21M.1.AHL.TZ2.10

pestleMathematicsAIHLPaper 121M· ahl-4-14-linear-transformation-of-a-single-rv-e(x)-and-var(x)-unbiased-estimatorssource ↗

A manufacturer of chocolates produces them in individual packets, claiming to have an average of $85$ chocolates per packet.

Talha bought $30$ of these packets in order to check the manufacturer’s claim.

Given that the number of individual chocolates is $x$ , Talha found that, from his $30$ packets, $Σ x = 2506$ and $Σ x^{2} = 209 738$ .

Find an unbiased estimate for the mean number $(μ)$ of chocolates per packet.

[1]

Use the formula $s_{n - 1}^{2} = \frac{Σ x^{2} - \frac{{(Σ x)}^{2}}{n}}{n - 1}$ to determine an unbiased estimate for the variance of the number of chocolates per packet.

[2]

Find a $95 %$ confidence interval for $μ$ . You may assume that all conditions for a confidence interval have been met.

[2]

Suggest, with justification, a valid conclusion that Talha could make.

[1]

Markscheme / solution

$\bar{x} = \frac{Σ x}{n} = \frac{2506}{30} = 83.5 (83.5333 \dots)$ A1

[1 mark]

$(s_{n - 1}^{2} = \frac{Σ x^{2} - \frac{{(Σ x)}^{2}}{n}}{n - 1} =) \frac{209738 - \frac{2506^{2}}{30}}{29}$ (M1)

$= 13.9 (13.9126 \dots)$ A1

[2 marks]

$(82.1, 84.9) (82.1405 \dots, 84.9261 \dots)$ A2

[2 marks]

$85$ is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect R1

Note: The conclusion must refer back to the original claim.

Allow use of a two sided $t$ -test giving a $p$ -value rounding to $0.04 < 0.05$ and therefore Talha would suggest that the manufacturer’s claims in incorrect.

[1 mark]

Examiners’ report

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