A2 October 2020 Q7

EdexcelCurrent spec15 marksIncludes hypothesis testingCentral Limit Theorem Quality of Tests

7. A six-sided die has sides labelled 1, 2, 3, 4, 5 and 6

The random variable \(S\) represents the score when the die is rolled.

Alicia rolls the die 45 times and the mean score, \(\overline{S}\), is calculated.

Assuming the die is fair and using a suitable approximation,

(a) find, to 3 significant figures, the value of \(k\) such that \(\mathrm{P}(\overline{S} \lt k) = 0.05\) (8)

(b) Explain the relevance of the Central Limit Theorem in part (a). (2)

Alicia considers the following hypotheses:

\(\mathrm{H}_0\): The die is fair

\(\mathrm{H}_1\): The die is not fair

If \(\overline{S} \lt 3.1\) or \(\overline{S} \gt 3.9\), then \(\mathrm{H}_0\) will be rejected.

Given that the true distribution of \(S\) has mean 4 and variance 3

(d) Describe what would happen to the power of this test if Alicia were to increase the number of rolls of the die.
Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
Realising \(S\) has a discrete uniform distribution over \(\{1, \ldots 6\}\)	M1	3.3
\(\mathrm{E}(S) = 1 \times \tfrac{1}{6} + 2 \times \tfrac{1}{6} + 3 \times \tfrac{1}{6} + 4 \times \tfrac{1}{6} + 5 \times \tfrac{1}{6} + 6 \times \tfrac{1}{6}\)	M1	1.1b
\(\mathrm{Var}(S) = \tfrac{6^2 - 1}{12}\) or \(1^2 \times \tfrac{1}{6} + 2^2 \times \tfrac{1}{6} + 3^2 \times \tfrac{1}{6} + 4^2 \times \tfrac{1}{6} + 5^2 \times \tfrac{1}{6} + 6^2 \times \tfrac{1}{6} - 3.5^2\)	M1	1.1b
\(\mathrm{E}(S) = 3.5\) and \(\mathrm{Var}(S) = \tfrac{35}{12}\)	A1	1.1b
\(\overline{S} \sim \mathrm{N}(3.5, \ldots)\)	M1	3.1a
\(\mathrm{Var}(\overline{S}) = \dfrac{\frac{35}{12}}{45} = \tfrac{7}{108}, \ \ \overline{S} \sim \mathrm{N}(3.5, 0.0648\ldots)\)	A1	1.1b
\(\mathrm{P}(\overline{S} \lt k) = 0.05 \rightarrow \dfrac{k - 3.5}{\sqrt{\frac{7}{108}}} = -1.6449\)	M1	3.4
\(k = 3.08122\ldots\) awrt 3.08	A1	1.1b
	(8)

Notes

M1: Setting up model for \(S\)

M1: Attempt at expression for \(\mathrm{E}(S)\)

M1: Attempt at expression for \(\mathrm{Var}(S)\)

A1: Correct mean and variance for \(S\) (may be implied by a correct distribution for \(\overline{S}\))

M1: Use of CLT to find distribution for \(\overline{S} \sim \mathrm{N}(\text{‘}3.5\text{’}, \ldots)\) f.t. their 3.5 but variance \(\neq \tfrac{35}{12}\)

A1: Correct distribution with correct variance, allow \(\sigma^2 =\) awrt 0.0648 or \(\sigma =\) awrt 0.255

M1: Standardising using their model and equating to a \(z\)-value \(1 \lt |z| \lt 2\)

A1: awrt 3.08

Mark scheme (b)

Scheme	Marks	AO
CLT applies since the sample size is large	B1	3.5b
CLT states that the sample mean/\(\overline{S}\) is (approximately) normally distributed	B1	3.5b
	(2)

Notes

B1: Correct explanation about appropriateness of the CLT given large sample size (allow > 30)

B1: Requires both sample and mean or \(\overline{S}\)

Mark scheme (c)

Scheme	Marks	AO
True \(\overline{S} \sim \mathrm{N}(4, \tfrac{3}{45})\)	M1	3.3
\(\mathrm{P}(\overline{S} \lt 3.1) + \mathrm{P}(\overline{S} \gt 3.9)\) or \(1 - \mathrm{P}(3.1 \lt \overline{S} \lt 3.9)\)	dM1	3.4
Power = awrt 0.651	A1	1.1b
	(3)

Notes

M1: Writing or using \(\overline{S} \sim \mathrm{N}(4, \tfrac{3}{45})\) allow \(\sigma^2 =\) awrt 0.0667 or \(\sigma =\) awrt 0.258

dM1: (dep on 1^st M1) correct probability statement for power

A1: awrt 0.651

Mark scheme (d)

Scheme	Marks	AO
E.g. The increase in sample size would decrease the variance of \(\overline{S}\) [leading to an increase in \(\mathrm{P}(\overline{S} \gt 3.9)\) and the decrease in \(\mathrm{P}(\overline{S} \lt 3.1)\) would be negligible]	B1	2.4
So the power would increase.	dB1	2.2a
	(2)
	(15 marks)

Notes

B1: Correct reasoning which refers to decrease in variance

dB1: (dep on 1^st B1) Correct deduction with no incorrect reasoning