## Part (i)
A t test might be used because:
- sample is small
- population variance is unknown
- Must assume background population is Normal.

| B1 | [3]
| B1 |
| B1 |

## Part (ii)
$H_0: \mu = 1.040$
$H_1: \mu \neq 1.040$

where $\mu$ is the mean specific gravity of the mixture.

| B1 | Both hypotheses. Hypotheses in words only must include "population". Do NOT allow "$\bar{X} = ...$" or similar unless $\bar{X}$ is clearly and explicitly stated to be a population mean. |
| B1 | For adequate verbal definition. Allow absence of "population" if correct notation is used. |

$\bar{x} = 1.0452$, $s_{n-1} = 0.007155$

Test statistic is $\frac{1.0452 - 1.040}{0.007155/\sqrt{9}} = 2.189(60)$.

| B1 | $s_n = 0.006746$ but do NOT allow this here or in construction of test statistic, but FT from there. |
| M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Allow alternative: $1.040 + (c's \cdot 1.860) \times \frac{0.007155}{\sqrt{9}}$ (= 1.0444) for subsequent comparison with $\bar{x}$. (Or $\bar{x} - (c's \cdot 860) \times \frac{0.007155}{\sqrt{9}}$ (= 1.0407) for comparison with 1.040.) c.a.o. but ft from here in any case if wrong. |

Refer to $t_8$.

| M1 | No ft from here if wrong. |

Double-tailed 10% point is 1.860.
Significant.
Seems mean specific gravity in the mixture does not meet the requirement.

| A1 | $P(r > 2.1896) = 0.05996.$ No ft from here if wrong. |
| A1 | If only c's test statistic. |
| A1 | If only c's test statistic. |

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## Part (iii)
CI is given by:
$$1.0452 \pm \frac{2.306 \times 0.007155}{\sqrt{9}} = 1.0452 \pm 0.0055 = (1.039(7), 1.050(7))$$

In repeated sampling, 95% of confidence intervals constructed in this way will contain the true population mean.

| M1 | |
| B1 | |
| M1 | |
| A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_8$ is OK. |
| E2 | E2, 1, 0. |

[6]

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