is normally distributed approximately. Hence no assumptions are needed about the population distribution of the masses of the biscuit bars in order for the test to be valid.

Full solutions

(i)

It means that each biscuit bar in the factory (the population) has an equal chance to be selected into the sample.

{\blacksquare}

(ii)

\begin{align*} & \textrm{Unbiased estimate of population mean} \\ & = \overline{x} \\ & = \frac{\sum (x-32)}{n} + 32 \\ & = \frac{-7.7}{40} + 32 \\ & = 31.808 \\ & = 31.8 \textrm{ (3 sf)} \; \blacksquare \end{align*}

\begin{align*} & \textrm{Unbiased estimate of population variance} \\ & = s^2 \\ & = \frac{1}{n-1}\left( \sum (x-32)^2 - \frac{\left(\sum (x-32)\right)^2}{n} \right) \\ & = \frac{1}{40-1}\left( 11.05 - \frac{\left(-7.7\right)^2}{40} \right) \\ & = 0.24533 \\ & = 0.245 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

Let

{\mu}

denote the population mean mass of biscuit bars.

{\textrm{H}_0: \mu = 32}

{\textrm{H}_1: \mu \neq 32}

Under

{\textrm{H}_0, }

test statistic

Z= \frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \sim N(0,1)

approximately by CLT since

{n=40}

is large

{p\textrm{-value} = 0.013970}

Since

{p\textrm{-value} > 0.01, }

we do not reject

{\textrm{H}_0}

There is insufficient evidence at the

{1\%}

level of significance to conclude whether the population mean mass of biscuit bars is 32 grams.

{\blacksquare}

(iv)

Since

{n=40}

is large, by the Central Limit Theorem, the sample mean

{\overline{X}}

is normally distributed approximately. Hence no assumptions are needed about the population distribution of the masses of the biscuit bars in order for the test to be valid.

{\blacksquare}