October 2020 Paper 2 Q9
9 A company supplies computers to businesses. In the past the company has found that computers are kept by businesses for a mean time of 5 years before being replaced. Claud, the manager of the company, thinks that the mean time before replacing computers is now different.
Claud decides to conduct a hypothesis test at the 5% level to test whether there is evidence to suggest that the mean time that businesses keep computers is not 5 years. He takes a random sample of 120 computers. Summary statistics for the length of time computers in this sample are kept are shown in Fig. 9.
| Statistics | |
|---|---|
| n | 120 |
| Mean | 4.8855 |
| σ | 2.6941 |
| s | 2.7054 |
| Σx | 586.2566 |
| Σx2 | 3735.1475 |
| Min | 0.1213 |
| Q1 | 2.5472 |
| Median | 4.8692 |
| Q3 | 7.0349 |
| Max | 9.9856 |
Fig. 9
- State the hypotheses for this test, explaining why the alternative hypothesis takes the form it does.
- Use a suitable distribution to carry out the test.
| Scheme | Marks | AO |
|---|---|---|
| eg randomly select \(N\) different businesses and then randomly select \(P\) computers from each business; may be implied by correct description | B1 | 2.4 |
| [1] |
Notes
B1: \(N \times P = 120\) where \(N\) and \(P\) are integers greater than 1.
eg 20 and 6 or 15 and 8
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H_0}: \mu = 5\) oe \(\mathrm{H_1}: \mu \ne 5\) oe | B1 | 1.1 |
| \(\mathrm{H_1}\) takes this form as Claud is testing whether the mean length of time is different to 5 oe | B1 | 2.4 |
| \(\mu\) is the population mean time for which computers are kept before being replaced | B1 | 2.5 |
| use of \(\mathrm{N}\left(5, \frac{2.7054^2}{120}\right)\) to find \(\mathrm{P}(\bar{X} < 4.8855)\) or \(\mathrm{invNorm}\left(p, 5, \frac{2.7054}{\sqrt{120}}\right)\) where \(p = 0.025\) or 0.05; may be implied by 0.3215 or 4.51595… or 4.59377… | M1 | 3.3 |
| \(\mathrm{P}(\bar{X} < 4.8855) =\) awrt 0.32 | A1 | 1.1 |
| \(0.32 > 0.025\) or \(4.8855 > 4.5(2)\) | M1 | 3.4 |
| not significant or accept \(\mathrm{H_0}\) or do not reject \(\mathrm{H_0}\) or reject \(\mathrm{H_1}\) | A1FT | 1.1 |
| insufficient evidence to suggest (at 5% level) that the (population) mean length of time (computers are kept) is not 5 years | A1FT | 2.2b |
| [8] |
Notes
B1: allow any parameter apart from \(\bar{x}\) for population mean as long as clearly defined as (population) mean
after B marks M1A1M1A1A1 may be earned if working with 2.7054 rounded to 2 or more sf
M1: condone use of 2.6941 (may be rounded) instead of 2.7054 for M1, may be implied by 0.32076 or CR is \(\bar{X} < 4.51797\ldots\), but A1 not available
or \(z = \dfrac{4.8855 - 5}{\frac{2.7054}{\sqrt{120}}}\) for M1
A1: or CR is \((\bar{X}) < 4.5 - 4.6\)
awrt \(-0.46\) A1 (may be implied by \(-0.466\) or \(-0.465567\ldots\) if 2.6941 used)
M1: comparison of their probability with 0.025 or comparison of 4.8855 with their critical value from use of 0.025, as long as previous M1 awarded
their \(-0.4636 > -1.96\) oe (corrected from the printed mark scheme: printed as \(-0.4436\))
M1 dep on award of previous M1
A1FT: may be embedded in conclusion in context
A1FT: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; A0 if answer spoiled