| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test application with standard hypothesis testing procedure. Part (i) requires routine calculation of differences, mean, standard deviation, and t-statistic comparison. Part (ii) adds a mild twist requiring iteration to find the threshold value of k, but the method remains mechanical. The question is slightly easier than average because it's well-structured with clear steps and no conceptual surprises, though it does require careful arithmetic. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Student | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) |
| 1st assessment | 124 | 121 | 111 | 113 | 118 | 119 |
| 2nd assessment | 127 | 119 | 114 | 110 | 120 | 122 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu_1 = \mu_2 \quad H_1: \mu_1 < \mu_2\) | B1 | Allow d=0 d\(>\)0 \(\mu_D\)=0 etc. If words, must have population. NOT \(\mu\), NOT \(\bar{d}\) |
| \(\bar{d} = 1\) | B1 | Allow \(\bar{d} = -1\) and consistent following work. NOT PAIRED allow B1,(hyps) B1 (\(118\frac{2}{3}\)-\(117\frac{2}{3}\)=1) B0 M1 for var based on both assessments including /6 A0 |
| Variance \(= \frac{6}{5}(\frac{44}{6} - 1^2) = 7.6\) | B1 | Or sd=2.757. B1 CV 2.015 or 1.812 M1 TS\(<\)CV B0 (assumption) |
| \(\frac{\text{"1"}}{\sqrt{\frac{\text{"7.6"}}{6}}} = 0.889\) | M1A1 | Allow M1 if \(\frac{6}{5}\) omitted. Must have /6. |
| \(CV = 2.015\) | B1 | Or p=0.2074/5 (\(>\)5%) |
| "\(0.889\)"\(<\)"2.015", do not reject \(H_0\) AG | M1 | Must be TS \(<\) CV (AG). CV must be from t, not z. |
| Differences are normally distributed. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New \(\bar{d} = 1 + k\) | B1 | |
| Var unchanged | B1ft | |
| \(\frac{\text{"1"}+k}{\sqrt{\frac{\text{"7.6"}}{6}}} > \text{"2.015"}\) | M1 | Must be same values as (i). Allow =, if inequality soi later. Must have /"6". SC If B1B1M0, allow B1 for 1.27 seen. |
| \(k > 1.27\) i.e. \(k = 2\) | A1 | NOT \(k = 1\) |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_1 = \mu_2 \quad H_1: \mu_1 < \mu_2$ | B1 | Allow d=0 d$>$0 $\mu_D$=0 etc. If words, must have population. NOT $\mu$, NOT $\bar{d}$ |
| $\bar{d} = 1$ | B1 | Allow $\bar{d} = -1$ and consistent following work. NOT PAIRED allow B1,(hyps) B1 ($118\frac{2}{3}$-$117\frac{2}{3}$=1) B0 M1 for var based on both assessments including /6 A0 |
| Variance $= \frac{6}{5}(\frac{44}{6} - 1^2) = 7.6$ | B1 | Or sd=2.757. B1 CV 2.015 or 1.812 M1 TS$<$CV B0 (assumption) |
| $\frac{\text{"1"}}{\sqrt{\frac{\text{"7.6"}}{6}}} = 0.889$ | M1A1 | Allow M1 if $\frac{6}{5}$ omitted. Must have /6. |
| $CV = 2.015$ | B1 | Or p=0.2074/5 ($>$5%) |
| "$0.889$"$<$"2.015", do not reject $H_0$ AG | M1 | Must be TS $<$ CV (AG). CV must be from t, not z. |
| Differences are normally distributed. | B1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New $\bar{d} = 1 + k$ | B1 | |
| Var unchanged | B1ft | |
| $\frac{\text{"1"}+k}{\sqrt{\frac{\text{"7.6"}}{6}}} > \text{"2.015"}$ | M1 | Must be same values as (i). Allow =, if inequality soi later. Must have /"6". SC If B1B1M0, allow B1 for 1.27 seen. |
| $k > 1.27$ i.e. $k = 2$ | A1 | NOT $k = 1$ |
---3 A tutor gave an assessment to 6 randomly chosen new eleven-year-old students. After each student had received 30 hours of tuition, they were given a second assessment. The scores are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Student & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\
\hline
1st assessment & 124 & 121 & 111 & 113 & 118 & 119 \\
\hline
2nd assessment & 127 & 119 & 114 & 110 & 120 & 122 \\
\hline
\end{tabular}
\end{center}
(i) Show that, at the $5 \%$ significance level, there is insufficient evidence that students' scores are higher, on average, after tuition than before tuition. State a necessary assumption.\\
(ii) Disappointed by this result, the tutor looked again at the first assessment. She discovered that the first assessment was too easy, in fact being a test for ten-year-olds, not eleven-year-olds. She decided to reduce each score for the first assessment by a constant integer $k$. Find the least value of $k$ for which there is evidence at the $5 \%$ significance level that the students' scores have, on average, improved.
\hfill \mbox{\textit{OCR S3 2015 Q3 [12]}}