2023 H2 Mathematics Paper 2 Question 8

Hypothesis Testing
Sampling Theory

Answers

(ai)

Let μ{{\mu}} denote the population mean mass, in kg, of the granulated sugar in a bag, and H0{{\mathrm{H}_0}} and H1{{\mathrm{H}_1}} be the null and alternative hypothesis respectively.

H0:μ=1{{\textrm{H}_0: \mu = 1}}
H1:μ≠1.{{\textrm{H}_1: \mu \neq 1.}}

(aii)

We do not know the distribution of the mass of granulated sugar in a bag. Moreover, the sample size of 10{{10}} is small, so the Central Limit Theorem cannot be used. Hence we are unable to know if the distribution of sample mean, X‾,{{\overline{X}, }} is normally distributed, and a z{{z}} -test is not suitable.

The sample taken is not random as the first bags are selected so not all bags have the same probability of being selected.

(bi)
H0:μ=2{{\textrm{H}_0: \mu = 2}}
H1:μ<2.{{\textrm{H}_1: \mu < 2.}}

{x‾∈R+:x‾≤1.98}.{{ \{ \overline{x} \in \mathbb{R}^+: \overline{x} \leq 1.98 \}.}}

(bii)

There is sufficient evidence at the 2.5%{{2.5\% }} level of significance to conclude that the mean mass of granulated sugar in a bag is less than 2{{2}} kg so the customers' suspicion is justified.

Full solutions

(ai)

Let μ{{\mu}} denote the population mean mass, in kg, of the granulated sugar in a bag, and H0{{\mathrm{H}_0}} and H1{{\mathrm{H}_1}} be the null and alternative hypothesis respectively.

H0:μ=1{{\textrm{H}_0: \mu = 1}}
H1:μ≠1  ■{{\textrm{H}_1: \mu \neq 1 \; \blacksquare}}

(aii)

We do not know the distribution of the mass of granulated sugar in a bag. Moreover, the sample size of 10{{10}} is small, so the Central Limit Theorem cannot be used. Hence we are unable to know if the distribution of sample mean, X‾,{{\overline{X}, }} is normally distributed, and a z{{z}} -test is not suitable   ■{{\; \blacksquare}}

The sample taken is not random as the first bags are selected so not all bags have the same probability of being selected   ■{{\; \blacksquare}}

(bi)
H0:μ=2{{\textrm{H}_0: \mu = 2}}
H1:μ<2  ■{{\textrm{H}_1: \mu < 2 \; \blacksquare}}
s2=1n−1(∑x2−(∑x)2n)=140−1(155.6746−(78.88)240)=0.00316\begin{align*} s^2 &= \frac{1}{n-1}\left( \sum x^2 - \frac{\left(\sum x\right)^2}{n} \right) \\ & = \frac{1}{40-1}\left( 155.6746 - \frac{\left(78.88\right)^2}{40} \right) \\ & = 0.00316 \end{align*}

Under H0,{{\mathrm{H}_0,}}

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

approximately by CLT since n=40{{n=40}} is large

For the critical region for this test (to reject the null hypothesis),

x‾−20.0031640≤−1.9600x‾≤1.98 (3 sf)\begin{align*} \frac{\overline{x}-2}{\sqrt{\frac{0.00316}{40}}} &\leq -1.9600 \\ \overline{x} &\leq 1.98 \textrm{ (3 sf)} \end{align*}

Critical region:

{x‾∈R+:x‾≤1.98}  ■{ \{ \overline{x} \in \mathbb{R}^+: \overline{x} \leq 1.98 \} \; \blacksquare }
(bii)
x‾=∑xn=78.8840=1.972\begin{align*} \overline{x} & = \frac{\sum x}{n} \\ & = \frac{78.88}{40} \\ & = 1.972 \end{align*}

Since x‾=1.972<1.98,{{\overline{x} = 1.972 < 1.98,}} the test statistic lies in the critical region and reject H0{{\mathrm{H}_0}}

There is sufficient evidence at the 2.5%{{2.5\% }} level of significance to conclude that the mean mass of granulated sugar in a bag is less than 2{{2}} kg so the customers' suspicion is justified   ■{{\; \blacksquare}}

Question Commentary

While we typically use the p -value method for questions with a structure like in part (b), the question leads us to use the critical region method. We last saw this in 2020 P2Q10.

For part (aii), the first reason is the rather common reason that the sample size is too small. The second reason is a bit trickier and involves thinking of the sampling theory topic and the ideas of random sampling.