A2 June 2022 Q2
2. A factory produces yellow tennis balls and white tennis balls. Independent samples, one of yellow tennis balls and one of white tennis balls, are taken. The table shows information about the weights of the yellow tennis balls, \(Y\) grams, and the weights of the white tennis balls, \(W\) grams.
| Sample size | Mean weight of random sample (grams) | Known population standard deviation of weights (grams) | |
|---|---|---|---|
| Yellow tennis balls | 120 | 57.2 | 1.2 |
| White tennis balls | 140 | 56.9 | 0.9 |
Jamie claims that the mean weight of the population of yellow tennis balls is greater than the mean weight of the population of white tennis balls. A test of Jamie’s claim is carried out.
You should state your hypotheses and the value of your test statistic clearly. (5)
| Scheme | Marks | AO |
|---|---|---|
| 1.96 | B1 | 3.3 |
| \(57.2 \pm \text{‘}1.96\text{’} \times \dfrac{1.2}{\sqrt{120}}\) | M1 | 2.1 |
| \((56.985\ldots,\ 57.414\ldots)\) awrt \((57.0,\ 57.4)\) | A1 | 1.1b |
| (3) |
Notes
B1: Understanding that sample mean can be modelled using a Normal distribution with \(z = 1.96\) Allow 1.959… or better from calculator
M1: Setting up confidence interval \(57.2 \pm z \times \dfrac{1.2}{\sqrt{120}}\)
A1: awrt (57.0, 57.4) Must come from correct \(z\) value = 1.96 (or better).
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\overline{Y} - \overline{W} \sim \mathrm{N}\ldots\ldots\) | M1 | 3.3 |
| \(\left(0,\ \dfrac{1.2^2}{120} + \dfrac{0.9^2}{140}\right)\) | A1A1 | 1.1b 1.1b |
| (3) | ||
| (ii) Central limit theorem applies so we do not need to know the distributions of \(Y\) and \(W\)/ Allows us to assume that sample means (\(\overline{Y}\) and \(\overline{W}\)) are normally distributed. | B1 | 2.4 |
| (1) |
Notes
M1: Translating context into a Normal distribution model
A1: Correct mean
A1: Correct variance (allow awrt 0.0178 or exact fraction \(\dfrac{249}{14000}\))
B1: Correct explanation about the distributions of \(Y\) and \(W\) or \(\overline{Y}\) and \(\overline{W}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \mu_{y(ellow)} = \mu_{w(hite)} \qquad \mathrm{H}_1 : \mu_{y(ellow)} \gt \mu_{w(hite)}\) | B1 | 2.5 |
| \(z = \dfrac{57.2 - 56.9}{\sqrt{\dfrac{1.2^2}{120} + \dfrac{0.9^2}{140}}} = 2.24949\ldots\) | M1 A1 | 3.1b 1.1b |
| CV = 1.6449 [ or \(p\)-value = 0.01224…] \(p\) = awrt 0.01 | B1 | 1.1b |
| [Reject \(\mathrm{H}_0\)] Significant evidence to support Jamie’s claim/mean weight of the population of yellow tennis balls is greater than mean weight of the population of white tennis balls. | A1 | 2.2b |
| (5) | ||
| (12 marks) |
Notes
B1: Both hypotheses (oe) correct with correct notation (if using \(\mu_x\) and \(\mu_y\) these must be defined).
M1: Standardising using normal distribution test statistic for difference of two means with known variance
A1: awrt 2.25
B1: Correct critical value 1.6449 or better, [or \(p\) = 0.01 or better from correct working]
A1: Drawing a correct inference in context. Do not allow contradictory statements, e.g. ‘Do not reject \(\mathrm{H}_0\), so Jamie’s claim is supported’