2021 H2 Mathematics Paper 2 Question 8

Hypothesis Testing

Answers

(i)

The manager should carry out a 1-tail test as he is only interested if the average life span of the tyres is more than 20,000{20,000} and not interested if the average life span of the tyres is less than 20,000.  ■{20,000. \; \blacksquare}

Let μ{\mu} denote the population mean life span of the tyres in thousand miles, X{X} denote the random variable of the life span of the tyres, and H0{\mathrm{H}_0} and H1{\mathrm{H}_1} be the null and alternative hypothesis respectively.

H0:μ=20{\textrm{H}_0: \mu = 20}
H1:μ>20  ■{\textrm{H}_1: \mu > 20 \; \blacksquare}

(ii)

Unbiased estimate of population mean =20.2{=20.2}
Unbiased estimate of population variance =0.755{=0.755}

(iii)

p-value=0.0630{p\textrm{-value} = 0.0630}
There is insufficient evidence at the 5%{5\%} level of significance to conclude whether mean life span of front tyres is more than 20,000{20,000} miles

(iv)

As we do not know the population distribution of the life span of front tyres, we will need a larger sample size of at least 30{30} to ensure that we can use the Central Limit Theorem to justify that the sample mean life span of front tyres X‾{\overline{X}} is normally distributed approximately.

Full solutions

(i)

The manager should carry out a 1-tail test as he is only interested if the average life span of the tyres is more than 20,000{20,000} and not interested if the average life span of the tyres is less than 20,000.  ■{20,000. \; \blacksquare}

Let μ{\mu} denote the population mean life span of the tyres in thousand miles, X{X} denote the random variable of the life span of the tyres, and H0{\mathrm{H}_0} and H1{\mathrm{H}_1} be the null and alternative hypothesis respectively.

H0:μ=20{\textrm{H}_0: \mu = 20}
H1:μ>20  ■{\textrm{H}_1: \mu > 20 \; \blacksquare}

(ii)

Unbiased estimate of population mean=x‾=∑(x−20)n+20=9.450+20=20.2  ■\begin{align*} & \textrm{Unbiased estimate of population mean} \\ & = \overline{x} \\ & = \frac{\sum (x-20)}{n} + 20 \\ & = \frac{9.4}{50} + 20 \\ & = 20.2 \; \blacksquare \end{align*}
Unbiased estimate of population variance=s2=1n−1(∑(x−20)2−(∑(x−20))2n)=150−1(38.76−(9.4)250)=0.75496=0.755 (3 sf)  ■\begin{align*} & \textrm{Unbiased estimate of population variance} \\ & = s^2 \\ & = \frac{1}{n-1}\left( \sum (x-20)^2 - \frac{\left(\sum (x-20)\right)^2}{n} \right) \\ & = \frac{1}{50-1}\left( 38.76 - \frac{\left(9.4\right)^2}{50} \right) \\ & = 0.75496 \\ & = 0.755 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

H0:μ=20{\textrm{H}_0: \mu = 20}
H1:μ>20{\textrm{H}_1: \mu > 20}
Under H0,{\textrm{H}_0, } test statistic
Z=X‾−μsn∼N(0,1)Z= \frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \sim N(0,1)
approximately by CLT since n=50{n=50} is large
p-value=0.063012{p\textrm{-value} = 0.063012}
Since p-value>0.05,{p\textrm{-value} > 0.05, } we do not reject H0{\textrm{H}_0}
There is insufficient evidence at the 5%{5\%} level of significance to conclude whether mean life span of front tyres is more than 20,000{20,000} miles ■{\blacksquare}

(iv)

As we do not know the population distribution of the life span of front tyres, we will need a larger sample size of at least 30{30} to ensure that we can use the Central Limit Theorem to justify that the sample mean life span of front tyres X‾{\overline{X}} is normally distributed approximately. ■{\blacksquare}