| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Spearman’s rank correlation coefficien |
| Type | Calculate Spearman's coefficient only |
| Difficulty | Easy -1.2 This is a straightforward calculation of Spearman's coefficient with only 4 data points and no ties, requiring simple arithmetic of differences squared and substitution into the formula. It's below average difficulty as it's purely mechanical with minimal computation. |
| Spec | 5.08e Spearman rank correlation |
| Position | 1st | 2nd | 3rd | 4th |
| Competition 1 | \(Q\) | \(R\) | \(S\) | \(P\) |
| Competition 2 | \(Q\) | \(P\) | \(R\) | \(S\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Attempt ranks 4 1 2 3 or 1 2 3 4 or 1 2 3 4 or 2 1 3 4; \(\Sigma d^2\) attempted (or 6); \(1 - \frac{6\Sigma d^2}{4(4^2-1)} = \frac{2}{5}\) | M1, A1, M1, M1, A1 | Ignore labels of rows or columns; No ranks seen, \(d = (0), \pm1, \pm1, \pm2\), or \(d^2 = (0), 1, 1, 4\) any order: M1A1M1; NOT \((\Sigma d)^2\); No wking, \(\Sigma d^2 = 6\): M1A1M1; No wking, ans \(\frac{2}{5}\): Full mks; Allow both sets of ranks reversed; NB incorrect method: 2 3 4 1 or 2 1 3 4 OR \(d = (0), \pm2, \pm1, \pm3\) any order OR \(d^2 = (0), 4, 1, 9\) any order (leading to \(\Sigma d^2 = 14\) and \(r_s = -\frac{2}{5}\)): M0A0M1M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((1 - 0.5565) \text{ or } 12 \times 0.85^{11} \times (1-0.85) + 0.85^{12}\) or \(1 - (1-0.85)^{12} - \ldots ^{12}C_{10} \times 0.85^{10}(1-0.85)^2\) ie 1 – (all 11 correct binomial terms) | M1, A1, 2 | or \(1 - 0.557\); NB \(1 - 0.4435\) (oe): M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(0.5565 - 0.2642\) or \(^{12}C_{10}(1-0.85)^7(0.85)^{10}\) | M1, 2, A1, 2 | or \(0.557 - 0.264\) |
| Answer | Marks |
|---|---|
| Answer: \(12 \times 0.85 \times (1-0.85)\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(1.53\) | A1 | \((\frac{3}{4})^2\) AND \(\frac{3}{4} \times \frac{1}{4}\) seen (possibly × 2); eg \((\frac{3}{4})^2 + \frac{3}{4} \times \frac{1}{4}\) or \(2 \times (\frac{3}{4})^2 + 2 \times \frac{3}{4} \times \frac{1}{4}\) or \(0.5625 + 0.1875\) or \(0.5625 + 0.375\); or eg \(\frac{9}{16}\) and \(\frac{3}{16}\) or \(\frac{9}{16}\) and \(\frac{3}{8}\) eg in table or list; or \((\frac{3}{4})^2 \times 2 \times \frac{3}{4} \times \frac{1}{4}\) oe or \(\frac{27}{128}\) or \(0.211\); Fully correct method; \(= \frac{27}{64}\) or \(0.422\) (3 sfs) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Method is either: Just \(4 \div 3\) or \(\frac{4}{3}\) or: Use of ratio of correct frequencies AND ratio of widths (correct or 4 and 2) | M1, M1 | Correct calc'n using 5.6, 28, 4, 5, 3 oe: M2; Correct calc'n using 5.6, 28, 4, 4, 2 oe: M1; ie fully correct method: M2; or: incorrect class widths, otherwise correct method: M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(5.6 \times \frac{4}{28} \times \frac{5}{3}\) or \(0.8 \times \frac{5}{3}\) or \((5.6 \div \frac{28}{5}) \times \frac{4}{3}\) or \(\frac{4}{3} \div 3\) oe | M2, M1, M1, M1 | \(\frac{4}{3}\) correctly obtained (or no wking) then further incorrect: M1M0A0; Use of ratio of widths OR freqs but not both: M0 eg \(5.6 \times \frac{4}{28}(= 0.8)\) or \(5.6 \times \frac{5}{3}(= 3.36)\): M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(= 1\frac{1}{3}\) or \(\frac{4}{3}\) or \(1.33\) (3 sf) oe | A1 | \(\frac{4}{3} = 2\): M0M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: 25 or 26 or 25.5; Med is 21st (or 22nd or 21.5th) in 31-35 class or "≈25 – 4"; Can be implied by calc'n; Med > 33 or "more than" | B1, B1, B1 | or 25 & 26; or med in last ≈ 7 in class or 33 ≈ 18th in class or 33 ≈ 18th in whole set; Can be implied by diagram; indep; The "≈" sign means ± 2; Allow on boundaries. Not class widths; Calc'ns need not be correct but need to contain relevant figures for gaining B1B1; Alternative Method: 33 ≈ 18th value B1; More values above 33 than below oe B1; Med > 33 B1; Ignore comment on skew; NB Use EITHER the main method OR the Alternative Method (above), not a mixture of the two. Choose the method that gives most marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: > 3 mid-pts attempted; \(\Sigma fx \div 50\) attempted \((= \frac{1819}{50})\); \(= 36.38\) or \(36.4\) (3 sf); \(\Sigma fx^2\) attempted \((= 68055.5)\); \(\sqrt{\frac{68055.5}{50} - (\frac{1819}{50})^2}\) or \(\sqrt{1361.11 - 36.38^2}\) \((= \sqrt{37.6056})\); \(= 6.13\) (3 sfs) | M1, M1, A1, M1, M1, A1 | seen or implied; > 3 terms. or 36 with correct working; Allow on boundaries. Not class widths; > 3 terms. Allow on boundaries. Not class widths (3364, 30492, 22963.5, 11236); completely correct method except midpts & ft their mean, dep not \(\sqrt{\text{neg}}\) Allow class widths for this mark only; NB mark is not just for "−mean"−, unlike q5(iii); \(\Sigma(fx)^2\): M0M0A0; If no wking for \(\Sigma fx^2\), check using their \(x\) and \(f\); If no wking or unclear wking: full mks for each correct ans for incorrect ans: 35.8 ≤ \(\mu\) ≤ 36.9 M0M1A0; 6.0 ≤ \(sd\) ≤ 6.25 M1M0A0; Alt for variance: \(\Sigma f(x - \bar{x})^2\) (= 1880.28); \(\sqrt{\frac{1880.28}{50}}\); \(= 6.13\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: (a) Decrease (b) Increase (c) Same (d) Same | B1B1, B1B1, 4 | Ignore other, eg "slightly" or "probably" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: All correct probs correctly placed, matching labels, if any | B2, 2 | B1 for 4 correct probs anywhere; Allow B2 with missing labels but only if probs consistently placed, ie R above B throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\frac{4}{10} \times \frac{9}{10} + \frac{6}{10} \times \frac{9}{10} + \frac{8}{10} \times \frac{9}{10} + \frac{10}{10} \times \frac{9}{10}\) or \(\frac{4}{15} + \frac{6}{15} + \frac{6}{6}\) \((= \frac{3}{5}\) (AG) | B2, 2, B2, 2 | B1: two of these products (or their results) added (not multiplied); or \(1 - (\frac{6}{10} \times \frac{9}{10} + \frac{8}{10} \times \frac{9}{10} + \frac{10}{10} \times \frac{9}{10})\) or \(1 - (\frac{1}{6} + \frac{10}{15})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: 5040 or 6! or 5!×6 or 720 | B1, 1, M1 | NOT 6! in denom |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\div 7!\) or ÷ "5040" or 1440 or \((5! \text{ or } 6!) \times 2\); \(= \frac{3}{7}\) oe or \(0.286\) (3 sf) | M1, M1, A1 | Any ÷ 7! or "5040" but NOT any × 2; eg \(\frac{6!}{5040}\) or \(\frac{1}{7}\) or \(0.143\) or \(\frac{1}{21}\) (3 sfs): M1M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: 3! × 4! alone or 144 | M1 | \(\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3!}{4}\) oe or \(7C3 \text{ or } 7C4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((÷ 7!\) or "5040") \(= \frac{1}{35}\) oe or \(0.0286\) (3sf) | A1, 2 | Not \(3! \times 4! \times \ldots\) (eg not \(3! \times 4! \times 5\)); not \(\frac{3!×4!}{...}\) ; not \(\frac{31×4!}{...}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: 5 seen or 5! seen; 3! × 4! × 5 or 5! × 3! or 720 or 5 × 144 | M1, M1 | or GGGBBBB, BGGGGBB, BRGGGBB, BBRGGGB, BBBRGG GBB or continuing pattern; or \(5 \times \frac{3!}{7} \times \frac{4!}{6}\) oe; \(= \frac{1}{7}\) oe or \(0.143\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(x\) | B1, 1 | Ignore explanations. "Neither" or "Both": B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Diag showing vertical differences only | B1 | Allow description instead of diag: "Distances from pts to line // to y-axis" oe; Allow ≥ one line, from a point to the line; State that sum of squares of these is min oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: –1; Ranks opposite or reversed or perfect neg corr'n between ranks oe | B1, B1dep 2 | Not approx –1; As \(x\) increases, \(y\) decreases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: "Negative" or "Not –1" | B1, 1 | eg "Strong neg" or any negative value > –1 or "Close to –1" |
**Answer:** Attempt ranks 4 1 2 3 or 1 2 3 4 or 1 2 3 4 or 2 1 3 4; $\Sigma d^2$ attempted (or 6); $1 - \frac{6\Sigma d^2}{4(4^2-1)} = \frac{2}{5}$ | **M1, A1, M1, M1, A1** | Ignore labels of rows or columns; No ranks seen, $d = (0), \pm1, \pm1, \pm2$, or $d^2 = (0), 1, 1, 4$ any order: M1A1M1; NOT $(\Sigma d)^2$; No wking, $\Sigma d^2 = 6$: M1A1M1; No wking, ans $\frac{2}{5}$: Full mks; Allow both sets of ranks reversed; NB incorrect method: 2 3 4 1 or 2 1 3 4 OR $d = (0), \pm2, \pm1, \pm3$ any order OR $d^2 = (0), 4, 1, 9$ any order (leading to $\Sigma d^2 = 14$ and $r_s = -\frac{2}{5}$): M0A0M1M1A0
## Question 3ia
**Answer:** $(1 - 0.5565) \text{ or } 12 \times 0.85^{11} \times (1-0.85) + 0.85^{12}$ or $1 - (1-0.85)^{12} - \ldots ^{12}C_{10} \times 0.85^{10}(1-0.85)^2$ ie 1 – (all 11 correct binomial terms) | **M1, A1, 2** | or $1 - 0.557$; NB $1 - 0.4435$ (oe): M0A0
## Question 3ib
**Answer:** $0.5565 - 0.2642$ or $^{12}C_{10}(1-0.85)^7(0.85)^{10}$ | **M1, 2, A1, 2** | or $0.557 - 0.264$
## Question 3ic
**Answer:** $12 \times 0.85 \times (1-0.85)$ | **M1** |
## Question 3ii
**Answer:** $1.53$ | **A1** | $(\frac{3}{4})^2$ AND $\frac{3}{4} \times \frac{1}{4}$ seen (possibly × 2); eg $(\frac{3}{4})^2 + \frac{3}{4} \times \frac{1}{4}$ or $2 \times (\frac{3}{4})^2 + 2 \times \frac{3}{4} \times \frac{1}{4}$ or $0.5625 + 0.1875$ or $0.5625 + 0.375$; or eg $\frac{9}{16}$ and $\frac{3}{16}$ or $\frac{9}{16}$ and $\frac{3}{8}$ eg in table or list; or $(\frac{3}{4})^2 \times 2 \times \frac{3}{4} \times \frac{1}{4}$ oe or $\frac{27}{128}$ or $0.211$; Fully correct method; $= \frac{27}{64}$ or $0.422$ (3 sfs) | **M1, M1, M1, A1** | Allow even if further incorrect wking; Ans 0.211: check wking but probably gets M1M1M0A0; Use of 0.85 instead of $\frac{1}{4}$: MR max M1M1M1A0
## Question 4i
**Answer:** Method is either: Just $4 \div 3$ or $\frac{4}{3}$ or: Use of ratio of correct frequencies AND ratio of widths (correct or 4 and 2) | **M1, M1** | Correct calc'n using 5.6, 28, 4, 5, 3 oe: M2; Correct calc'n using 5.6, 28, 4, 4, 2 oe: M1; ie fully correct method: M2; or: incorrect class widths, otherwise correct method: M1
## Question 4ii
**Answer:** $5.6 \times \frac{4}{28} \times \frac{5}{3}$ or $0.8 \times \frac{5}{3}$ or $(5.6 \div \frac{28}{5}) \times \frac{4}{3}$ or $\frac{4}{3} \div 3$ oe | **M2, M1, M1, M1** | $\frac{4}{3}$ correctly obtained (or no wking) then further incorrect: M1M0A0; Use of ratio of widths OR freqs but not both: M0 eg $5.6 \times \frac{4}{28}(= 0.8)$ or $5.6 \times \frac{5}{3}(= 3.36)$: M0
## Question 4iii
**Answer:** $= 1\frac{1}{3}$ or $\frac{4}{3}$ or $1.33$ (3 sf) oe | **A1** | $\frac{4}{3} = 2$: M0M0A0
## Question 4ii (median)
**Answer:** 25 or 26 or 25.5; Med is 21st (or 22nd or 21.5th) in 31-35 class or "≈25 – 4"; Can be implied by calc'n; Med > 33 or "more than" | **B1, B1, B1** | or 25 & 26; or med in last ≈ 7 in class or 33 ≈ 18th in class or 33 ≈ 18th in whole set; Can be implied by diagram; indep; The "≈" sign means ± 2; Allow on boundaries. Not class widths; Calc'ns need not be correct but need to contain relevant figures for gaining B1B1; **Alternative Method:** 33 ≈ 18th value B1; More values above 33 than below oe B1; Med > 33 B1; Ignore comment on skew; NB Use EITHER the main method OR the Alternative Method (above), not a mixture of the two. Choose the method that gives most marks.
## Question 4iii
**Answer:** > 3 mid-pts attempted; $\Sigma fx \div 50$ attempted $(= \frac{1819}{50})$; $= 36.38$ or $36.4$ (3 sf); $\Sigma fx^2$ attempted $(= 68055.5)$; $\sqrt{\frac{68055.5}{50} - (\frac{1819}{50})^2}$ or $\sqrt{1361.11 - 36.38^2}$ $(= \sqrt{37.6056})$; $= 6.13$ (3 sfs) | **M1, M1, A1, M1, M1, A1** | seen or implied; > 3 terms. or 36 with correct working; Allow on boundaries. Not class widths; > 3 terms. Allow on boundaries. Not class widths (3364, 30492, 22963.5, 11236); completely correct method except midpts & ft their mean, dep not $\sqrt{\text{neg}}$ Allow class widths for this mark only; NB mark is not just for "−mean"−, unlike q5(iii); $\Sigma(fx)^2$: M0M0A0; If no wking for $\Sigma fx^2$, check using their $x$ and $f$; If no wking or unclear wking: full mks for each correct ans for incorrect ans: 35.8 ≤ $\mu$ ≤ 36.9 M0M1A0; 6.0 ≤ $sd$ ≤ 6.25 M1M0A0; **Alt for variance:** $\Sigma f(x - \bar{x})^2$ (= 1880.28); $\sqrt{\frac{1880.28}{50}}$; $= 6.13$ (3 sf) | M1, M1, A1
## Question 4iv
**Answer:** (a) Decrease (b) Increase (c) Same (d) Same | **B1B1, B1B1, 4** | Ignore other, eg "slightly" or "probably"
## Question 5 (if done with replacement)
No marks in any part of this question.
## Question 5i
**Answer:** All correct probs correctly placed, matching labels, if any | **B2, 2** | B1 for 4 correct probs anywhere; Allow B2 with missing labels but only if probs consistently placed, ie R above B throughout
## Question 5ii
**Answer:** $\frac{4}{10} \times \frac{9}{10} + \frac{6}{10} \times \frac{9}{10} + \frac{8}{10} \times \frac{9}{10} + \frac{10}{10} \times \frac{9}{10}$ or $\frac{4}{15} + \frac{6}{15} + \frac{6}{6}$ $(= \frac{3}{5}$ (AG) | **B2, 2, B2, 2** | B1: two of these products (or their results) added (not multiplied); or $1 - (\frac{6}{10} \times \frac{9}{10} + \frac{8}{10} \times \frac{9}{10} + \frac{10}{10} \times \frac{9}{10})$ or $1 - (\frac{1}{6} + \frac{10}{15})$ | B1: 1 – two of these products (or results) added (not multiplied); NB incorrect methods can lead to correct ans; AG so no wking no mks; No ft from tree in (i)
## Question 6ia
**Answer:** 5040 or 6! or 5!×6 or 720 | **B1, 1, M1** | NOT 6! in denom
## Question 6ib
**Answer:** $\div 7!$ or ÷ "5040" or 1440 or $(5! \text{ or } 6!) \times 2$; $= \frac{3}{7}$ oe or $0.286$ (3 sf) | **M1, M1, A1** | Any ÷ 7! or "5040" but NOT any × 2; eg $\frac{6!}{5040}$ or $\frac{1}{7}$ or $0.143$ or $\frac{1}{21}$ (3 sfs): M1M1A0
## Question 6iia
**Answer:** 3! × 4! alone or 144 | **M1** | $\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3!}{4}$ oe or $7C3 \text{ or } 7C4$
## Question 6iib
**Answer:** $(÷ 7!$ or "5040") $= \frac{1}{35}$ oe or $0.0286$ (3sf) | **A1, 2** | Not $3! \times 4! \times \ldots$ (eg not $3! \times 4! \times 5$); not $\frac{3!×4!}{...}$ ; not $\frac{31×4!}{...}$
## Question 6b
**Answer:** 5 seen or 5! seen; 3! × 4! × 5 or 5! × 3! or 720 or 5 × 144 | **M1, M1** | or GGGBBBB, BGGGGBB, BRGGGBB, BBRGGGB, BBBRGG GBB or continuing pattern; or $5 \times \frac{3!}{7} \times \frac{4!}{6}$ oe; $= \frac{1}{7}$ oe or $0.143$ (3 sf) | **A1, 3** | NB no mark for ÷ 7! or "5040" in this part
## Question 7i
**Answer:** $x$ | **B1, 1** | Ignore explanations. "Neither" or "Both": B0
## Question 7ii
**Answer:** Diag showing vertical differences only | **B1** | Allow description instead of diag: "Distances from pts to line // to y-axis" oe; Allow ≥ one line, from a point to the line; State that sum of squares of these is min oe | **B1, 2** | dep vert or horiz lines (not both) drawn or described
## Question 7iii
**Answer:** –1; Ranks opposite or reversed or perfect neg corr'n between ranks oe | **B1, B1dep 2** | Not approx –1; As $x$ increases, $y$ decreases | B1 | Allow eg: –1 because neg corr'n so ranks must be reversed; Ignore other; NOT neg corr'n or strong neg rel'nship oe; NOT comment about "disagreement" or "agreement"
## Question 7iv
**Answer:** "Negative" or "Not –1" | **B1, 1** | eg "Strong neg" or any negative value > –1 or "Close to –1"
2 The orders in which 4 contestants, $P , Q , R$ and $S$, were placed in two competitions are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Position & 1st & 2nd & 3rd & 4th \\
\hline
Competition 1 & $Q$ & $R$ & $S$ & $P$ \\
\hline
Competition 2 & $Q$ & $P$ & $R$ & $S$ \\
\hline
\end{tabular}
\end{center}
Calculate Spearman's rank correlation coefficient between these two orders.
\hfill \mbox{\textit{OCR S1 2011 Q2 [5]}}