF transformations · One-sample, paired and pooled t inference · Chi-squared tests and rank tests · Probability generating functions
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Question 1A continuous random variable X has density ax for 0<x<1 and a(2-x) for 1<=x<2, and zero elsewhere. Find a, E(X) and Var(X). For Y=X^2, derive the complete cumulative distribution function and density of Y, and find P(Y<=1).7 indicative marks
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A continuous random variable X has density ax for 0<x<1 and a(2-x) for 1<=x<2, and zero elsewhere. Find a, E(X) and Var(X). For Y=X^2, derive the complete cumulative distribution function and density of Y, and find P(Y<=1).
Cambridge 2027: Direct conceptual match
Integrated both density pieces and both second-moment pieces separately; used monotonicity of x^2 on the non-negative support to transform the CDF and checked continuity at y=1.
The two triangular density pieces each integrate to a/2; hence a=1. Reflection x -> 2-x leaves the density unchanged, so the mean is 1.
E(X^2)=integral_0^1 x^3 dx + integral_1^2 x^2(2-x) dx = 1/4+11/12=7/6. Subtracting the squared mean gives 1/6.
X is non-negative, so Y<=y is equivalent to X<=sqrt(y) when y>=0. Integrating the first density piece gives F_X(x)=x^2/2 up to 1; on 1<x<2 the complementary upper triangle gives F_X(x)=1-(2-x)^2/2.
Substitute sqrt(y) into these CDF pieces, attach the outside-support values, then differentiate on the open pieces. There are no point masses at the joins, and P(Y<=1)=1/2.
Self-marking points
Normalisation gives a(1/2+1/2)=1, so a=1.
Symmetry about 1, or integration, gives E(X)=1.
Integrating x^2 f(x) gives E(X^2)=7/6 and Var(X)=1/6.
For 0<=y<=1, F_Y(y)=y/2; for 1<y<4, F_Y(y)=1-(2-sqrt(y))^2/2.
F_Y(y)=0 for y<0 and 1 for y>=4; these pieces are continuous at 0, 1 and 4.
Differentiation gives f_Y(y)=1/2 for 0<y<1 and (2-sqrt(y))/(2sqrt(y)) for 1<y<4, zero otherwise.
P(Y<=1)=F_Y(1)=1/2.
Examiner tip
Transform the CDF before differentiating; the support of Y is [0,4], not [0,2].
Question 2A random sample of 9 observations from a normal population with unknown variance has mean 12 and sample standard deviation 3 (computed with denominator n-1). Test H0: mu=10 against H1: mu>10 at 5%, using the given critical value t_(8,0.95)=1.860. Construct a 95% confidence interval using t_(8,0.975)=2.306. Separately, 9 independent paired differences, assumed normally distributed, have mean 2 and sample standard deviation 3. Explain and carry out the corresponding one-sided test of zero mean difference.6 indicative marks
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A random sample of 9 observations from a normal population with unknown variance has mean 12 and sample standard deviation 3 (computed with denominator n-1). Test H0: mu=10 against H1: mu>10 at 5%, using the given critical value t_(8,0.95)=1.860. Construct a 95% confidence interval using t_(8,0.975)=2.306. Separately, 9 independent paired differences, assumed normally distributed, have mean 2 and sample standard deviation 3. Explain and carry out the corresponding one-sided test of zero mean difference.
Cambridge 2027: Direct conceptual match
Verified both statistics from n=9 and s=3, used df=8 and explicitly distinguished one-sided 1.860 from two-sided 2.306; checked the paired-difference independence assumption.
The population standard deviation is not known, so a normal critical value is not appropriate for this small sample. Under H0 the statistic follows t with n-1=8 degrees of freedom.
The observed value 2 lies beyond the supplied one-sided 5% threshold. The two-sided 95% interval uses the different threshold 2.306. Its inclusion of 10 is consistent with a significant one-sided 5% test but not a two-sided 5% test.
For matched measurements, form each difference first. The mean and standard deviation given for the differences produce the same numerical statistic. A paired test does not treat the 18 raw measurements as independent.
Self-marking points
For the first sample the estimated standard error is 3/sqrt(9)=1.
The test statistic is (12-10)/1=2, with 8 degrees of freedom.
Since 2>1.860, reject H0 at 5%; there is evidence that mu>10.
The 95% confidence interval is 12 +/- 2.306, namely [9.694,14.306].
Analyse the paired differences as a single sample: t=(2-0)/(3/sqrt(9))=2, with 8 degrees of freedom.
Reject zero mean difference in favour of a positive mean difference; independence is required across pairs, not between the two measurements within each pair.
Examiner tip
State the direction of the alternative before choosing the tail; the one-sided test and two-sided interval use different quantiles.
Question 3Independent random samples from two normal populations have n1=n2=6, sample means 20 and 16, and sample standard deviations 2 and 4 (denominator n-1). Assume the population variances are equal but unknown. Test equality of means against a two-sided alternative at 5%, using t_(10,0.975)=2.228. Give a 95% confidence interval for mu1-mu2 and explain the effect of dropping the equal-variance assumption.6 indicative marks
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Independent random samples from two normal populations have n1=n2=6, sample means 20 and 16, and sample standard deviations 2 and 4 (denominator n-1). Assume the population variances are equal but unknown. Test equality of means against a two-sided alternative at 5%, using t_(10,0.975)=2.228. Give a 95% confidence interval for mu1-mu2 and explain the effect of dropping the equal-variance assumption.
Cambridge 2027: Direct conceptual match
Computed pooled variance from independent sums of squares, SE sqrt(10/3), t=2.19089 and the interval using the supplied quantile; the question explicitly supplies the normality and common-variance assumptions.
The two within-sample sums of squared deviations are 5*4 and 5*16. Their sum 100 is divided by the combined degrees of freedom 10.
Under the equal-variance model, the mean difference has estimated standard error sqrt(10/3), so t is approximately 2.191. It falls just short of the supplied two-sided threshold.
The interval contains zero, consistent with the test decision. Failure to reject does not establish equal means. Unequal sample standard deviations do not by themselves logically refute the model assumption, but the pooled method depends on it; a different procedure would need separate justification.
Self-marking points
The pooled variance is ((6-1)*2^2+(6-1)*4^2)/(6+6-2)=10.
The estimated standard error of the difference is sqrt(10*(1/6+1/6))=sqrt(10/3).
The statistic is 4/sqrt(10/3)=2.19089..., with 10 degrees of freedom.
Since |t|<2.228, do not reject equality of means at the two-sided 5% level.
The confidence interval is 4 +/- 2.228sqrt(10/3), approximately [-0.0678,8.0678].
Pooling relies on equal population variances; without that assumption this pooled t procedure is not justified.
Examiner tip
Use the pooled variance, not the mean of the two standard deviations, and do not interpret non-significance as proof of equality.
Question 4A die is rolled independently 60 times with observed counts 8,12,10,9,11,10. Test the fully specified fair-die model at 5%, using the given chi-squared critical value 11.070 for 5 degrees of freedom. Separately, an independent random sample is classified by two binary attributes with observed 2x2 table [[20,10],[10,20]]. Test independence at 5%, using the given critical value 3.841 for 1 degree of freedom. Use Pearson's statistic without a continuity correction and state why the expected-count approximation is acceptable.7 indicative marks
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A die is rolled independently 60 times with observed counts 8,12,10,9,11,10. Test the fully specified fair-die model at 5%, using the given chi-squared critical value 11.070 for 5 degrees of freedom. Separately, an independent random sample is classified by two binary attributes with observed 2x2 table [[20,10],[10,20]]. Test independence at 5%, using the given critical value 3.841 for 1 degree of freedom. Use Pearson's statistic without a continuity correction and state why the expected-count approximation is acceptable.
Cambridge 2027: Direct conceptual match
Independently totalled all six Pearson contributions and computed all four independence expectations from margins; checked df=5 and 1 and expected counts 10 and 15 respectively.
For a fully specified uniform die, the six expectations are fixed before seeing the outcomes. Calculate each contribution (O-E)^2/E; their sum is 1. The total-count constraint removes one degree of freedom.
For independence, fit expectations using row total * column total / grand total. All four fitted values equal 15 and all four absolute residuals equal 5, giving 100/15=20/3.
Use the appropriate degrees of freedom and supplied critical value in each case. The second result indicates association, not causation. Counts here satisfy the usual expected-count rule; grouping would need consideration if small expected counts occurred.
Self-marking points
Under the fair-die model each expected count is 10.
The goodness-of-fit statistic is (4+4+0+1+1+0)/10=1, with 6-1=5 degrees of freedom.
Since 1<11.070, do not reject the specified fair-die model; this does not prove fairness.
In the 2x2 table both row and column totals are 30 and the total is 60, so each fitted expected count is 15.
The independence statistic is 4*(5^2/15)=20/3, with (2-1)(2-1)=1 degree of freedom.
Since 20/3>3.841, reject independence; there is evidence of association in the sampled population.
All expected counts are at least 5 and the observations are independent; no parameters were estimated for the fair-die probabilities.
Examiner tip
Degrees of freedom differ between a specified goodness-of-fit model and a contingency table; base approximation checks on expected, not observed, counts.
Question 5Nine independent, non-zero paired changes are -1,2,3,4,5,6,7,8,9. Test whether the population median change is positive, first using a one-sided sign test and then a one-sided Wilcoxon signed-rank test at 5%. For the signed-rank test additionally assume a continuous distribution symmetric about its location. There are no ties. Calculate exact tail probabilities by counting equally likely sign assignments under the respective null hypotheses. Explain the differing assumptions.6 indicative marks
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Nine independent, non-zero paired changes are -1,2,3,4,5,6,7,8,9. Test whether the population median change is positive, first using a one-sided sign test and then a one-sided Wilcoxon signed-rank test at 5%. For the signed-rank test additionally assume a continuous distribution symmetric about its location. There are no ties. Calculate exact tail probabilities by counting equally likely sign assignments under the respective null hypotheses. Explain the differing assumptions.
Cambridge 2027: Direct conceptual match
Enumerated the sign-test count C(9,8)+1 and the two subsets of ranks with sum at most 1; checked paired-change interpretation, absence of zeros/ties and the extra signed-rank symmetry assumption.
For the sign test, a continuous distribution with median zero has equal positive and negative probabilities. Exactly eight or nine positives can occur in C(9,8)+C(9,9)=10 of the 512 assignments.
For signed ranks, order the absolute differences. Only the smallest magnitude is negative, hence W-=1. Under the symmetry null the 512 sign patterns on ranks 1,...,9 are equally likely. Only no negative rank or rank 1 alone gives W-<=1.
The exact one-sided probabilities are both below .05. The signed-rank result uses magnitudes through ranks and has stronger assumptions; it is not an assumption-free replacement for a t test. These data are paired changes, not two independent sets of raw observations.
Self-marking points
There are 8 positive and 1 negative changes, so under the sign-test null the number positive is Bin(9,1/2).
The one-sided sign-test probability is P(B>=8)=(9+1)/2^9=10/512=0.01953125.
Ranks of absolute changes are 1 through 9; the negative rank sum W-=1 and positive rank sum W+=44.
For a positive-location alternative the relevant signed-rank tail is W-<=1; only the empty negative-rank set and {1} satisfy this.
The exact signed-rank probability is 2/512=0.00390625; both tests reject their null at 5%.
The sign test does not require symmetry; the signed-rank calculation requires independent continuous differences symmetric under the null. Pairing is handled by testing the changes, with independence across pairs.
Examiner tip
Choose the signed-rank tail to match the alternative and state the symmetry requirement; do not double a one-sided probability.
Question 6X is Bernoulli with P(X=1)=0.3. Independently, T counts the number of trials up to and including the first success, with success probability 0.5 on each independent trial. Derive the probability generating function of W=X+T. Use its first two derivatives at 1 to find E(W) and Var(W), and check those moments by independence.6 indicative marks
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X is Bernoulli with P(X=1)=0.3. Independently, T counts the number of trials up to and including the first success, with success probability 0.5 on each independent trial. Derive the probability generating function of W=X+T. Use its first two derivatives at 1 to find E(W) and Var(W), and check those moments by independence.
Cambridge 2027: Direct conceptual match
Derived the geometric PGF from its support beginning at 1; differentiated the product and independently added Bernoulli and geometric means and variances.
Because T counts the success trial its support starts at 1, giving the geometric series sum s/(2-s). For |s|<2 this series converges and in particular contains a neighbourhood of s=1.
Products of PGFs describe sums of independent random variables. Differentiate the two factors: G_X'=0.3, G_X''=0, G_T'=2/(2-s)^2 and G_T''=4/(2-s)^3.
At 1 the first derivative of the product is 2.3 and the second is 5.2. The second derivative gives E[W(W-1)], not E(W^2). Adding the mean and subtracting its square yields 2.21, also obtained by adding the independent variances.
Self-marking points
G_X(s)=0.7+0.3s and G_T(s)=sum_(k>=1)(s/2)^k=s/(2-s), valid for |s|<2.
Independence gives G_W(s)=(0.7+0.3s)s/(2-s).
G_T'(1)=2 and G_T''(1)=4; hence G_W'(1)=0.3+2=2.3.
