A2 June 2023 Q8

EdexcelCurrent spec12 marksConfidence Intervals

8. A bag contains a large number of marbles of which an unknown proportion, \(p\), is yellow.

Three random samples of size \(n\) are taken, and the number of yellow marbles in each sample, \(Y_1\), \(Y_2\) and \(Y_3\), is recorded.

Two estimators \(\hat{p}_1\) and \(\hat{p}_2\) are proposed to estimate the value of \(p\)

\[\hat{p}_1 = \frac{Y_1 + 3Y_2 - 2Y_3}{2n}\]\[\hat{p}_2 = \frac{2Y_1 + 3Y_2 + Y_3}{6n}\]

(a) Show that \(\hat{p}_1\) and \(\hat{p}_2\) are both unbiased estimators of \(p\) (3)

(b) Find the variance of \(\hat{p}_1\) (2)

The variance of \(\hat{p}_2\) is \(\dfrac{7p(1-p)}{18n}\)

The estimator \(\hat{p}_3 = \dfrac{Y_1 + aY_2 + 3Y_3}{bn}\) where \(a\) and \(b\) are positive integers.

(d) Find the pair of values of \(a\) and \(b\) such that \(\hat{p}_3\) is a better unbiased estimator of \(p\) than both \(\hat{p}_1\) and \(\hat{p}_2\)
You must show all stages of your working. (5)

Mark scheme (a)

Scheme	Marks	AO
\(Y \sim \mathrm{B}(n, p) \quad \mathrm{E}(Y) = np\)	M1	3.3
\(\mathrm{E}(\hat{p}_1) = \dfrac{np + 3np - 2np}{2n} = p\)	M1	3.4
\(\mathrm{E}(\hat{p}_2) = \dfrac{2np + 3np + np}{6n} = p\), therefore, both are unbiased.	A1cso	1.1b
	(3)

Notes

M1: use of mean of \(np\)

M1: use of binomial model (may be implied by use of \(np\)) to find either expected value

A1cso: correct solution with both expectations evaluated to \(p\) and correct conclusion

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{Var}(\hat{p}_1) = \dfrac{np(1-p) + 9np(1-p) + 4np(1-p)}{4n^2}\)	M1	2.1
\(\mathrm{Var}(\hat{p}_1) = \dfrac{7p(1-p)}{2n}\)	A1	1.1b
	(2)

Notes

M1: use of sum of variances, all terms positive and sight or use of 3² and 2²

A1: correct expression for \(\mathrm{Var}(\hat{p}_1)\)

Mark scheme (c)

Scheme	Marks	AO
[Both are unbiased so…] as \(\mathrm{Var}(\hat{p}_2) \lt \mathrm{Var}(\hat{p}_1)\)	M1	2.4
\(\hat{p}_2\) is the better estimator.	A1ft	2.2a
	(2)

Notes

M1: correct explanation, ft their \(\mathrm{Var}(\hat{p}_1)\)

A1ft: correct deduction, consistent with their \(\mathrm{Var}(\hat{p}_1)\)

Mark scheme (d)

Scheme	Marks	AO
\(\hat{p}_3\) is unbiased so \(b = a + 4\)	M1	1.1b
Require \(\mathrm{Var}(\hat{p}_3) \lt \mathrm{Var}(\hat{p}_2)\) so \(\dfrac{a^2 + 10}{b^2} \lt \dfrac{7}{18}\)	M1	2.1
\(18(a^2 + 10) \lt 7(a + 4)^2 \to 11a^2 - 56a + 68 \lt 0\)	M1	1.1b
\(2 \lt a \lt \frac{34}{11}\)	A1ft	1.1b
\(a = 3\) and \(b = 7\)	A1	2.2a
	(5)
	(12 marks)

Notes

M1: use of \(\mathrm{E}(\hat{p}_3) = p\) to set up an equation in \(a\) and \(b\) only

M1: setting up an inequality \(\mathrm{Var}(\hat{p}_3) \lt \mathrm{Var}(\hat{p}_2)\) using a correct expression for \(\mathrm{Var}(\hat{p}_3)\)

M1: eliminating one variable and setting up 3TQ inequality

A1ft: correct solution to their inequality, must choose “inside” region

A1: correctly deducing the values of \(a\) and \(b\)