Question 1
Topic – AHL 4.12
The purpose of this question is to help a company decide whether or not they should use a new technique to make a component.A factory produces components for a tractor. They have designed a new technique to produce one of their components that they hope will increase its useful lifespan.
They test 120 components made with the new technique and 240 with the technique they currently use. At the end of 250 hours of use, they check the components, and record whether they have no cracks, minor cracks or major cracks.
The data from the trial are given in the table.
In total 141 components had no cracks.
(a) (i) Show that the value of a is 53.
(ii) Find the value of b.
One of the components from the trial is selected at random.
(b) Given that this component had minor cracks find the probability that it was produced by the new technique.
(c) A test for independence is performed at the 5% significance level to determine whether a component having no cracks, minor cracks or major cracks is independent of the production technique used.
(i) State the null and alternative hypotheses.
(ii) Find the p-value.
(iii) State the conclusion of the test in context, justifying your answer.
(d) For the components in the trial that were made with the current technique, show that the proportion which developed cracks is
As an alternative measure, the researchers decide to let be the probability that a component, made with the new technique, develops cracks. They then test the following hypotheses:
$H_0: p = \frac{19}{30}$
$H_1: p < \frac{19}{30}$
In a randomly selected sample of 120 components made with the new technique let X be the number which developed cracks. The researchers assume that, under the null hypothesis,
$X \sim B(120, \frac{19}{30})$
(e) State one additional assumption that the researchers are making in choosing this distribution.
(f) Use appropriate data from the trial to perform the test proposed by the researchers, at the 5% significance level. State the conclusion of the test, justifying your answer.
(g) In comparison with the test in part (c), state one mathematical reason why
(i) the test in part (f) might be preferred.
(ii) the test in part (f) might not be preferred.
For these components, the researchers also consider the mean time taken until cracks develop. It is hoped that using the new technique will increase this value. A second trial is carried out and the times, in hours, taken for cracks to appear is recorded.
The mean time taken for cracks to appear ($\bar{t}$) and the value of $s_{n-1}$ for each technique are given in the following table.
(h) Perform an appropriate test at the 5% significance level to determine whether the new technique increases the mean time taken for cracks to appear.
The company decides to go ahead with the new technique and publishes the following statement: “Statistical tests show the new technique will significantly increase the time before components crack and need to be replaced.”
(i) Comment on this statement.
▶️Answer/Explanation
Solution: –
(a) (i) 141-88 = 53
(ii) (120-53-54) = 13
(b) Restricting the size of the sample space to 150 (54+96)
$\frac{54}{150} \times \frac{25}{240} = 0.36$
(c) (i) H$_0$: The development of cracks and the technique used are independent
H$_1$: The development of cracks and the technique used are not independent
(ii) ($p$-value $\rightarrow$) 0.0170 (0.0169864…)
(iii) 0.0170 < 0.05
hence there is sufficient evidence to reject the null hypothesis that the development of cracks and the technique used are independent.
(d) $\frac{96+56}{240} = \frac{152}{240} = \frac{19}{30}$
(e) $EITHER$
the probability of each component developing cracks is independent of
all the other components in the sample.
$OR$
the development of cracks can be partitioned into two clear groups
(f) $67 \text{seen}$
$EITHER$
attempt to find a probability ≤ 67 (condone strict inequality for (M1))
(P(X≤67)→) 0.0549 (0.0549093…)
0.0549 > 0.05
$OR$
attempt to find the critical region
critical region is X ≤ 66
66 < 67 or ’67 is not in the critical region’
$THEN$
$EITHER$
do not reject the null hypothesis (as there is insufficient evidence that
the new technique reduces the number of cracks).
$OR$
do not accept the alternative hypothesis (as there is insufficient evidence
that the new technique reduces the number of cracks).
(g) (i) the test for a proportion is directional and so considers whether the new
treatment reduces the number of components developing cracks.
(ii) EITHER
there could be variation in the value of p chosen for the null hypothesis /
the value of p from the sample might not be a representative of
the current technique
OR
the test in (f) does not treat minor and major cracks as different attributes /
the test in (c) does treat minor and major cracks as different attributes
OR
the test in (f) has to make an additional assumption (for example ‘independence’)
(h) EITHER
let $\mu_{1}$ be the mean length of time before cracks appear with the new technique
and $\mu_{2}$ be the mean length with the current technique
H$_0$: $\mu_{1}=\mu_{2}$
H$_1$: $\mu_{1}>\mu_{2}$
OR
H$_0$: the POPULATION mean length of time before cracks appear is the same
for both groups
H$_1$: the new technique increases the POPULATION mean length of time before cracks appear.
OR
H$_0$: the mean length of time before cracks appear in ALL components
made with the new technique is the same as for ALL components made
with the current technique.
H$_1$: the mean length of time before cracks appear in ALL components
made with the new technique is greater than the mean for ALL components
made with the current technique.
THEN
recognition of the need to use of a two-sample test
p-value = 0.0162 (0.0162328…)
0.0162 < 0.05
reject the null hypothesis (OR accept the alternative hypothesis)
(there is sufficient evidence to that the new technique increases the mean
length of time before the cracks appear)
(i) EITHER
(though statistically significant) the new technique only seems to increase the time before
cracks appear by 1 hour out of 250, so it is not a significant increase
(i.e. the effect size is small)
OR
the minimum time (not mean time) before cracks appear should be considered
given the context / An appropriate confidence interval should be considered,
and not simply the mean.