Edexcel S3 2014 June — Question 7 10 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2014
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyStandard +0.3 This is a straightforward application of standard hypothesis testing procedures with clearly defined steps: part (a) requires inverse normal calculation (routine), and part (b) is a textbook one-sample z-test with known variance. The question provides all necessary information explicitly and follows a standard template, making it slightly easier than average despite being from Further Maths Statistics.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.05e Hypothesis test for normal mean: known variance

7. A machine fills packets with \(X\) grams of powder where \(X\) is normally distributed with mean \(\mu\). Each packet is supposed to contain 1 kg of powder. To comply with regulations, the weight of powder in a randomly selected packet should be such that \(\mathrm { P } ( X < \mu - 30 ) = 0.0005\)
  1. Show that this requires the standard deviation to be 9.117 g to 3 decimal places. A random sample of 10 packets is selected from the machine. The weight, in grams, of powder in each packet is as follows 999.8991 .61000 .31006 .11008 .2997 .0993 .21000 .0997 .11002 .1
  2. Assuming that the standard deviation of the population is 9.117 g , test, at the \(1 \%\) significance level, whether or not the machine is delivering packets with mean weight of less than 1 kg . State your hypotheses clearly.

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z = \pm 3.2905\)B1
\(\sigma = \dfrac{30}{3.2905}\)M1 M1 for \(\dfrac{30}{\text{their }
\(\sigma = 9.117\)A1cso A1 cso as given answer
(3 marks)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: \mu = 1000\), \(H_1: \mu < 1000\)B1 1st B1 both hypotheses correct. Accept 1kg if consistent units used
mean weight \(= 999.54\)B1 2nd B1: 999.54 (g) or 0.99954 (kg)
\(z = \dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} = \dfrac{(999.54-1000)}{\frac{9.117}{\sqrt{10}}} = -0.160\) or \(\dfrac{c-1000}{\sqrt{\frac{83.12}{10}}} = -2.3263 \therefore \text{CR } c < 993.29\)M1A1 1st M1 for standardising using their mean, allow \(\pm\), 1000 and \(\frac{9.117}{\sqrt{10}}\). 1st A1 awrt \(-0.160\) unless clearly using \(
1% critical value \(= -2.3263\)B1 3rd B1: \(\pm 2.3263\) sign consistent with test statistic or \(p = 0.4364 > 0.01\); NB \(p = 0.5636 < 0.99\)
\(-2.3263 \not< -0.160\)
Accept \(H_0\) / not in critical regiondM1 2nd dM1 dependent upon 1st M for correct statement linking test statistic and cv. Contradictory statements score M0
There is no evidence that the machine is delivering packets of mean weight less than 1 kgA1ft 2nd A1 for correct conclusion in context. Must mention 'machine' and 'packets'
(7 marks) — Total 10
## Question 7:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = \pm 3.2905$ | B1 | |
| $\sigma = \dfrac{30}{3.2905}$ | M1 | M1 for $\dfrac{30}{\text{their } |z|}$, $|z| > 1$ |
| $\sigma = 9.117$ | A1cso | A1 cso as given answer |

**(3 marks)**

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 1000$, $H_1: \mu < 1000$ | B1 | 1st B1 both hypotheses correct. Accept 1kg if consistent units used |
| mean weight $= 999.54$ | B1 | 2nd B1: 999.54 (g) or 0.99954 (kg) |
| $z = \dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} = \dfrac{(999.54-1000)}{\frac{9.117}{\sqrt{10}}} = -0.160$ or $\dfrac{c-1000}{\sqrt{\frac{83.12}{10}}} = -2.3263 \therefore \text{CR } c < 993.29$ | M1A1 | 1st M1 for standardising using their mean, allow $\pm$, 1000 and $\frac{9.117}{\sqrt{10}}$. 1st A1 awrt $-0.160$ unless clearly using $|z|$, then accept 0.160 or CR awrt 993. Condone $-0.16$ if fully correct expression seen |
| 1% critical value $= -2.3263$ | B1 | 3rd B1: $\pm 2.3263$ sign consistent with test statistic or $p = 0.4364 > 0.01$; NB $p = 0.5636 < 0.99$ |
| $-2.3263 \not< -0.160$ | | |
| Accept $H_0$ / not in critical region | dM1 | 2nd dM1 dependent upon 1st M for correct statement linking test statistic and cv. Contradictory statements score M0 |
| There is no evidence that the machine is delivering packets of mean weight less than 1 kg | A1ft | 2nd A1 for correct conclusion in context. Must mention 'machine' and 'packets' |

**(7 marks) — Total 10**

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7. A machine fills packets with $X$ grams of powder where $X$ is normally distributed with mean $\mu$. Each packet is supposed to contain 1 kg of powder.

To comply with regulations, the weight of powder in a randomly selected packet should be such that $\mathrm { P } ( X < \mu - 30 ) = 0.0005$
\begin{enumerate}[label=(\alph*)]
\item Show that this requires the standard deviation to be 9.117 g to 3 decimal places.

A random sample of 10 packets is selected from the machine. The weight, in grams, of powder in each packet is as follows

999.8991 .61000 .31006 .11008 .2997 .0993 .21000 .0997 .11002 .1
\item Assuming that the standard deviation of the population is 9.117 g , test, at the $1 \%$ significance level, whether or not the machine is delivering packets with mean weight of less than 1 kg . State your hypotheses clearly.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S3 2014 Q7 [10]}}