| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Challenging +1.2 This is a multi-part question combining continuous probability distributions with chi-squared testing. Parts (i)-(iii) involve standard A-level techniques (sketching, integration, using normalization), while parts (iv)-(v) require calculating expected frequencies and performing a routine goodness-of-fit test. The integration is straightforward polynomial work, and the hypothesis test follows a standard template. More demanding than average due to the combination of pure and applied statistics and the multi-step nature, but all techniques are standard S3 material with no novel insights required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.06b Fit prescribed distribution: chi-squared test |
| Time (minutes) | \(0 \leqslant x < 1\) | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) |
| Observed frequency | 5 | 15 | 23 | 11 | 6 |
| Expected frequency | 17.76 | 9.12 | 1.632 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve through the origin and in the first quadrant only | G1 | |
| A single maximum; curve returns to \(y=0\); nothing to the right of \(x=5\) | G1 | |
| No turning point at \(x=0\); turning point at \(x=5\); \((5,0)\) labelled (p.i. by an indicated scale) | G1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(x) = k\int_0^x t(t-5)^2\, dt\) | M1 | Correct integral for \(F(x)\) with limits (which may appear later) |
| \(= k\left[\dfrac{t^4}{4} - \dfrac{10t^3}{3} + \dfrac{25t^2}{2}\right]_0^x\) | M1 | Correctly integrated |
| \(= k\left(\dfrac{x^4}{4} - \dfrac{10x^3}{3} + \dfrac{25x^2}{2}\right)\) | A1 | Limits used correctly to obtain expression. Condone absence of "\(-0\)". Do not require complete definition of \(F(x)\). Dependent on both M1s |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F(5) = 1\) | ||
| \(\therefore k\left(\frac{5^4}{4} - \frac{10 \times 5^3}{3} + \frac{25 \times 5^2}{2}\right) = 1\) | M1 | Substitute \(x = 5\) and equate to 1. |
| \(\therefore k\left(\frac{1875 - 5000 + 3750}{12}\right) = 1\) | Expect to see evidence of at least this line of working (oe) for A1. | |
| \(\therefore k \times \frac{625}{12} = 1\) | ||
| \(\therefore k = \frac{12}{625}\) | A1 | Convincingly shown. Beware printed answer. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(0 \leq x < 1\), Expected \(f = 60 \times F(1)\) | M1 | Use of \(60 \times F(x)\) with correct \(k\). |
| \(= 60 \times \frac{12}{625}\left(\frac{1^4}{4} - \frac{10 \times 1^3}{3} + \frac{25 \times 1^2}{2}\right) = 10.848\) | A1 | Allow also \(31.488\) – frequency for \(1 \leq x < 2\) provided that one found using \(F(x)\). Allow either frequency found by integration. |
| For \(1 \leq x < 2\), Expected \(f = 60 - \Sigma(\text{the rest}) = 20.64\) | B1 | FT \(31.488\) – previous answer. Or allow \(60 \times (F(2) - F(1))\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): The model is suitable / fits the data. \(H_1\): The model is not suitable / does not fit the data. | B1 | Both hypotheses. Must be the right way round. Do not accept "data fit model" oe. |
| Merge last 2 cells: Obs \(f = 17\), Exp \(f = 10.752\) | M1 | |
| \(X^2 = 3.1525 + 1.5411 + 1.5460 + 3.6307 = 9.870\) | M1 | Calculation of \(X^2\). |
| A1 | c.a.o. | |
| Refer to \(\chi^2_3\). | M1 | Allow correct df (= cells \(- 1\)) from wrongly grouped table and ft. Otherwise, no ft if wrong. |
| Upper \(2.5\%\) point is \(9.348\). | A1 | No ft from here if wrong. \(P(X^2 > 9.870) = 0.0197\). |
| Significant. | A1 | ft only c's test statistic. |
| Sufficient evidence to suggest that the model is not suitable in this context. | A1 | ft only c's test statistic. Conclusion in context. Do not accept "data do not fit model" oe. |
| [8] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve through the origin and in the first quadrant only | G1 | |
| A single maximum; curve returns to $y=0$; nothing to the right of $x=5$ | G1 | |
| No turning point at $x=0$; turning point at $x=5$; $(5,0)$ labelled (p.i. by an indicated scale) | G1 | |
**[3]**
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = k\int_0^x t(t-5)^2\, dt$ | M1 | Correct integral for $F(x)$ with limits (which may appear later) |
| $= k\left[\dfrac{t^4}{4} - \dfrac{10t^3}{3} + \dfrac{25t^2}{2}\right]_0^x$ | M1 | Correctly integrated |
| $= k\left(\dfrac{x^4}{4} - \dfrac{10x^3}{3} + \dfrac{25x^2}{2}\right)$ | A1 | Limits used correctly to obtain expression. Condone absence of "$-0$". Do not require complete definition of $F(x)$. Dependent on both M1s |
**[3]**
# Question 3:
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $F(5) = 1$ | | |
| $\therefore k\left(\frac{5^4}{4} - \frac{10 \times 5^3}{3} + \frac{25 \times 5^2}{2}\right) = 1$ | M1 | Substitute $x = 5$ and equate to 1. |
| $\therefore k\left(\frac{1875 - 5000 + 3750}{12}\right) = 1$ | | Expect to see evidence of at least this line of working (oe) for A1. |
| $\therefore k \times \frac{625}{12} = 1$ | | |
| $\therefore k = \frac{12}{625}$ | A1 | Convincingly shown. Beware printed answer. |
| **[2]** | | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| For $0 \leq x < 1$, Expected $f = 60 \times F(1)$ | M1 | Use of $60 \times F(x)$ with correct $k$. |
| $= 60 \times \frac{12}{625}\left(\frac{1^4}{4} - \frac{10 \times 1^3}{3} + \frac{25 \times 1^2}{2}\right) = 10.848$ | A1 | Allow also $31.488$ – frequency for $1 \leq x < 2$ provided that one found using $F(x)$. Allow either frequency found by integration. |
| For $1 \leq x < 2$, Expected $f = 60 - \Sigma(\text{the rest}) = 20.64$ | B1 | FT $31.488$ – previous answer. Or allow $60 \times (F(2) - F(1))$ |
| **[3]** | | |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: The model is suitable / fits the data. $H_1$: The model is not suitable / does not fit the data. | B1 | Both hypotheses. Must be the right way round. Do not accept "data fit model" oe. |
| Merge last 2 cells: Obs $f = 17$, Exp $f = 10.752$ | M1 | |
| $X^2 = 3.1525 + 1.5411 + 1.5460 + 3.6307 = 9.870$ | M1 | Calculation of $X^2$. |
| | A1 | c.a.o. |
| Refer to $\chi^2_3$. | M1 | Allow correct df (= cells $- 1$) from wrongly grouped table and ft. Otherwise, no ft if wrong. |
| Upper $2.5\%$ point is $9.348$. | A1 | No ft from here if wrong. $P(X^2 > 9.870) = 0.0197$. |
| Significant. | A1 | ft only c's test statistic. |
| Sufficient evidence to suggest that the model is not suitable in this context. | A1 | ft only c's test statistic. Conclusion in context. Do not accept "data do not fit model" oe. |
| **[8]** | | |
---3 The random variable $X$ has the following probability density function, $\mathrm { f } ( x )$.
$$f ( x ) = \begin{cases} k x ( x - 5 ) ^ { 2 } & 0 \leqslant x < 5 \\ 0 & \text { elsewhere } \end{cases}$$
(i) Sketch $\mathrm { f } ( x )$.\\
(ii) Find, in terms of $k$, the cumulative distribution function, $\mathrm { F } ( x )$.\\
(iii) Hence show that $k = \frac { 12 } { 625 }$.
The random variable $X$ is proposed as a model for the amount of time, in minutes, lost due to stoppages during a football match. The times lost in a random sample of 60 matches are summarised in the table. The table also shows some of the corresponding expected frequencies given by the model.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Time (minutes) & $0 \leqslant x < 1$ & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ \\
\hline
Observed frequency & 5 & 15 & 23 & 11 & 6 \\
\hline
Expected frequency & & & 17.76 & 9.12 & 1.632 \\
\hline
\end{tabular}
\end{center}
(iv) Find the remaining expected frequencies.\\
(v) Carry out a goodness of fit test, using a significance level of $2.5 \%$, to see if the model might be suitable in this context.
\hfill \mbox{\textit{OCR MEI S3 2013 Q3 [19]}}