A2 June 2019 Q5
5. Alexa believes that students are equally likely to achieve the same percentage score on each of two tests, paper I and paper II. She randomly selects 8 students and gives them each paper I and paper II. The percentage scores for each paper are recorded.
The following paired data are collected.
| Student | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
|---|---|---|---|---|---|---|---|---|
| Paper I (%) | 70 | 70 | 84 | 80 | 64 | 65 | 65 | 90 |
| Paper II (%) | 64 | 76 | 72 | 74 | 68 | 64 | 58 | 76 |
Test, at the 1% significance level, whether or not there is evidence to support Alexa’s belief.
State your hypotheses clearly and show your working. (7)
| Scheme | Marks | AO |
|---|---|---|
| \(d\): 6 –6 12 6 –4 1 7 14 | M1 | 3.1b |
| \(\bar{d} = \pm 4.5 \qquad s_\mathrm{d} = \sqrt{50.285\ldots} = 7.09\ldots\) | M1 | 1.1b |
| \(\mathrm{H}_0 : \mu_\mathrm{d} = 0 \qquad \mathrm{H}_1 : \mu_\mathrm{d} \neq 0\) | B1 | 3.3 |
| \(t = \pm\dfrac{\text{“}4.5\text{”}\sqrt{8}}{\text{“}7.09\ldots\text{”}}\) oe | M1 | 1.1b |
| \(= \pm 1.7948\ldots\) awrt \(\pm 1.79/1.8\) | A1 | 1.1b |
| Critical value \(t_7 = \pm 3.499\) | B1 | 1.1b |
| There is insufficient evidence that the papers are of a different level of difficulty or Alexa’s belief is correct | A1ft | 2.2b |
| (7) | ||
| (7 marks) |
Notes
M1: for realising that the model to use is the paired \(t\)-test and finding the differences \((\pm)\) At least 3 correct
M1: correct method for finding \(\bar{d}\) and \(s_\mathrm{d}\).
B1: Using a correct model for difference and both hypotheses correct using the notation \(\mu_\mathrm{d}\) or \(\mu\)
Condone \(\mu_I = \mu_{II}\) and \(\mu_I \neq \mu_{II}\)
M1: Using the correct method to find test statistics ie \(t = \pm\dfrac{\text{“}\text{their } 4.5\text{”}\sqrt{8}}{\text{“}\text{their } 7.09\ldots\text{”}}\)
A1: awrt 1.79 or 1.8
B1: for correct critical value \(t = \pm 3.499\) with compatible sign
A1ft: Drawing a correct inference in context using their CV and their value of \(t\)
NB difference of means test gets M0M0B1M0A0B0A0