OCR S2 2013 June — Question 6 11 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2013
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyStandard +0.3 This is a straightforward two-tail hypothesis test with clearly stated hypotheses, given summary statistics, and standard procedure. Students must calculate the sample mean (36.68), estimate variance, compute a z-statistic, and compare to critical values at 1% level. While it requires multiple steps and careful calculation, it follows a completely standard template taught in S2 with no conceptual challenges or novel elements, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

6 The random variable \(X\) denotes the yield, in kilograms per acre, of a certain crop. Under the standard treatment it is known that \(\mathrm { E } ( X ) = 38.4\). Under a new treatment, the yields of 50 randomly chosen regions can be summarised as $$n = 50 , \quad \sum x = 1834.0 , \quad \sum x ^ { 2 } = 70027.37 .$$ Test at the \(1 \%\) level whether there has been a change in the mean crop yield.

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu = 38.4\) [Allow E(\(X\)) both times]; \(H_1: \mu \neq 38.4\)B2 Both correct: B2. One error e.g. no or different symbols, one-tail etc: B1. But \(\bar{x}\), \(x\), \(t\) etc: B0
\(\hat{\mu} = \bar{x} = 36.68\)B1 36.68 seen anywhere
\(\hat{\sigma}^2 = \dfrac{50}{49}\!\left(\dfrac{70027.37}{50} - 36.68^2\right) = 56.25\)M1 M1 A1 Use biased variance formula \([55.125]\). Multiply by 50/49. 56.25. Allow rounded if clearly correct
\(\alpha\): \(z = \dfrac{36.68 - 38.4}{\sqrt{56.25/50}} = -1.62\)M1 A1 Standardise using \(\sqrt{50}\) or 50. \(z\), a.r.t. \(-1.62\) or \(p = 0.0525\). If 50 missing, no more marks
\(> -2.576\) [or \(0.0525 > .005\)]A1 ft Compare \(-z\) with \(-2.576\) or \(+z\) with 2.576. \(p\) in range \([0.052, 0.053]\)
\(\beta\): CV is \(38.4 - 2.576\sqrt{\dfrac{56.25}{50}} = 35.6677\); \(36.68 > 35.6677\)M1 A1 A1 ft CV \(38.4 - z\sigma/\sqrt{50}\), ignore \(38.4 +\) anything. A.r.t 35.7. CV ft and correct comparison
Do not reject \(H_0\). Insufficient evidence of a change in crop yieldM1 A1 ft Correct first conclusion, needs correct method & comparison if seen. Contextualised, "evidence" somewhere. *Not* "evidence of no change"
Total: 11
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 38.4$ [Allow E($X$) both times]; $H_1: \mu \neq 38.4$ | B2 | Both correct: B2. One error e.g. no or different symbols, one-tail etc: B1. But $\bar{x}$, $x$, $t$ etc: B0 |
| $\hat{\mu} = \bar{x} = 36.68$ | B1 | 36.68 seen anywhere |
| $\hat{\sigma}^2 = \dfrac{50}{49}\!\left(\dfrac{70027.37}{50} - 36.68^2\right) = 56.25$ | M1 M1 A1 | Use biased variance formula $[55.125]$. Multiply by 50/49. 56.25. Allow rounded if clearly correct |
| $\alpha$: $z = \dfrac{36.68 - 38.4}{\sqrt{56.25/50}} = -1.62$ | M1 A1 | Standardise using $\sqrt{50}$ or 50. $z$, a.r.t. $-1.62$ or $p = 0.0525$. If 50 missing, no more marks |
| $> -2.576$ [or $0.0525 > .005$] | A1 ft | Compare $-z$ with $-2.576$ or $+z$ with 2.576. $p$ in range $[0.052, 0.053]$ |
| $\beta$: CV is $38.4 - 2.576\sqrt{\dfrac{56.25}{50}} = 35.6677$; $36.68 > 35.6677$ | M1 A1 A1 ft | CV $38.4 - z\sigma/\sqrt{50}$, ignore $38.4 +$ anything. A.r.t 35.7. CV ft and correct comparison |
| Do not reject $H_0$. Insufficient evidence of a change in crop yield | M1 A1 ft | Correct first conclusion, needs correct method & comparison if seen. Contextualised, "evidence" somewhere. *Not* "evidence of no change" |
| | **Total: 11** | |

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6 The random variable $X$ denotes the yield, in kilograms per acre, of a certain crop. Under the standard treatment it is known that $\mathrm { E } ( X ) = 38.4$. Under a new treatment, the yields of 50 randomly chosen regions can be summarised as

$$n = 50 , \quad \sum x = 1834.0 , \quad \sum x ^ { 2 } = 70027.37 .$$

Test at the $1 \%$ level whether there has been a change in the mean crop yield.

\hfill \mbox{\textit{OCR S2 2013 Q6 [11]}}