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9231 P44 - Nov 2025 - Q3 - 10 marks
6623
Two different types of juice extractor, machine \(X\) and machine \(Y\), are being compared. The manufacturer claims that machine \(Y\) extracts more juice per orange on average than machine \(X\).
A random sample of 20 oranges is selected and each carefully measured. Oranges of a similar size and mass are then paired and numbered from 1 to 10. One orange from each pair is randomly allocated to machine \(X\), with the other allocated to machine \(Y\).
The amount of juice, in ml, extracted from each orange is recorded in the following table.
Pair
1
2
3
4
5
6
7
8
9
10
Machine \(X\)
65
73
58
61
72
79
64
65
69
71
Machine \(Y\)
68
72
64
63
75
82
63
63
72
74
(a) Use a \(t\)-test to test the manufacturer's claim at the \(1\%\) significance level.
(b) State an assumption required for the test in part (a) to be valid.
The manufacturer notices that the amounts of juice extracted from the oranges in pair 4 were recorded incorrectly. In fact, the amount of juice extracted by machine \(X\) was \(60\text{ ml}\) and by machine \(Y\) was \(62\text{ ml}\).
(c) Explain, with justification, whether the conclusion of the test in part (a) remains the same.
Solution
Answer:
There is insufficient evidence at the \(1\%\) significance level to support the manufacturer’s claim that machine \(Y\) extracts more juice per orange on average than machine \(X\).
A required assumption is that the population distribution of the paired differences \(Y-X\) is normal.
The conclusion remains the same. The corrected values for pair 4 leave the paired difference unchanged, so the paired \(t\)-test is unchanged.
Since the oranges were paired by similar size and mass, use a paired \(t\)-test. Let \(d=Y-X\), so positive values support the manufacturer’s claim.
The hypotheses are \(H_0:\mu_d=0\) and \(H_1:\mu_d\gt0\), where \(\mu_d\) is the population mean paired difference.
Pair
1
2
3
4
5
6
7
8
9
10
\(d=Y-X\)
3
\(-1\)
6
2
3
3
\(-1\)
\(-2\)
3
3
For these differences, \(\sum d=19\) and \(\sum d^2=91\). Hence \(\bar d=\frac{19}{10}=1.9\).
The unbiased sample variance of the differences is \(s_d^2=\frac{1}{10-1}\left(91-\frac{19^2}{10}\right)=\frac{54.9}{9}=6.1\).
The test statistic is therefore \(t=\frac{\bar d-0}{s_d/\sqrt{n}}=\frac{1.9}{\sqrt{6.1/10}}=2.43\) to 3 significant figures.
There are \(10-1=9\) degrees of freedom. For a one-tailed test at the \(1\%\) significance level, the critical value is \(2.821\).
Since \(2.43\lt2.821\), the result is not significant at the \(1\%\) level. We do not reject \(H_0\). Therefore, there is insufficient evidence to support the manufacturer’s claim that machine \(Y\) extracts more juice per orange on average than machine \(X\).
The paired \(t\)-test assumes that the population distribution of the paired differences \(Y-X\) is normal.
Originally, the difference for pair 4 was \(63-61=2\).
After correction, the difference for pair 4 is \(62-60=2\).
So the paired difference for pair 4 has not changed, and hence the full set of paired differences is unchanged. Therefore \(\bar d\), \(s_d\), the test statistic and the conclusion are all unchanged. The conclusion remains the same.
