AQA S3 2014 June — Question 2 6 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2014
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 test for difference of means with known variances, requiring straightforward application of the normal distribution test. The question provides all necessary information clearly, and students need only to calculate the test statistic using the formula for difference of means, find the critical value, and make a conclusion. While it requires careful handling of the variance of the difference (σ²_B/n_B + σ²_G/n_G), this is a routine S3 procedure with no conceptual challenges beyond textbook application.
Spec2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean

2 Each household within a district council's area has two types of wheelie-bin: a black one for general refuse and a green one for garden refuse. Each type of bin is emptied by the council fortnightly. The weight, in kilograms, of refuse emptied from a black bin can be modelled by the random variable \(B \sim \mathrm {~N} \left( \mu _ { B } , 0.5625 \right)\). The weight, in kilograms, of refuse emptied from a green bin can be modelled by the random variable \(G \sim \mathrm {~N} \left( \mu _ { G } , 0.9025 \right)\). The mean weight of refuse emptied from a random sample of 20 black bins was 21.35 kg . The mean weight of refuse emptied from an independent random sample of 15 green bins was 21.90 kg . Test, at the \(5 \%\) level of significance, the hypothesis that \(\mu _ { B } = \mu _ { G }\).
[0pt] [6 marks]

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
\(H_0: \mu_B = \mu_G\), \(H_1: \mu_B \neq \mu_G\)B1 Both hypotheses correct
\(\bar{B} - \bar{G} = 21.35 - 21.90 = -0.45\)M1 Correct difference of means
\(\text{Var}(\bar{B} - \bar{G}) = \frac{0.5625}{20} + \frac{0.9025}{15}\)M1 Correct variance formula for difference of sample means
\(= 0.028125 + 0.060167 = 0.088292\)A1 Correct variance value
\(z = \frac{-0.45}{\sqrt{0.088292}} = \frac{-0.45}{0.29714} = -1.514\)M1 Correct test statistic calculation
Critical value \(z = \pm 1.96\) (5%, two-tailed)B1 Correct critical value stated
\(-1.514 < 1.96\), do not reject \(H_0\)
# Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_B = \mu_G$, $H_1: \mu_B \neq \mu_G$ | B1 | Both hypotheses correct |
| $\bar{B} - \bar{G} = 21.35 - 21.90 = -0.45$ | M1 | Correct difference of means |
| $\text{Var}(\bar{B} - \bar{G}) = \frac{0.5625}{20} + \frac{0.9025}{15}$ | M1 | Correct variance formula for difference of sample means |
| $= 0.028125 + 0.060167 = 0.088292$ | A1 | Correct variance value |
| $z = \frac{-0.45}{\sqrt{0.088292}} = \frac{-0.45}{0.29714} = -1.514$ | M1 | Correct test statistic calculation |
| Critical value $z = \pm 1.96$ (5%, two-tailed) | B1 | Correct critical value stated |
| $|-1.514| < 1.96$, do not reject $H_0$ | A1 | Correct conclusion — insufficient evidence that $\mu_B \neq \mu_G$ |

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2 Each household within a district council's area has two types of wheelie-bin: a black one for general refuse and a green one for garden refuse. Each type of bin is emptied by the council fortnightly.

The weight, in kilograms, of refuse emptied from a black bin can be modelled by the random variable $B \sim \mathrm {~N} \left( \mu _ { B } , 0.5625 \right)$.

The weight, in kilograms, of refuse emptied from a green bin can be modelled by the random variable $G \sim \mathrm {~N} \left( \mu _ { G } , 0.9025 \right)$.

The mean weight of refuse emptied from a random sample of 20 black bins was 21.35 kg . The mean weight of refuse emptied from an independent random sample of 15 green bins was 21.90 kg .

Test, at the $5 \%$ level of significance, the hypothesis that $\mu _ { B } = \mu _ { G }$.\\[0pt]
[6 marks]

\hfill \mbox{\textit{AQA S3 2014 Q2 [6]}}