SCSA Mathematics Specialist Statistical inference

Using the property of a PDF
$\int_{-\infty}^{\infty} p(x) dx = 1$
Using $n=5$ in the given PDF
$\int_0^\infty \frac{k^5 t^4}{4!} e^{-\frac{t}{3}} dt = 1$

Solving the equation: $k = \frac{1}{3}$
Mean of distribution for waiting time until 5th call, $\mu$
$E(X) = \int_{-\infty}^{\infty} x p(x) dx$
$\mu = \int_0^\infty t \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = \int_0^\infty \frac{\left(\frac{1}{3}\right)^5 t^5}{4!} e^{-\frac{t}{3}} dt$
$= 15 \text{ minutes}$

Variance of distribution for 5th call
$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x) dx$
$= \int_0^\infty (t-15)^2 \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = 45 \text{ minutes}^2$
$\therefore \sigma = \sqrt{45} \text{ minutes}$

Consider the distribution of the sample mean of the waiting time until the 5th phone call is received, $\bar{T}$.
As the sample size is large, the distribution of $\bar{T}$ can be considered normal.
$\mu_{\bar{T}} = 15$ and $\sigma_{\bar{T}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{45}}{\sqrt{80}} = 0.75$

Using normal cdf on GDC: $P(\bar{T} > 16) \approx 0.09$

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 $x_1, x_2, \dots, x_{100}$.

Given $n=40, \bar{x}=9.31$ and $s=0.52$

Using GDC

$CI(95\%) = (9.15, 9.47)$ hours

Using GDC

$CI(99\%) = (9.10, 9.52)$ hours

The 95% confidence interval does not include the claimed mean battery life of 9.5 hours, although the 99% CI does.

So the comment is not reasonable.

$\text{For a total of 365 g, the sample mean } \overline{M} = \frac{365}{50} = 7.3 \text{ grams per biscuit}$

$\text{Require } P\left(\overline{M} < 7.3\right) = 0.1729... = 0.17$

$\sigma_{\overline{M}} = \frac{1.5}{\sqrt{n}}$

$\text{We require } P(-k < z < k) = 0.98 \text{ this gives } k = 2.326$

$\text{Hence } 2.326\left(\frac{1.5}{\sqrt{n}}\right) < 0.2 \implies \text{Solving gives } n > 304.32$

i.e. we require at least 305 biscuits to have the sample mean differ by less than 0.2 grams

Let $\mu_Y$ be the population mean for the mass of chocolate per biscuit for the YBC company (grams).

For the YBC total of 1090 grams, this gives $\overline{M} = 7.56944...$ grams.

$\text{The distribution for } \overline{M} \sim N\left(7.56944, \sigma_{\overline{M}}^2\right) \text{ where } \sigma_{\overline{M}} = \frac{1.8}{\sqrt{144}} = 0.15$

$\text{Confidence Interval for } \mu_Y \text{ 95\% level :}$

$\text{Confidence Interval for } \mu_Y \text{ 99\% level :}$

$\text{The BAU population mean } \mu = 7.5$ is WITHIN the confidence interval using $\overline{M} = 7.56944 \text{ and } \sigma = 1.8$. i.e. the claim is NOT vindicated.

i.e. the YBC company are NOT using significantly more chocolate per biscuit than compared to BAU.

Given $\mu_{\bar{x}} = 25.2$
$\sigma_{\bar{x}_1} = \frac{\sigma}{\sqrt{n}} = \frac{4.7}{\sqrt{120}}$
$= 0.429 \text{ minutes}$

Using GDC
$P(\bar{X}_1 \le 25) = 0.32$

$P(\bar{X}_1 > k) = 0.9$
Using GDC
$k = 24.65$ minutes

$P(z \le z_1) \approx 0.4 \Rightarrow z_1 = -0.253$
$z = \frac{\bar{X}_2 - \mu}{\frac{\sigma}{\sqrt{n}}}$
$-0.253 = \frac{25 - 25.2}{\frac{4.7}{\sqrt{n}}}$

Using GDC
$n \approx 35.3$

The sample size is 35.

$95\% \text{ CI: } 400 - \frac{200}{2} \le \mu \le 400 + \frac{200}{2}$

$\text{i.e. } 300 \le \mu \le 500$

Margin of error $= 100$

i.e. $100 = 1.96 \times \sigma\left(\overline{X}\right) \implies \therefore \sigma\left(\overline{X}\right) = 51.02$

The interval width is reduced by a factor of 4, so the sample size needs to increase by a factor of $4^2 = 16$

i.e. a sample size of $16n$ is required.

$\overline{X} \sim N\left(\mu, \frac{s^2}{2n}\right) \implies \text{i.e. } \overline{X} \sim N\left(\mu, \frac{51.02^2}{2}\right) \implies \therefore \sigma\left(\overline{X}\right) = 36.0768...$

Since the true value of $\mu$ is unknown, we CANNOT determine which interval contains the true mean. This is due to the inherent nature of random sampling.

Confidence interval A will have the smaller width since the level of confidence 95% is less than that of B 99%.

Need to compare the standard deviation of the sample means:

$C: \quad \sigma\left(\overline{X}\right) = \frac{s}{\sqrt{2n}} = 0.707\left(\frac{s}{\sqrt{n}}\right)$
$D: \quad \sigma\left(\overline{X}\right) = \frac{0.8s}{\sqrt{n}} = 0.8\left(\frac{s}{\sqrt{n}}\right)$

Hence confidence interval C will have the smaller width.

Sample 1: $n = 1$
$P(X > 83.2) = 0.145$
$P\left(z > \frac{83.2-\mu}{\sigma}\right) = 0.145$
$\frac{83.2 - \mu}{\sigma} = 1.058$
$\mu = 83.2 - 1.058\sigma \quad ... (1)$

Sample 2: $n = 12$
$P(\bar{X} < 74.1) = 0.079$
$P\left(z < \frac{74.1-\mu}{\frac{\sigma}{\sqrt{12}}}\right) = 0.079$
$\frac{74.1 - \mu}{\frac{\sigma}{\sqrt{12}}} = -1.412$
$\mu = 74.1 + \frac{1.412\sigma}{\sqrt{12}} \quad ... (2)$

Using graph facility of GDC to solve (1) and (2)
$\mu = 76.63 \text{ kg}, \sigma = 6.21 \text{ kg}$

Sample 3: Consider the 90% CI
Since $\bar{x} = 79.1, \mu = 76.63$ can only lie in an interval below the lower bound of CI.
Determining $n$ where the lower bound of CI $= \mu$
$\bar{x} - z\frac{s}{\sqrt{n}} = 76.63$
$79.1 - 1.64 \times \frac{6.21}{\sqrt{n}} = 76.63$
Using solve facility of GDC, $n \approx 17.1$
As $\mu$ must lie in an interval below the lower bound of CI, the range of values is $n \ge 18$ where $n \in Z$.

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

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 $\frac{1}{\sqrt{2}}$: