Specimen 2017 H2 Mathematics Paper 2 Question 10

Hypothesis Testing

Answers

(i)

He should carry out a 2-tail test because in testing whether the alternative process is different, the time required could be higher or lower.

(ii)

Since n=50{n=50} is large, by the Central Limit Theorem, the sample mean time taken will be normally distributed approximately so it is not necessary to know anything about the distribution of the times taken.

(iii)

p-value=0.155.{p\textrm{-value}=0.155.} Do not reject H0{\textrm{H}_0}
There is insufficient evidence at the 10%{10\%} level of significance to conclude whether the average time taken for the manufacture of a control panel is different using the alternative process

(iv)

The alternative process might be able to produce electronic control panels of higher quality which will justify the use of a process that takes a longer average time.

(v)

If σ2<2.19,{\sigma^2 < 2.19,} then the Finance Manager will conclude that the average time taken for the manufacture of a control panel is shorter using the alternative process.

Full solutions

(i)

He should carry out a 2-tail test because in testing whether the alternative process is different, the time required could be higher or lower. ■{\blacksquare}

(ii)

Since n=50{n=50} is large, by the Central Limit Theorem, the sample mean time taken will be normally distributed approximately so it is not necessary to know anything about the distribution of the times taken. ■{\blacksquare}

(iii)

H0:μ=17{\textrm{H}_0: \mu = 17}
H1:μ≠17{\textrm{H}_1: \mu \neq 17}
Under H0,{\textrm{H}_0, } test statistic
Z=T‾−μs/n∼N(0,1)Z= \frac{\overline{T} - \mu}{s/\sqrt{n}} \sim N(0,1)
approximately by CLT since n=50{n=50} is large
p-value=0.15539{p\textrm{-value} = 0.15539}
Since p-value>0.1,{p\textrm{-value} > 0.1, } we do not reject H0{\textrm{H}_0}
There is insufficient evidence at the 10%{10\%} level of significance to conclude whether the average time taken for the manufacture of a control panel is different using the alternative process ■{\blacksquare}

(iv)

The alternative process might be able to produce electronic control panels of higher quality which will justify the use of a process that takes a longer average time. ■{\blacksquare}

(v)

H0:μ=17{\textrm{H}_0: \mu = 17}
H1:μ<17{\textrm{H}_1: \mu < 17}
t‾=16.7{\overline{t} = 16.7}

Under H0,{\textrm{H}_0, } test statistic

Z=T‾−μσ/n∼N(0,1)Z= \frac{\overline{T} - \mu}{\sigma/\sqrt{n}} \sim N(0,1)
approximately by CLT since n=40{n=40} is large

For a left-tail test at 10%{10\%} level of significance,

zcrit=−1.2816z_{\textrm{crit}} = -1.2816
If the manager is to conclude that the average time taken is shorter (reject H0,{\textrm{H}_0,})
16.7−17σ/40<−1.2816σ<1.4805σ2<2.19\begin{align*} \frac{16.7-17}{\sigma/\sqrt{40}} &< -1.2816 \\ \sigma &< 1.4805 \\ \sigma^2 &< 2.19 \end{align*}
If σ2<2.19,{\sigma^2 < 2.19,} then the Finance Manager will conclude that the average time taken for the manufacture of a control panel is shorter using the alternative process ■{\blacksquare}