2020 H2 Maths Paper 2 Q10 — Worked Solution

Part (i)

4 marks

Let $X\%$ be the percentage of carbon, by weight, in a round steel bar, so $X \sim N\left( 1.5,\ {0.09}^{2} \right)$ . The production manager tests

H_{0}:\mu = 1.5\quad\text{against}\quad H_{1}:\mu \neq 1.5

at the $5\%$ level of significance, where $\mu$ is the population mean percentage of carbon in round steel bars. Under $H_{0}$ , since $X$ is normally distributed with known variance ${0.09}^{2}$ , the sample mean of $n = 15$ bars is exactly normal:

\overline{X} \sim N\left( 1.5,\ \frac{{0.09}^{2}}{15} \right).

Standardising, under $H_{0}$ the test statistic

Z = \frac{\overline{X} - 1.5}{0.09/\sqrt{15}} \sim N(0,\ 1).

Since this is a two-tailed test at the $5\%$ level, $H_{0}$ is rejected if $|Z| > 1.9600$ (by GC, the upper $2.5\%$ point of $N(0,1)$ ), i.e.

\left| \overline{X} - 1.5 \right| > 1.9600 \times \frac{0.09}{\sqrt{15}}

= 0.0455\ \ \text{(by GC, unrounded }0.045546...\text{).}

So the critical region is $\overline{X} < 1.4544...$ or $\overline{X} > 1.5455...$ , i.e.

\overline{x} < 1.45\quad\text{or}\quad\overline{x} > 1.55\ \ \left( \text{to }3\text{ s.f.} \right).

$\therefore$ the critical region is $\overline{X} < 1.45$ or $\overline{X} > 1.55$ .

Part (ii)

2 marks

Explanation: In the earlier test, the percentage of carbon in round steel bars is given to be normally distributed, so the sample mean $\overline{X}$ is exactly normally distributed whatever the sample size, and a small sample of $15$ bars is enough for the test to be valid. For the new flat bars, the distribution of the percentage of carbon is not stated to be normal, so the production manager needs a large sample (here $n = 40$ ) so that, by the Central Limit Theorem, the sample mean $\overline{X}$ is approximately normally distributed even though the underlying distribution of carbon percentage in a single bar is unknown.

Part (iii)

2 marks

Since $n = 40$ , $\Sigma x = 10.16$ and $\Sigma x^{2} = 2.586342$ , the unbiased estimate of the population mean is

\overline{x} = \frac{\Sigma x}{n} = \frac{10.16}{40} = 0.254.

The unbiased estimate of the population variance is

s^{2}

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

= \frac{1}{39}\left( 2.586342 - \frac{{10.16}^{2}}{40} \right)

= 0.000146\ \ \text{(3 s.f., unrounded }0.0001462...\text{).}

$\therefore$ the unbiased estimates are $\overline{x} = 0.254$ and $s^{2} = 0.000146$ .

Part (iv)

5 marks

Let $\mu$ be the population mean percentage of carbon, by weight, in flat bars. The production manager tests

H_{0}:\mu = 0.25\quad\text{against}\quad H_{1}:\mu > 0.25

at the $2\frac{1}{2}\%$ level of significance.

Approach

The underlying distribution of carbon percentage in a flat bar is not given as normal, but $n = 40$ is large, so we use the Central Limit Theorem (part (ii)) with the unbiased variance estimate from part (iii) in place of the population variance, rather than an exact normal test.

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

\overline{X} \sim N\left( 0.25,\ \frac{s^{2}}{n} \right)\text{ approximately, where }s^{2} = 0.0001462...\ \text{ from part (iii).}

Using the given sample data,

z

= \frac{\overline{x} - 0.25}{\sqrt{s^{2}/n}}

= \frac{0.254 - 0.25}{\sqrt{0.0001462.../40}}

= 2.09\ \ \text{(by GC, unrounded }2.0922...\text{).}

By GC, the $p$ -value is

P(Z \geq 2.0922...) = 0.0182.

Since $0.0182 < 0.025$ , i.e. the test statistic $z = 2.09$ lies beyond the critical value $z = 1.960$ (the upper $2\frac{1}{2}\%$ point of $N(0,1)$ ), we reject $H_{0}$ at the $2\frac{1}{2}\%$ level of significance.

To check your answer: the sample mean

0.254

is only

0.004

above

0.25

, but the standard error

\sqrt{s^{2}/n} \approx 0.00191

is small, so even this small difference is several standard errors out — consistent with rejecting

H_{0}

. ✓

There is sufficient evidence, at the $2\frac{1}{2}\%$ level of significance, that the mean percentage of carbon in the flat bars is more than $0.25\%$ , supporting the production manager’s suspicion that too much carbon has been added.

2020 H2 Maths Paper 2, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

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