2007 H2 Mathematics Paper 2 Question 7

Hypothesis Testing

Answers

Unbiased estimate of population mean

{=30.84}

Unbiased estimate of population variance

{=33.7}

{p\textrm{-value} = 0.0382}

There is sufficient evidence at the

{5\%}

level of significance to conclude that the population mean time for a student to complete the project exceeds 30 hours.

Since

{n=150}

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

{\overline{X}}

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

\begin{align*} & \textrm{Unbiased estimate of population mean} \\ & = \overline{x} \\ & = \frac{\sum x}{n} \\ & = \frac{4626}{150} \\ & = 30.84 \; \blacksquare \end{align*}

\begin{align*} & \textrm{Unbiased estimate of population variance} \\ & = s^2 \\ & = \frac{1}{n-1}\left( \sum x^2 - \frac{\left(\sum x\right)^2}{n} \right) \\ & = \frac{1}{150-1}\left( 147691 - \frac{\left(4626\right)^2}{150} \right) \\ & = 33.726 \\ & = 33.7 \textrm{ (3 sf)} \; \blacksquare \end{align*}

Let

{\mu}

denote the population mean time for a student to complete the project.

{\textrm{H}_0: \mu = 30}

{\textrm{H}_1: \mu > 30}

Under

{\textrm{H}_0, }

test statistic

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

approximately by CLT since

{n=150}

is large

{p\textrm{-value} = 0.038238}

Since

{p < 0.05, }

we reject

{\textrm{H}_0}

There is sufficient evidence at the

{5\%}

level of significance to conclude that the population mean time for a student to complete the project exceeds 30 hours.

{\blacksquare}

Since

{n=150}

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

{\overline{X}}

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

{\blacksquare}