| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard S3 content: verifying a pdf, finding mean/variance through integration, and performing a chi-squared goodness of fit test with given probabilities. All parts follow routine procedures with no novel insight required, though the integration and test execution require careful work. Slightly easier than average due to the scaffolded structure and given probabilities. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.06b Fit prescribed distribution: chi-squared test |
| \(0 < X \leqslant \mu - \sigma\) | \(\mu - \sigma < X \leqslant \mu\) | \(\mu < X \leqslant \mu + \sigma\) | \(\mu + \sigma < X < \lambda\) |
| 0.18573 | 0.25871 | 0.36983 | 0.18573 |
| 4 | 11 | 20 | 15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) > 0\) for all \(x\) in the domain | E1 | |
| \(\int_0^\lambda \frac{2x}{\lambda^2}\,dx = \left[\frac{x^2}{\lambda^2}\right]_0^\lambda = \frac{\lambda^2}{\lambda^2} = 1\) | M1 | Correct integral with limits |
| A1 | Shown equal to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu = \int_0^\lambda \frac{2x^2}{\lambda^2}\,dx = \left[\frac{2x^3/3}{\lambda^2}\right]_0^\lambda = \frac{2\lambda}{3}\) | M1 | Correct integral with limits |
| A1 | c.a.o. | |
| \(P(X < \mu) = \int_0^\mu \frac{2x}{\lambda^2}\,dx = \left[\frac{x^2}{\lambda^2}\right]_0^\mu\) | M1 | Correct integral with limits |
| \(= \frac{\mu^2}{\lambda^2} = \frac{4\lambda^2/9}{\lambda^2} = \frac{4}{9}\) | ||
| which is independent of \(\lambda\) | A1 | Answer plus comment. ft c's \(\mu\) provided the answer does not involve \(\lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Given \(E(X^2) = \frac{\lambda^2}{2}\) | ||
| \(\sigma^2 = \frac{\lambda^2}{2} - \frac{4\lambda^2}{9} = \frac{\lambda^2}{18}\) | M1 | Use of \(\text{Var}(X) = E(X^2) - (E(X))^2\) |
| A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expected frequencies table: Prob: \(0.18573, 0.25871, 0.36983, 0.18573\); Expected f: \(9.2865, 12.9355, 18.4915, 9.2865\) | M1 | Probs \(\times 50\) for expected frequencies |
| A1 | All correct | |
| \(X^2 = 3.0094 + 0.2896 + 0.1231 + 3.5152 = 6.937(3)\) | 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. \(P(X^e > 6.937) = 0.0739\) |
| Upper 5% point is 7.815 | A1 | No ft from here if wrong |
| Not significant | A1 | ft only c's test statistic |
| Suggests model fits the data for these jars | A1 | ft only c's test statistic |
| But with a 10% significance level (cv = 6.251) a different conclusion would be reached | E1 | Any valid comment which recognises that the test statistic is close to the critical values |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) > 0$ for all $x$ in the domain | E1 | |
| $\int_0^\lambda \frac{2x}{\lambda^2}\,dx = \left[\frac{x^2}{\lambda^2}\right]_0^\lambda = \frac{\lambda^2}{\lambda^2} = 1$ | M1 | Correct integral with limits |
| | A1 | Shown equal to 1 |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu = \int_0^\lambda \frac{2x^2}{\lambda^2}\,dx = \left[\frac{2x^3/3}{\lambda^2}\right]_0^\lambda = \frac{2\lambda}{3}$ | M1 | Correct integral with limits |
| | A1 | c.a.o. |
| $P(X < \mu) = \int_0^\mu \frac{2x}{\lambda^2}\,dx = \left[\frac{x^2}{\lambda^2}\right]_0^\mu$ | M1 | Correct integral with limits |
| $= \frac{\mu^2}{\lambda^2} = \frac{4\lambda^2/9}{\lambda^2} = \frac{4}{9}$ | | |
| which is independent of $\lambda$ | A1 | Answer plus comment. ft c's $\mu$ provided the answer does not involve $\lambda$ |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Given $E(X^2) = \frac{\lambda^2}{2}$ | | |
| $\sigma^2 = \frac{\lambda^2}{2} - \frac{4\lambda^2}{9} = \frac{\lambda^2}{18}$ | M1 | Use of $\text{Var}(X) = E(X^2) - (E(X))^2$ |
| | A1 | c.a.o. |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expected frequencies table: Prob: $0.18573, 0.25871, 0.36983, 0.18573$; Expected f: $9.2865, 12.9355, 18.4915, 9.2865$ | M1 | Probs $\times 50$ for expected frequencies |
| | A1 | All correct |
| $X^2 = 3.0094 + 0.2896 + 0.1231 + 3.5152 = 6.937(3)$ | 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. $P(X^e > 6.937) = 0.0739$ |
| Upper 5% point is 7.815 | A1 | No ft from here if wrong |
| Not significant | A1 | ft only c's test statistic |
| Suggests model fits the data for these jars | A1 | ft only c's test statistic |
| But with a 10% significance level (cv = 6.251) a different conclusion would be reached | E1 | Any valid comment which recognises that the test statistic is close to the critical values |4 A random variable $X$ has probability density function $\mathrm { f } ( x ) = \frac { 2 x } { \lambda ^ { 2 } }$ for $0 < x < \lambda$, where $\lambda$ is a positive constant.\\
(i) Show that, for any value of $\lambda , \mathrm { f } ( x )$ is a valid probability density function.\\
(ii) Find $\mu$, the mean value of $X$, in terms of $\lambda$ and show that $\mathrm { P } ( X < \mu )$ does not depend on $\lambda$.\\
(iii) Given that $\mathrm { E } \left( X ^ { 2 } \right) = \frac { \lambda ^ { 2 } } { 2 }$, find $\sigma ^ { 2 }$, the variance of $X$, in terms of $\lambda$.
The random variable $X$ is used to model the depth of the space left by the filling machine at the top of a jar of jam. The model gives the following probabilities for $X$ (whatever the value of $\lambda$ ).
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$0 < X \leqslant \mu - \sigma$ & $\mu - \sigma < X \leqslant \mu$ & $\mu < X \leqslant \mu + \sigma$ & $\mu + \sigma < X < \lambda$ \\
\hline
0.18573 & 0.25871 & 0.36983 & 0.18573 \\
\hline
\end{tabular}
\end{center}
A sample of 50 random observations of $X$, classified in the same way, is summarised by the following frequencies.
\begin{center}
\begin{tabular}{ | l | l | l | l | }
\hline
4 & 11 & 20 & 15 \\
\hline
\end{tabular}
\end{center}
(iv) Carry out a suitable test at the $5 \%$ level of significance to assess the goodness of fit of $X$ to these data. Explain briefly how your conclusion may be affected by the choice of significance level.
\hfill \mbox{\textit{OCR MEI S3 2009 Q4 [18]}}