Question 5 - A-Level Maths

OCR S2 2012 June — Question 5 11 marks

Exam Board	OCR
Module	S2 (Statistics 2)
Year	2012
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Z-tests (known variance)
Type	Two-tail z-test
Difficulty	Moderate -0.3 This is a straightforward two-tail hypothesis test with summary statistics provided. Students must calculate sample mean and variance, then perform a standard z-test procedure. While it requires multiple steps (calculate statistics, set up hypotheses, find critical values, conclude), each step follows a routine algorithm with no conceptual challenges or novel problem-solving required. Slightly easier than average due to being a textbook application of the hypothesis testing framework.
Spec	5.05b Unbiased estimates: of population mean and variance 5.05c Hypothesis test: normal distribution for population mean

5 The acidity $A$ (measured in pH ) of soil of a particular type has a normal distribution. The pH values of a random sample of 80 soil samples from a certain region can be summarised as $$\Sigma a = 496 , \quad \Sigma a ^ { 2 } = 3126 .$$ Test, at the $10 \%$ significance level, whether in this region the mean pH of soil is 6.1 .

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Question 5:

Answer	Marks	Guidance
Answer	Marks	Guidance
$H_0: \mu = 6.1$, $H_1: \mu \neq 6.1$	B2	Both: B2. One error, B1, but $\bar{x}$, $x$, $r$ etc: 0. 6.2: B0
$\hat{\mu} = \bar{x} = 6.2$	B1	6.2 $[31/5]$ seen somewhere (other than hypotheses)
$\hat{\sigma}^2 = \dfrac{80}{79}\!\left(\dfrac{3126}{80} - 6.2^2\right) = 0.643$	M1	Correct formula for biased estimate $[0.635$ or $127/200]$. If single formula used, M2 or, if wrong, allow M1 for divisor 79 anywhere
M1	Divide by 79 somewhere
A1	Variance estimate, a.r.t. 0.643, can be implied. $[254/395$ leading to $127/15800]$
$\alpha$: $z = \dfrac{6.2 - 6.1}{\sqrt{0.643/80}} = 1.115$	M1	Standardise their 6.2 with reasonable variance attempt, needs 80, allow cc. 80 needed, otherwise M0 and no more marks. If clearly $\mu = 6.2$ used, no more marks
$[1 - \Phi(1.115) = 0.1325 > 0.05]$	A1	$z \in [1.11, 1.12]$ (not $-$) or $p \in [0.1323, 0.1333]$. A1 uses number used for comparison
$1.115 < 1.645$	A1	Compare $z$ with 1.645 (allow $-1.645$ if $z < 0$) or $p\ (< 0.5)$ with 0.05. Withhold if inequality incorrect or if 1-tailed. Must be consistent signs/tails and like-with-like
$\beta$: CV $6.1 + 1.645 \times \sqrt{\dfrac{0.643}{80}}$, $= 6.247$ and $6.2 < 6.247$	M1	$6.1 + z\sqrt{(\sigma^2/80)}$, allow $\pm$, $\sqrt{}$ errors. Allow $6.2 - (\text{or} \pm)$ but no more marks afterwards
A1	CV, a.r.t. 6.25, needs $z = 1.645$, allow biased $\hat{\sigma}^2$. If no 79 earlier but used here, recovers M1A1
$A1\sqrt{}$	Compare 6.2 with CV from $+$ sign, $\sqrt{}$ on $z$ (but not $\sigma$)
Do not reject $H_0$. Insufficient evidence that pH value is not 6.1	M1	Needs essentially correct method and comparison, needs 80 but no need for correct variance. First conclusion wrong: M0A0 even if second correct.
$A1\sqrt{}$	Needs context and "evidence" or equivalent, ft on their $z/p$/CV. "1.115 $> $ 1.645 so do not reject $H_0$" etc: (A0)M1A1
[11]
Notes: Biased estimate used: typically gets B2B1 M1M0A0 M1A0A1 M1A1 [total 8]		$\bar{x}$ and $\mu$ interchanged: allow final M1A1 if anywhere right, but if always wrong (in hypotheses and $z$) M0A0. Typically gets B0B0B1 M1M1A1 M1A0A0 M0A0 [total 5]

## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 6.1$, $H_1: \mu \neq 6.1$ | B2 | Both: B2. One error, B1, but $\bar{x}$, $x$, $r$ etc: 0. 6.2: B0 |
| $\hat{\mu} = \bar{x} = 6.2$ | B1 | 6.2 $[31/5]$ seen somewhere (other than hypotheses) |
| $\hat{\sigma}^2 = \dfrac{80}{79}\!\left(\dfrac{3126}{80} - 6.2^2\right) = 0.643$ | M1 | Correct formula for biased estimate $[0.635$ or $127/200]$. If single formula used, M2 or, if wrong, allow M1 for divisor 79 anywhere |
| | M1 | Divide by 79 somewhere |
| | A1 | Variance estimate, a.r.t. 0.643, can be implied. $[254/395$ leading to $127/15800]$ |
| $\alpha$: $z = \dfrac{6.2 - 6.1}{\sqrt{0.643/80}} = 1.115$ | M1 | Standardise their 6.2 with reasonable variance attempt, needs 80, allow cc. *80 needed, otherwise M0 and no more marks. If clearly $\mu = 6.2$ used, no more marks* |
| $[1 - \Phi(1.115) = 0.1325 > 0.05]$ | A1 | $z \in [1.11, 1.12]$ (not $-$) or $p \in [0.1323, 0.1333]$. A1 uses number used for comparison |
| $1.115 < 1.645$ | A1 | Compare $z$ with 1.645 (allow $-1.645$ if $z < 0$) or $p\ (< 0.5)$ with 0.05. Withhold if inequality incorrect or if 1-tailed. Must be consistent signs/tails and like-with-like |
| $\beta$: CV $6.1 + 1.645 \times \sqrt{\dfrac{0.643}{80}}$, $= 6.247$ and $6.2 < 6.247$ | M1 | $6.1 + z\sqrt{(\sigma^2/80)}$, allow $\pm$, $\sqrt{}$ errors. Allow $6.2 - (\text{or} \pm)$ but no more marks afterwards |
| | A1 | CV, a.r.t. 6.25, needs $z = 1.645$, allow biased $\hat{\sigma}^2$. If no 79 earlier but used here, recovers M1A1 |
| | $A1\sqrt{}$ | Compare 6.2 with CV from $+$ sign, $\sqrt{}$ on $z$ (but not $\sigma$) |
| Do not reject $H_0$. Insufficient evidence that pH value is not 6.1 | M1 | Needs essentially correct method and comparison, needs 80 but no need for correct variance. First conclusion wrong: M0A0 even if second correct. |
| | $A1\sqrt{}$ | Needs context and "evidence" or equivalent, ft on their $z/p$/CV. "1.115 $> $ 1.645 so do not reject $H_0$" etc: (A0)M1A1 |
| | **[11]** | |
| **Notes:** Biased estimate used: typically gets B2B1 M1M0A0 M1A0A1 M1A1 [total 8] | | $\bar{x}$ and $\mu$ interchanged: allow final M1A1 if *anywhere* right, but if always wrong (in hypotheses and $z$) M0A0. Typically gets B0B0B1 M1M1A1 M1A0A0 M0A0 [total 5] |

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5 The acidity $A$ (measured in pH ) of soil of a particular type has a normal distribution. The pH values of a random sample of 80 soil samples from a certain region can be summarised as

$$\Sigma a = 496 , \quad \Sigma a ^ { 2 } = 3126 .$$

Test, at the $10 \%$ significance level, whether in this region the mean pH of soil is 6.1 .

\hfill \mbox{\textit{OCR S2 2012 Q5 [11]}}

This paper (8 questions)

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Q1 2 Q2 6 Q3 7 Q4 11 Q5 11 Q6 11 Q7 12 Q8 12

Answer	Marks	Guidance
Answer	Marks	Guidance
\(H_0: \mu = 6.1\), \(H_1: \mu \neq 6.1\)	B2	Both: B2. One error, B1, but \(\bar{x}\), \(x\), \(r\) etc: 0. 6.2: B0
\(\hat{\mu} = \bar{x} = 6.2\)	B1	6.2 \([31/5]\) seen somewhere (other than hypotheses)
\(\hat{\sigma}^2 = \dfrac{80}{79}\!\left(\dfrac{3126}{80} - 6.2^2\right) = 0.643\)	M1	Correct formula for biased estimate \([0.635\) or \(127/200]\). If single formula used, M2 or, if wrong, allow M1 for divisor 79 anywhere
M1	Divide by 79 somewhere
A1	Variance estimate, a.r.t. 0.643, can be implied. \([254/395\) leading to \(127/15800]\)
\(\alpha\): \(z = \dfrac{6.2 - 6.1}{\sqrt{0.643/80}} = 1.115\)	M1	Standardise their 6.2 with reasonable variance attempt, needs 80, allow cc. 80 needed, otherwise M0 and no more marks. If clearly \(\mu = 6.2\) used, no more marks
\([1 - \Phi(1.115) = 0.1325 > 0.05]\)	A1	\(z \in [1.11, 1.12]\) (not \(-\)) or \(p \in [0.1323, 0.1333]\). A1 uses number used for comparison
\(1.115 < 1.645\)	A1	Compare \(z\) with 1.645 (allow \(-1.645\) if \(z < 0\)) or \(p\ (< 0.5)\) with 0.05. Withhold if inequality incorrect or if 1-tailed. Must be consistent signs/tails and like-with-like
\(\beta\): CV \(6.1 + 1.645 \times \sqrt{\dfrac{0.643}{80}}\), \(= 6.247\) and \(6.2 < 6.247\)	M1	\(6.1 + z\sqrt{(\sigma^2/80)}\), allow \(\pm\), \(\sqrt{}\) errors. Allow \(6.2 - (\text{or} \pm)\) but no more marks afterwards
A1	CV, a.r.t. 6.25, needs \(z = 1.645\), allow biased \(\hat{\sigma}^2\). If no 79 earlier but used here, recovers M1A1
\(A1\sqrt{}\)	Compare 6.2 with CV from \(+\) sign, \(\sqrt{}\) on \(z\) (but not \(\sigma\))
Do not reject \(H_0\). Insufficient evidence that pH value is not 6.1	M1	Needs essentially correct method and comparison, needs 80 but no need for correct variance. First conclusion wrong: M0A0 even if second correct.
\(A1\sqrt{}\)	Needs context and "evidence" or equivalent, ft on their \(z/p\)/CV. "1.115 \(> \) 1.645 so do not reject \(H_0\)" etc: (A0)M1A1
[11]
Notes: Biased estimate used: typically gets B2B1 M1M0A0 M1A0A1 M1A1 [total 8]		\(\bar{x}\) and \(\mu\) interchanged: allow final M1A1 if anywhere right, but if always wrong (in hypotheses and \(z\)) M0A0. Typically gets B0B0B1 M1M1A1 M1A0A0 M0A0 [total 5]