$\overline{x} = \frac{10.2+25.4}{2} = 17.8$

Hence the mean personal loan was ＄17 800.

Half-width $w = 17.8 - 10.2 = 7.6$
$\therefore 7.6 = 2.5758 \times \sigma(\overline{X})$

i.e. $\sigma(\overline{X}) = 2.95$ (2 d.p.)$
i.e. standard deviation is ＄2950

Ali is not correct. It can be said that:

1. A single confidence interval either contains $\mu$ or it doesn't.
2. The value of $\mu$ is unknown so we do not know if any given CI contains $\mu$.
3. If we repeatedly take samples of size $n$ then we will find that approximately 99% of these intervals will contain the true value of $\mu$.
4. The value of $n$ may be less than 30, meaning that the distribution of the sample mean may not be distributed, hence the 99% confidence interval may not be valid.

The confidence interval changes is two ways.

1. The whole interval is translated upwards by 2.
2. The standard error $\sigma(\overline{X})$ and consequently the width of the interval is scaled by a factor of $\frac{1}{\sqrt{2}}$:

For 99% confidence using $k = 2.576$
$2 = (2.576)\left(\frac{10.98}{\sqrt{n}}\right) \quad \text{Solving gives } n = 200.0029... \quad \text{i.e. } n = 200$

1. The intervals may have different midpoints, as the sample means from the two sample may be different.
2. The intervals may have different widths, as the sample standard deviations may be different.

No James is NOT correct.
The interval is wider because it is based on data with more variation in it. Consequently, the sample mean has more variation. That is, the location of the interval is more variable. We cannot be sure that any confidence interval contains the true mean.

Since $n = 100 > 30$, then $\overline{X} \sim N\left(\mu, \sigma_{\overline{X}}^2\right)$

i.e. normally distributed and centred with a mean $\mu$ and estimated standard deviation of the sample mean
$\sigma\left(\overline{X}\right) = \frac{1}{\sqrt{100}} = 0.1 \text{ mm}.$

$P\left(\left|\overline{X} - \mu\right| > 0.2\right) = P(|z| > 2) = 2(0.023) = 0.046$

Width $w = 2(k)\left(\sigma\left(\overline{X}\right)\right) = 2(1.96)(0.1) = 0.392$

Any 2 of the following:

1. The idea simply replicates (repeats) the data in sample $R_1$. As such the sample $R_2$ is no longer random. Therefore the assumptions for using the normal distribution for the sample mean does not hold anymore.
2. Repeating the data values in sample $R_1$ will not reflect the true random variation in the data (manufacturing process). The confidence interval will therefore not be a true representation of the variation in the sample mean.
3. The sample mean and standard deviation will change if a new larger sample is taken. However, with Lillian's idea these will not change. This will affect both the width and location of the confidence interval.
4. Replicating the sample does NOT decrease the width of the confidence interval. Let the sample observations for $R_1$ be $x_1, x_2, \dots, x_{100}$.