2020 H2 Mathematics Paper 2 Question 10

Hypothesis Testing

Answers

(i)

{x‾∈R+:x‾<1.45 or x‾>1.55}{\{ \overline{x} \in \mathbb{R}^+: \overline{x} < 1.45} \allowbreak {\textrm{ or } \overline{x} > 1.55 \}}

(ii)

In the earlier test, it is known that the percentage of carbon in the steel bars is distributed normally with known standard distribution. In the test for the new flat bars, both the population distribution and the population variance is unknown.
Hence a large sample size of 40≥30{40 \geq 30} so that the Central Limit Theorem applies and the test statistic of the sample mean X‾{\overline{X}} will then be normally distributed approximately and a Z−{Z\mathrm{-}}test can be carried out.

(iii)

Unbiased estimate of population mean =0.254{=0.254}
Unbiased estimate of population variance =0.000146{=0.000146}

(iv)

p-value=0.0182{p\textrm{-value} = 0.0182}
There is sufficient evidence at the 2.5%{2.5\%} level of significance to conclude that mean amount of carbon in the flat bars is more than 0.25%.{0.25\%.}

Full solutions

(i)

H0:μ=1.5{\textrm{H}_0: \mu = 1.5}
H1:μ≠1.5  ■{\textrm{H}_1: \mu \neq 1.5 \; \blacksquare}
Under H0,{\mathrm{H}_0,}
Z=X‾−μσn∼N(0,1)Z= \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1)
For the critical region for this test (to reject the null hypothesis),
x‾−1.50.0915<−1.9600 or x‾−1.50.0915>1.9600x‾<1.45 (3 sf) or x‾>1.55 (3 sf)\begin{gather*} \frac{\overline{x}-1.5}{\frac{0.09}{\sqrt{15}}} < -1.9600 &\textrm{ or } & \frac{\overline{x}-1.5}{\frac{0.09}{\sqrt{15}}} > 1.9600 \\ \overline{x} < 1.45 \textrm{ (3 sf)} &\textrm{ or } & \overline{x} > 1.55 \textrm{ (3 sf)} \end{gather*}
Critical region:
{x‾∈R+:x‾<1.45 or x‾>1.55}  ■\{ \overline{x} \in \mathbb{R}^+: \overline{x} < 1.45 \textrm{ or } \overline{x} > 1.55 \} \; \blacksquare

(ii)

In the earlier test, it is known that the percentage of carbon in the steel bars is distributed normally with known standard distribution. In the test for the new flat bars, both the population distribution and the population variance is unknown.
Hence a large sample size of 40≥30{40 \geq 30} so that the Central Limit Theorem applies and the test statistic of the sample mean X‾{\overline{X}} will then be normally distributed approximately and a Z−{Z\mathrm{-}}test can be carried out. ■{\blacksquare}

(iii)

Unbiased estimate of population mean=x‾=∑xn=10.1640=0.254  ■\begin{align*} & \textrm{Unbiased estimate of population mean} \\ & = \overline{x} \\ & = \frac{\sum x}{n} \\ & = \frac{10.16}{40} \\ & = 0.254 \; \blacksquare \end{align*}
Unbiased estimate of population variance=s2=1n−1(∑x2−(∑x)2n)=140−1(2.586342−(10.16)240)=0.00014621=0.000146 (3 sf)  ■\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}{40-1}\left( 2.586342 - \frac{\left(10.16\right)^2}{40} \right) \\ & = 0.00014621 \\ & = 0.000146 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

H0:μ=0.25{\textrm{H}_0: \mu = 0.25}
H1:μ>0.25{\textrm{H}_1: \mu > 0.25}
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=40{n=40} is large
p-value=0.018209{p\textrm{-value} = 0.018209}
Since p<0.025,{p < 0.025, } we reject H0{\textrm{H}_0}
There is sufficient evidence at the 2.5%{2.5\%} level of significance to conclude that mean amount of carbon in the flat bars is more than 0.25%.{0.25\%.} ■{\blacksquare}