S2 June 2014 (R) Q5
5. Sammy manufactures wallpaper. She knows that defects occur randomly in the manufacturing process at a rate of 1 every 8 metres. Once a week the machinery is cleaned and reset. Sammy then takes a random sample of 40 metres of wallpaper from the next batch produced to test if there has been any change in the rate of defects.
Thomas claims that his new machine would reduce the rate of defects and invites Sammy to test it. Sammy takes a random sample of 200 metres of wallpaper produced on Thomas’ machine and finds 19 defects.
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \lambda = \tfrac{1}{8}\) (or \(\lambda = 5\)) \(\mathrm{H}_1 : \lambda \ne \tfrac{1}{8}\) (or \(\lambda \ne 5\)) allow \(\lambda\) or \(\mu\) | B1 |
| \(X \sim \mathrm{Po}(5)\), \(\mathrm{P}(X \leqslant 1) = 0.0404\) or \(\mathrm{P}(X \geqslant 10) = 0.0318\) or \(\mathrm{P}(X \geqslant 9) = 0.0681\) | M1 |
| Critical Regions: \(X \leqslant 1\) or \(X \geqslant 10\) | A1, A1 |
| (4) |
Notes
B1 for suitable hypotheses
M1 for correct use of Po(5). Award if one relevant probability is seen or a correct CR. Allow if a correct CR written as a Probability statement
1st A1 for \(X \leqslant 1\) or \(X \lt 2\) or \(0 \lt X \lt 2\) or \(0 \leqslant X \lt 2\) or \(0 \lt X \leqslant 1\) oe. Allow any letter
2nd A1 for \(X \geqslant 10\) or \(X \gt 9\) or \(10 \leqslant x \leqslant 40\) or \(9 \lt x \leqslant 40\) oe. Allow any letter
Ignore any \(\cup\) or \(\cap\) signs
Do not allow CR written as probability statements
| Scheme | Marks |
|---|---|
| \(0.0404 + 0.0318 = 0.0722\) (or 7.22% significance level) | M1A1 |
| (2) |
Notes
M1 for adding their probabilities of ‘their’ critical regions if sum gives a probability less than 1 or award if a correct answer given
A1 for awrt 0.0722 (o.e)
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \lambda = \tfrac{1}{8}\) (or \(\lambda = 25\)) \(\mathrm{H}_1 : \lambda \lt \tfrac{1}{8}\) (or \(\lambda \lt 25\)) allow \(\lambda\) or \(\mu\) | B1 |
| [\(Y\) = no. of defects in 200m of wallpaper] \(Y \sim \mathrm{Po}(25)\) \(Y \approx\sim \mathrm{N}\left(25, \sqrt{25}^{\,2}\right)\) | M1A1 |
| \(\mathrm{P}(Y \leqslant 19) \approx \mathrm{P}\left(Z \lt \dfrac{19.5 - 25}{\sqrt{25}}\right)\) or \(\pm\dfrac{x - 0.5 - 25}{5} = 1.96\) | M1M1 |
| \(= \left[\mathrm{P}(Z \lt -1.1)\right] = 0.1357\) (or 0.13566… from calc) \(x = 35.3\) | A1 |
| [> 0.05] not significant, there is insufficient evidence to support Thomas’ claim. Or The number/rate/amount of defects is not decreased/less/reduced | A1cso |
| (7) | |
| (13 marks) |
Notes
B1 for suitable hypotheses
1st M1 for normal approximation
1st A1 for mean = 25 and variance = 25 or sd = 5 may be seen in the standardisation formula or implied by a correct answer
2nd M1 for attempting a continuity correction (Method 1: \(19 \pm 0.5\) / Method 2: \(x \pm 0.5\))
3rd M1 for standardising using their mean and their standard deviation and using either Method 1 [19.5, 19, 18.5 accept \(\pm z\).] Method 2 [\((x \pm 0.5)\) and equal to a \(\pm z\) value]
2nd A1 for awrt 0.136 or 35.3 or \(-1.1 \gt -1.96\)
3rd A1 for a correct contextualised conclusion. cao for a one tailed test, must come from correct working. Condone incorrect hypotheses.
NB if finding \(\mathrm{P}(X = 19)\) ie \(\mathrm{P}(X \leqslant 19.5) - \mathrm{P}(X \leqslant 18.5)\) they can get B1 M1 A1 M1 M1 A0 A0
(Note: the printed critical-value method uses 1.96, 35.3 and −1.96, which are as printed in the mark scheme. For this one-tailed 5% test the critical value is \(z = -1.6449\), which gives \(\dfrac{x + 0.5 - 25}{5} = -1.6449\), \(x = 16.3\); 19 is not in the critical region, so the conclusion is the same.)