A2 June 2022 Q7
7. A machine fills bags with flour. The weight of flour delivered by the machine into a bag, \(X\) grams, is normally distributed with mean \(\mu\) grams and standard deviation 30 grams.
To check if there is any change to the mean weight of flour delivered by the machine into each bag, Olaf takes a random sample of 10 bags. The weight of flour, \(x\) grams, in each bag is recorded and \(\bar{x} = 1020\)
Olaf decides to alter the test so that the hypotheses are \(\mathrm{H}_0: \mu = 1000\) and \(\mathrm{H}_1: \mu \gt 1000\) but keeps the level of significance at 5%
He takes a second sample of size \(n\) and finds the critical region, \(\overline{X} \gt c\)
When the true value of \(\mu\) is 1020 grams, the probability of making a Type II error is 0.0050, to 2 significant figures.
| Scheme | Marks | AO |
|---|---|---|
| \(\overline{X} \sim \mathrm{N}(1000, 90)\) (May be implied by correct prob or \(z\) value seen) | M1 | 3.3 |
| \(\mathrm{P}(\overline{X} \gt 1020) = 0.0175\ldots\) or \(z = 2.108\) | A1 | 3.4 |
| \(0.0175\ldots \lt 0.025\) or \(z = 2.108\ldots \gt 1.96\) therefore reject \(\mathrm{H}_0\). | M1 | 1.1b |
| There is evidence that the mean weight of the flour in a bag is not 1000 g or evidence of a change in mean weight of flour in a bag | A1 cso | 2.2b |
| (4) |
Notes
1st M1 Setting up the correct model. Normal with \(\mu = 1000, \sigma^2 = 90\) or \(\sigma = \sqrt{90}\) or awrt 9.49
1st A1 Using the model to find the correct \(z\) value or \(\mathrm{P}(\overline{X} \gt 1020) =\) awrt 0.0175
Allow CR \(\overline{X} \geqslant 1018.59..\) awrt 1019 [ > is OK] Ignore lower CR provided < 1000 (corrected from the printed mark scheme, which has \(\overline{C}\) for \(\overline{X}\))
2nd M1 Correct comparison or non-contextual conclusion. Allow comparison of 1020 with critical region. Dep on \(\mathrm{P}(\overline{X} \gt 1020)\) M0 if there are contradictory statements.
2nd A1 cso dep on M1A1M1 for a correct conclusion in context with underlined words
Do NOT accept “mean weight has increased”
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\overline{Y} \sim \mathrm{N}\left(1000, \dfrac{900}{n}\right) \Rightarrow\right]\ \dfrac{c - 1000}{30/\sqrt{n}} = 1.6449\) | M1 | 3.4 |
| \(c = 1000 + \dfrac{49.347}{\sqrt{n}}\) | A1 | 1.1b |
| (2) |
Notes
M1 For Finding the CR using the Normal distribution. Condone \(\sigma = \sqrt{\frac{30}{n}}\) to score M1
\(\dfrac{c - 1000}{30/\sqrt{n}} = z\) where \(|z| \gt 1.5\)
Allow any inequality or = for M1 in (b) and M1 A1ft M1 in (c)
A1 A correct equation in the form \(c = \ldots\) and for use of awrt 1.6449 (implied by awrt 49.3[4]) Condone \(\overline{X}\) used for \(c\) (o.e.)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\text{“}1000 + \frac{49.347}{\sqrt{n}}\text{”} - 1020}{30/\sqrt{n}} = -2.5758\) | M1 A1ft | 3.4 1.1b |
| \(\dfrac{126.621}{\sqrt{n}} = 20\) or \(\dfrac{49.34\ldots}{c - 1000} = \dfrac{-77.274}{c - 1020}\) (Allow 2sf accuracy) | dM1 | 1.1b |
| \(n = \underline{\mathbf{40}}\) | A1 | 2.1 |
| \(c = 1007.8\ldots\) awrt 1010 | A1 | 1.1b |
| (5) | ||
| (11 marks) |
Notes
1st M1 Standardising using their \(c\) (letter or expression) and equating to \(z\) (\(|z| \gt 2\)) to form an equation in \(n\) or \(n\) and \(c\). Can ft their \(\sigma\) used in (b) for M1A1ft here
1st A1ft Ft their “\(c\)” for a correct equation with \(-2.58\) (or 1.64 or 1.65 used in (b))
2nd dM1 Dependent 1st M1. Isolating or eliminating either \(\sqrt{n}\) or \(n\) or eliminating \(c\) leading to an equation for \(n\) or \(c\)
2nd A1 For 40 (allow 41) Must be an integer. With correct working. e.g. Check correct \(\sigma\) has been used
3rd A1 For awrt 1010 from correct working