Edexcel S4 2009 June — Question 1 8 marks

Exam BoardEdexcel
ModuleS4 (Statistics 4)
Year2009
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyStandard +0.3 This is a straightforward one-tail z-test with given summary statistics. Students must calculate sample mean and variance, perform a standard hypothesis test procedure, and compare to critical value. While it requires multiple steps (calculate statistics, set up hypotheses, find test statistic, conclude), each step follows a standard algorithm taught in S4 with no novel problem-solving required. Slightly above average difficulty due to being Further Maths content and requiring careful execution of the full hypothesis testing framework.
Spec5.05c Hypothesis test: normal distribution for population mean

  1. A company manufactures bolts with a mean diameter of 5 mm . The company wishes to check that the diameter of the bolts has not decreased. A random sample of 10 bolts is taken and the diameters, \(x \mathrm {~mm}\), of the bolts are measured. The results are summarised below.
$$\sum x = 49.1 \quad \sum x ^ { 2 } = 241.2$$ Using a \(1 \%\) level of significance, test whether or not the mean diameter of the bolts is less than 5 mm .
(You may assume that the diameter of the bolts follows a normal distribution.)

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 5\); \(H_1: \mu < 5\)B1 both
CR: \(t_9(0.01) > 2.821\)B1
\(\bar{x} = 4.91\)B1
\(s^2 = \frac{1}{9}\left(241.2 - \frac{49.1^2}{10}\right) = 0.0132222\)M1 A1 s = awrt 0.115
\(t = \frac{4.91 - 5 }{\frac{\sqrt{0.013222}}{\sqrt{10}}} = \pm 2.475\)
Since 2.475 is not in the critical region there is insufficient evidence to reject \(H_0\) and conclude that the mean diameter of the bolts is not less than (not equal to) 5mm.A1ft
# Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 5$; $H_1: \mu < 5$ | B1 | both |
| CR: $t_9(0.01) > 2.821$ | B1 | |
| $\bar{x} = 4.91$ | B1 | |
| $s^2 = \frac{1}{9}\left(241.2 - \frac{49.1^2}{10}\right) = 0.0132222$ | M1 A1 | s = awrt 0.115 |
| $t = \frac{|4.91 - 5|}{\frac{\sqrt{0.013222}}{\sqrt{10}}} = \pm 2.475$ | M1 A1 | 2.47 – 2.48 |
| Since 2.475 is not in the critical region there is insufficient evidence to reject $H_0$ and conclude that the mean diameter of the bolts is not less than (not equal to) 5mm. | A1ft | |

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\begin{enumerate}
  \item A company manufactures bolts with a mean diameter of 5 mm . The company wishes to check that the diameter of the bolts has not decreased. A random sample of 10 bolts is taken and the diameters, $x \mathrm {~mm}$, of the bolts are measured. The results are summarised below.
\end{enumerate}

$$\sum x = 49.1 \quad \sum x ^ { 2 } = 241.2$$

Using a $1 \%$ level of significance, test whether or not the mean diameter of the bolts is less than 5 mm .\\
(You may assume that the diameter of the bolts follows a normal distribution.)\\

\hfill \mbox{\textit{Edexcel S4 2009 Q1 [8]}}