Edexcel S4 2005 June — Question 3 8 marks

Exam BoardEdexcel
ModuleS4 (Statistics 4)
Year2005
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward one-sample t-test with clear hypotheses (H₀: μ = 1010, H₁: μ < 1010), small sample requiring calculation of sample mean and standard deviation, then comparison with critical value. While it requires multiple computational steps and understanding of hypothesis testing framework, it follows a standard S4 template with no conceptual surprises or novel problem-solving required—slightly easier than average due to its routine nature.
Spec5.05c Hypothesis test: normal distribution for population mean

3. A machine is set to fill bags with flour such that the mean weight is 1010 grams. To check that the machine is working properly, a random sample of 8 bags is selected. The weight of flour, in grams, in each bag is as follows. $$\begin{array} { l l l l l l l l } 1010 & 1015 & 1005 & 1000 & 998 & 1008 & 1012 & 1007 \end{array}$$ Carry out a suitable test, at the \(5 \%\) significance level, to test whether or not the mean weight of flour in the bags is less than 1010 grams. (You may assume that the weight of flour delivered by the machine is normally distributed.)
(Total 8 marks)

AnswerMarks Guidance
Answer/WorkingMarks Guidance
Let \(x\) = net weight of flour. \(\sum x = 1055\); \(\bar{x} = \frac{1006.875}{8}\)B1 AWRT 1006.9
\(\sum x^2 = 8110611\); \(s^2 = \frac{1}{7}(8110611 - \frac{1055^2}{8}) = 33.2875...\)M1 A1 Allow from calculator; AWRT 33.7 or AWRT 5.77
\(\therefore s = 5.767...\)A1
\(H_0: \mu = 1010\); \(H_1: \mu < 1010\)B1 Both
CV: \(\t\ = 1.895\)
\(t = \frac{\1006.875 - 1010\ }{5.767.../\sqrt{7}} = \pm 1.5324\)
Since -1.53 is not in the critical region (\(t<-1.895\)), there is insufficient evidence to reject \(H_0\) and we conclude that the mean weight of flour delivered by the machine is 1010g.A1N (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $x$ = net weight of flour. $\sum x = 1055$; $\bar{x} = \frac{1006.875}{8}$ | B1 | AWRT 1006.9 |
| $\sum x^2 = 8110611$; $s^2 = \frac{1}{7}(8110611 - \frac{1055^2}{8}) = 33.2875...$ | M1 A1 | Allow from calculator; AWRT 33.7 or AWRT 5.77 |
| $\therefore s = 5.767...$ | A1 | |
| $H_0: \mu = 1010$; $H_1: \mu < 1010$ | B1 | Both |
| CV: $\|t\| = 1.895$ | B1 | |
| $t = \frac{\|1006.875 - 1010\|}{5.767.../\sqrt{7}} = \pm 1.5324$ | M1 A1 | Use $\bar{x} - \mu_0$ / $s/\sqrt{n}$; AWRT -1.53 |
| Since -1.53 is not in the critical region ($t<-1.895$), there is insufficient evidence to reject $H_0$ and we conclude that the mean weight of flour delivered by the machine is 1010g. | A1N (a) | |

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3. A machine is set to fill bags with flour such that the mean weight is 1010 grams.

To check that the machine is working properly, a random sample of 8 bags is selected. The weight of flour, in grams, in each bag is as follows.

$$\begin{array} { l l l l l l l l } 
1010 & 1015 & 1005 & 1000 & 998 & 1008 & 1012 & 1007
\end{array}$$

Carry out a suitable test, at the $5 \%$ significance level, to test whether or not the mean weight of flour in the bags is less than 1010 grams. (You may assume that the weight of flour delivered by the machine is normally distributed.)\\
(Total 8 marks)\\

\hfill \mbox{\textit{Edexcel S4 2005 Q3 [8]}}