AQA S3 2008 June — Question 3 6 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2008
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypePooled variance estimation
DifficultyStandard +0.3 This is a standard two-sample z-test for difference in means with known population standard deviations. It requires straightforward application of the test statistic formula for independent normal samples, calculation of the standard error, and comparison with critical values. The question is slightly easier than average because the standard deviations are given (not estimated), making it a z-test rather than t-test, and the procedure is routine for S3 level.
Spec5.05c Hypothesis test: normal distribution for population mean

3 Pitted black olives in brine are sold in jars labelled " 340 grams net weight". Two machines, A and B, independently fill these jars with olives before the brine is added. The weight, \(X\) grams, of olives delivered by machine A may be modelled by a normal distribution with mean \(\mu _ { X }\) and standard deviation 4.5. The weight, \(Y\) grams, of olives delivered by machine B may be modelled by a normal distribution with mean \(\mu _ { Y }\) and standard deviation 5.7. The mean weight of olives from a random sample of 10 jars filled by machine A is found to be 157 grams, whereas that from a random sample of 15 jars filled by machine \(B\) is found to be 162 grams. Test, at the \(1 \%\) level of significance, the hypothesis that \(\mu _ { X } = \mu _ { Y }\).
(6 marks)

AnswerMarks Guidance
Main Solution
\(H_0: \mu_x = \mu_y\); \(H_1: \mu_x \neq \mu_y\)B1 Both
\(\text{SL } \alpha = 0.01(1\%)\); \(\text{CV } z = (\pm) 2.57 \text{ to } 2.58\)B1 AWFW (2.5758)
\(z = \frac{\157 - 162\ }{\sqrt{\frac{4.5^2}{10} + \frac{5.7^2}{15}}} = \)
M1Denominator
\((\pm) 2.44 \text{ to } 2.445\)A1 AWFW (2.4424)
No evidence, at 1% level, to reject hypothesis that \(\mu_x = \mu_y\)A1↑ ft on \(z\), CV and signs; or equivalent; 6 marks
Total: 6 marks
| **Main Solution** |
|---|
| $H_0: \mu_x = \mu_y$; $H_1: \mu_x \neq \mu_y$ | B1 | Both |
| $\text{SL } \alpha = 0.01(1\%)$; $\text{CV } z = (\pm) 2.57 \text{ to } 2.58$ | B1 | AWFW (2.5758) |
| $z = \frac{\|157 - 162\|}{\sqrt{\frac{4.5^2}{10} + \frac{5.7^2}{15}}} = $ | M1 | Numerator |
| | M1 | Denominator |
| $(\pm) 2.44 \text{ to } 2.445$ | A1 | AWFW (2.4424) |
| No evidence, at 1% level, to reject hypothesis that $\mu_x = \mu_y$ | A1↑ | ft on $z$, CV and signs; or equivalent; 6 marks |

**Total: 6 marks**

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3 Pitted black olives in brine are sold in jars labelled " 340 grams net weight". Two machines, A and B, independently fill these jars with olives before the brine is added.

The weight, $X$ grams, of olives delivered by machine A may be modelled by a normal distribution with mean $\mu _ { X }$ and standard deviation 4.5.

The weight, $Y$ grams, of olives delivered by machine B may be modelled by a normal distribution with mean $\mu _ { Y }$ and standard deviation 5.7.

The mean weight of olives from a random sample of 10 jars filled by machine A is found to be 157 grams, whereas that from a random sample of 15 jars filled by machine $B$ is found to be 162 grams.

Test, at the $1 \%$ level of significance, the hypothesis that $\mu _ { X } = \mu _ { Y }$.\\
(6 marks)

\hfill \mbox{\textit{AQA S3 2008 Q3 [6]}}