S2 June 2014 Q5

EdexcelOld spec11 marksIncludes hypothesis testingBinomial Distribution Normal Distribution

(a) State the conditions under which the normal distribution may be used as an approximation to the binomial distribution. (2)

A company sells seeds and claims that 55% of its pea seeds germinate.

(b) Write down a reason why the company should not justify their claim by testing all the pea seeds they produce. (1)

To test the company’s claim, a random sample of 220 pea seeds was planted.

Given that 135 of the 220 pea seeds germinated,

(d) use a normal approximation to test, at the 5% level of significance, whether or not the company’s claim is justified. (7)

Scheme	Marks
\(n\) is large and \(p\) close to 0.5	B1B1
	(2)

Scheme	Marks
There would be no pea seeds left	B1
	(1)

Scheme	Marks
\(\mathrm{H}_0 : p = 0.55 \quad \mathrm{H}_1 : p \ne 0.55\)	B1
	(1)

Mark scheme (d)

Scheme	Marks
\(X \sim \mathrm{N}(121, 54.45)\)	B1
\(\mathrm{P}(X \geqslant 134.5) = \mathrm{P}\left(Z \geqslant \dfrac{134.5 - 121}{\sqrt{54.45}}\right)\) or \(\pm\dfrac{x - 0.5 - 121}{\sqrt{54.45}} = 1.96\)	M1M1A1
\(= \mathrm{P}(Z \geqslant 1.8295..)\) \(= 1 - 0.9664\)
\(= 0.0336/0.0337\) \(x = 135.96\)	A1
Accept \(\mathrm{H}_0\) not in CR, not significant	M1
The company’s claim is justified or 55% of its pea seeds germinate	A1cso
	(7)
	(11 marks)

Notes

Alternative

Scheme	Marks
\(X \sim \mathrm{N}(99, 54.45)\)	B1
\(\mathrm{P}(X \leqslant 85) = P\left(Z \leqslant \dfrac{85.5 - 99}{\sqrt{54.45}}\right)\) or \(\pm\dfrac{x + 0.5 - 99}{\sqrt{54.45}} = 1.96\)	M1 M1 A1
\(= \mathrm{P}(Z \geqslant 1.8295..)\) \(= 1 - 0.9664\)
\(= 0.0336/0.0337\) \(x = 84.04\)	A1
Accept \(\mathrm{H}_0\) not in CR, not significant	M1
The company’s claim is justified or 55% of its pea seeds germinate	A1cso

(corrected from the printed mark scheme: the Alternative prints \(x = 107.5\); solving \(\dfrac{x + 0.5 - 99}{\sqrt{54.45}} = -1.96\) gives \(x = 84.04\))

B1 correct mean and Var, may be seen in the standardiation formula as 121 and \(\sqrt{54.45}\) or 7.38 to 2dp or implied by a correct answer

M1 for attempting a continuity correction (Method 1:135/85 \(\pm\) 0.5 / Method 2:\(x \pm 0.5\))

M1 for standardising using their mean and their standard deviation and using either Method 1 [134.5, 135, 135.5, 85, 85.5 or 84.5 accept \(\pm z\).] Method 2 [ (\(x \pm 0.5\)) and equal to a \(\pm z\) value]

A1 correct \(z\) value awrt \(\pm 1.83\) or \(\pm\dfrac{134.5 - 121}{\sqrt{54.45}}\) \(\left(\dfrac{85.5 - 99}{\sqrt{54.45}}\right)\) or \(\pm\dfrac{x - 0.5 - 121}{\sqrt{54.45}} = 1.96\) \(\left(\pm\dfrac{x + 0.5 - 99}{\sqrt{54.45}} = 1.96\right)\) or(allow 1.6449 if 1 tail test in (c))

A1 awrt 0.0336/0.0337 or awrt 136 (allow 126 if one tail test in (c)) or a comparison of awrt1.83 with 1.96 (1.6449)

M1 A correct statement. Accept \(\mathrm{H}_0\), oe if a 2-tailed test in (c), reject \(\mathrm{H}_0\), oe if a 1-tailed test in (c). Allow for a correct contextual statement. Do not allow contradictions of non-contextual statements.

A1 A correct contextual statement to include words in bold/underlined for a 2-tailed test. This is not a follow through mark.

NB if finding \(\mathrm{P}(X = 135)\) they can get B1 M1 M1 A0 A0 M0 A0