| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with binomial distribution. Part (i) is simple arithmetic (finding sample mean), part (ii) is routine binomial probability calculation, part (iii) tests understanding of the expected frequency rule (≥5), and part (iv) is a standard hypothesis test procedure. All steps are textbook applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of home wins | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Number of weeks | 0 | 1 | 2 | 8 | 8 | 9 | 7 | 1 | 2 | 0 | 0 |
| Number of home wins | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Totals |
| Observed number of weeks | 0 | 1 | 2 | 8 | 8 | 9 | 7 | 1 | 2 | 0 | 0 | 38 |
| Expected number of weeks | 0.096 | 0.788 | 2.899 | 6.326 | 9.058 | 8.893 | 6.064 | 2.835 | 0.870 | 0.158 | 0.013 | 38 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(171 \div 38\) | M1 | Attempt at \(\Sigma\ fx/38\). Allow divisions in either order, or combined. |
| \(\div 10\) | M1 | Use \(\mu = np\) |
| 0.45 AG | A1 | CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^{10}C_7 \times 0.45^7 \times (1-0.45)^3\) | M1 | \(= 0.0746\). Or tables: 0.9726-0.8980 |
| \(\times 38\) | M1 | |
| 2.835 AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Some Es \(< 5\) | B1 | Must have 'expected' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| HW: 0–3, 4, 5, 6–10; Obs: 11, 8, 9, 10; Exp: 10.109, 9.058, 8.893, 9.940 | B1 | May be implied. |
| \(\frac{(11 - \text{"10.109"})^2}{\text{"10.109"}} + \ldots\) | M1 | |
| 0.204 | A1 | Allow 0.2 |
| \(CV = 5.991\) | B1ft | ft correct CV for df = classes \(- 2\). B0M1A0B1ftM1A0 max for n(classes)\(\neq\)4. |
| TS \(<\) CV, do not reject NH | M1 | Allow if incorrect CV from correct tail. |
| There is insufficient evidence that the data is not B(10, 0.45) | A1 |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $171 \div 38$ | M1 | Attempt at $\Sigma\ fx/38$. Allow divisions in either order, or combined. |
| $\div 10$ | M1 | Use $\mu = np$ |
| 0.45 AG | A1 | CWO |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^{10}C_7 \times 0.45^7 \times (1-0.45)^3$ | M1 | $= 0.0746$. Or tables: 0.9726-0.8980 |
| $\times 38$ | M1 | |
| 2.835 AG | A1 | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Some Es $< 5$ | B1 | Must have 'expected' |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| HW: 0–3, 4, 5, 6–10; Obs: 11, 8, 9, 10; Exp: 10.109, 9.058, 8.893, 9.940 | B1 | May be implied. |
| $\frac{(11 - \text{"10.109"})^2}{\text{"10.109"}} + \ldots$ | M1 | |
| 0.204 | A1 | Allow 0.2 |
| $CV = 5.991$ | B1ft | ft correct CV for df = classes $- 2$. B0M1A0B1ftM1A0 max for n(classes)$\neq$4. |
| TS $<$ CV, do not reject NH | M1 | Allow if incorrect CV from correct tail. |
| There is insufficient evidence that the data is not B(10, 0.45) | A1 | |
---6 In each of 38 randomly selected weeks of the English Premier Football League there were 10 matches. Table 1 summarises the number of home wins in 10 matches, $X$, and the corresponding number of weeks.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Number of home wins & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Number of weeks & 0 & 1 & 2 & 8 & 8 & 9 & 7 & 1 & 2 & 0 & 0 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
A researcher investigates whether $X$ can be modelled by the distribution $\mathrm { B } ( 10 , p )$. He calculates the expected frequencies using a value of $p$ obtained from the sample mean.\\
(i) Show that $p = 0.45$.
Table 2 shows the observed and expected number of weeks.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Number of home wins & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & Totals \\
\hline
Observed number of weeks & 0 & 1 & 2 & 8 & 8 & 9 & 7 & 1 & 2 & 0 & 0 & 38 \\
\hline
Expected number of weeks & 0.096 & 0.788 & 2.899 & 6.326 & 9.058 & 8.893 & 6.064 & 2.835 & 0.870 & 0.158 & 0.013 & 38 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2\\
(ii) Show how the value of 2.835 for 7 home wins is obtained.}
\end{center}
\end{table}
The researcher carries out a test, at the $5 \%$ significance level, of whether the distribution $\mathrm { B } ( 10 , p )$ fits the data.\\
(iii) Explain why it is necessary to combine classes.\\
(iv) Carry out the test.
\hfill \mbox{\textit{OCR S3 2015 Q6 [13]}}