2017 H2 Maths Paper 2 Q7 — Worked Solution

Part (i)

1 mark

A sample is random in this context if every biscuit bar produced that day has an equal chance of being selected for the sample, and bars are selected independently of one another.

Part (ii)

2 marks

Since $n = 40$ , $\Sigma(x - 32) = - 7.7$ and $\Sigma(x - 32)^{2} = 11.05$ ,

\overline{x} = 32 + \frac{\Sigma(x - 32)}{n}

= 32 + \frac{- 7.7}{40}

= 31.8075

s^{2} = \frac{1}{n - 1}\left\lbrack \Sigma(x - 32)^{2} - \frac{\left( \Sigma(x - 32) \right)^{2}}{n} \right\rbrack

= \frac{1}{39}\left\lbrack 11.05 - \frac{( - 7.7)^{2}}{40} \right\rbrack

= \frac{9.5675}{39}

= 0.245\ \left( 3\ \text{s.f.} \right)

$\therefore$ the unbiased estimate of the population mean is $31.8075\ g$ , and the unbiased estimate of the population variance is $0.245\ g^{2}$ (3 s.f.).

Part (iii)

5 marks

Let $\mu$ be the population mean mass, in grams, of biscuit bars produced that day.

We test

H_{0}:\mu = 32\quad\quad H_{1}:\mu \neq 32

at the $1\%$ significance level.

Approach

The population distribution of masses is not given, but $n = 40$ is large, so by the Central Limit Theorem $\overline{X}$ is approximately normal even though the population variance is unknown (and is estimated by $s^{2}$ ). This justifies a $z$ -test.

Under $H_{0}$ , since $n = 40$ is large, by the Central Limit Theorem

\overline{X} \sim N\left( 32,\ \frac{s^{2}}{n} \right)\ \text{approximately, where}\ s^{2} = 0.24533\ \left( 5\ \text{s.f.} \right)

Using a $z$ -test,

z = \frac{\overline{x} - 32}{\sqrt{s^{2}/n}}

= \frac{31.8075 - 32}{\sqrt{0.24533/40}}

= - 2.458\ \left( 4\ \text{s.f.} \right)\ \ \text{(by GC)}

By GC, the $p$ -value is $0.0140$ (3 s.f.).

Since $p\text{-value} = 0.0140 > 0.01$ , we do not reject $H_{0}$ .

$\therefore$ at the $1\%$ level of significance, there is insufficient evidence that the mean mass of biscuit bars differs from $32$ grams, i.e. no evidence against the claim on the packets.

Part (iv)

2 marks

Explanation: No need to know the population distribution because $n = 40$ is large, so by the Central Limit Theorem the sample mean $\overline{X}$ is approximately normally distributed whatever the underlying distribution of the individual bar masses is. This means the test in (iii) is valid (approximately) regardless of the population distribution.

2017 H2 Maths Paper 2, Question 7

Part (i)

Part (ii)

Part (iii)

Part (iv)

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