9758/2020/P2/Q10

Carbon steel is made by adding carbon to iron; this makes the iron stronger, though less flexible. A steel manufacturing company makes carbon steel in the form of round bars. The process is designed to manufacture bars in which the amount of carbon in each bar, by weight, is $1.5 \%$. It is known that the percentage of carbon in the steel bars is distributed normally, and that the standard deviation is $0.09 \%$

After comments from customers, the production manager wishes to test, at the $5 \%$ level of significance, if the percentage of carbon in the steel bars is, in fact, $1.5 \%$. He examines a random sample of 15 bars to determine the percentage of carbon in each bar.

(i) Find the critical region for this test.

[4]

The company recently launched a new line of flat bars made from mild steel. In mild steel the amount of carbon in each bar, by weight, is $0.25 \%$.

Comments from customers suggest that these bars are not sufficiently flexible, which makes the production manager suspect that too much carbon has been added. He decides to perform a hypothesis test on a random sample of 40 of the new flat bars to find out if this is the case.

(ii) Explain why the production manager takes a sample of 40 flat bars for this test when he only took a sample of 15 round bars in his earlier test.

[2]

The amounts of carbon, $x \%$, in a random sample of 40 flat bars are summarised as follows.

$$
n=40 \quad \Sigma x=10.16 \quad \Sigma x^{2}=2.586342
$$

(iii) Calculate unbiased estimates of the population mean and variance for the percentage amount of carbon in the flat bars.

[2]

(iv) Test, at the $2 \frac{1}{2} \%$ level of significance, whether the mean amount of carbon in the flat bars is more than $0.25 \%$. You should state your hypotheses and define any symbols that you use.

[5]

(i) Let $\bar{x}$ be the sample mean amount of carbon in each bar.
Since this is a 2 tailed-test, there are 2 critical regions for the test.
For $\mathrm{H}_0$ to be rejected,
$\frac{\bar{x}-1.5}{\frac{0.09}{\sqrt{15}}} \leqslant-1.95996$ or $\frac{\bar{x}-1.5}{\frac{0.09}{\sqrt{15}}} \geqslant 1.95996$
$\begin{array}{ll}\bar{x} \leqslant 1.4545 & \bar{x} \geqslant 1.5455 \\ \bar{x} \leqslant 1.45 & \bar{x} \geqslant 1.55\end{array}$
$\bar{x} \leqslant 1.45 \quad \bar{x} \geqslant 1.55$
$\therefore \bar{x} \in[0,1.45] \cup[1.55,100]$

(ii) Since this is a new line of flat bars made, the underlying distribution may not be known.

[A1]

By taking a sample size of 40 which is considered sufficiently large, by Central Limit Theorem the distribution of the sample mean mass weight of the flat bars will be approximately normal.

[Al]

(iii) Unbiased estimate of the population mean,
$$
\begin{aligned}
\bar{x} &=\frac{10.16}{40} \\
&=0.254
\end{aligned}
$$

[A1]

Unbiased estimate of the population variance,
$$
\begin{aligned}
s^2 &=\frac{1}{39}\left[2.586342-\frac{(10.16)^2}{40}\right] \\
&=0.00014621 \\
& \approx 0.000146
\end{aligned}
$$

[Al]

(iv) Let $\mu$ be the population mean amount of carbon in the flat bars, in percentage.
$$
\begin{array}{l}
\mathrm{H}_0: \mu=0.25 \\
\mathrm{H}_1: \mu>0.25
\end{array}
$$

[A1]

At the $2.5 \%$ level of significance,
Under $\mathrm{H}_0$,
since $n=40$ is large, by Central Limit Theorem,
$\bar{X} \sim \mathrm{N}\left(0.25, \frac{s^2}{40}\right)$ approximately

[B1]

Test Statistic, $Z=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}$
Given $m=0.25, \bar{x}=0.254, s^2=0.00014621$, $n=40$.
From GC,
$p$-value $\approx 0.0182$

[B2]

Since $p$-value $=0.0182<0.025$, we reject $\mathrm{H}_0$ at the $2.5 \%$ level of significance and conclude that there is sufficient evidence that the mean amount of carbon is more than $0.25 \%$.

[B I]