| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with expected frequencies provided, followed by routine exponential distribution calculations. Part (i) requires combining cells (standard procedure), calculating test statistic, and comparing to critical value. Parts (ii)-(v) involve straightforward application of exponential distribution formulas with no conceptual challenges. All techniques are textbook exercises requiring minimal problem-solving. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.06b Fit prescribed distribution: chi-squared test |
| Number of call-outs | 0 | 1 | 2 | 3 | 4 | 5 or more |
| Frequency (days) | 145 | 79 | 22 | 6 | 3 | 0 |
| Number of call-outs | 0 | 1 | 2 | 3 | 4 | 5 or more |
| Expected frequency | 139.947 | 83.968 | 25.190 | 5.038 | 0.756 | 0.101 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): The model for the number of callouts fits the data. \(H_1\): The model for the number of callouts does not fit the data. | B1 B1 | Do not allow "Data fit the model" o.e. for either hypothesis. |
| Merge last 3 cells: Obs \(= 9\), Exp \(= 5.895\) | M1 | Calculation of \(X^2\) |
| \(X^2 = 0.1824 + 0.2939 + 0.4040 + 1.6355 = 2.515(8)\) | M1 A1 | Cao. Require 3/4 sf; condone up to 6. |
| Refer to \(\chi^2_2\) | M1 | Allow correct df (\(= \text{cells} - 2\)) from wrongly grouped table and ft. Otherwise no ft if wrong. \(P(X^2 > 2.5158) = 0.2842\) |
| Upper 5% point is 5.991. | A1 | No ft from here if wrong. |
| Not significant. | A1 | ft only c's test statistic |
| Suggests it is reasonable to suppose that the model fits the data. | A1 | ft only c's test statistic. "Non-assertive" conclusion in words (+context). Do not allow "Data fit model" o.e. |
| [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= \frac{5}{3}\) \(\therefore \lambda = 0.6\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(t) = \int_0^t 0.6e^{-0.6x}\,dx\) | M1 | Correct integral with limits (may be implied subsequently). Allow use of "\(+ c\)" accompanied by valid attempt to evaluate it. |
| \(= \left[-e^{-0.6x}\right]_0^t\) | A1 | Correctly integrated. |
| \(= \left(-e^{-0.6t} - (-e^0)\right) = 1 - e^{-0.6t}\) | A1 | Limits used or \(c\) evaluated correctly. Accept unsimplified form. If final answer given in terms of \(\lambda\) then allow max M1A1A0. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(T > 1) = 1 - F(1)\) | M1 | ft c's \(F(t)\) |
| \(= 1 - (1 - e^{-0.6}) = 0.5488\) | A1 | cao. Allow any exact form of the correct answer. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(m) = \frac{1}{2}\) \(\therefore 1 - e^{-0.6m} = \frac{1}{2}\) | M1 | Use of definition of median. Allow use of c's \(F(t)\). |
| \(\therefore e^{-0.6m} = \frac{1}{2}\) \(\therefore -0.6m = -\ln 2\) \(\therefore m = \frac{\ln 2}{0.6}\) | M1 | Convincing attempt to rearrange to "\(m = \ldots\)", to include use of logs. |
| \(m = 1.155\) (days) | A1 | Cao obtained only from correct \(F(t)\). Must be evaluated. Require 2 to 4 sf; condone 5. |
| [3] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: The model for the number of callouts fits the data. $H_1$: The model for the number of callouts does not fit the data. | B1 B1 | Do not allow "Data fit the model" o.e. for either hypothesis. |
| Merge last 3 cells: Obs $= 9$, Exp $= 5.895$ | M1 | Calculation of $X^2$ |
| $X^2 = 0.1824 + 0.2939 + 0.4040 + 1.6355 = 2.515(8)$ | M1 A1 | Cao. Require 3/4 sf; condone up to 6. |
| Refer to $\chi^2_2$ | M1 | Allow correct df ($= \text{cells} - 2$) from wrongly grouped table and ft. Otherwise no ft if wrong. $P(X^2 > 2.5158) = 0.2842$ |
| Upper 5% point is 5.991. | A1 | No ft from here if wrong. |
| Not significant. | A1 | ft only c's test statistic |
| Suggests it is reasonable to suppose that the model fits the data. | A1 | ft only c's test statistic. "Non-assertive" conclusion in words (+context). Do not allow "Data fit model" o.e. |
| **[9]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= \frac{5}{3}$ $\therefore \lambda = 0.6$ | B1 | |
| **[1]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(t) = \int_0^t 0.6e^{-0.6x}\,dx$ | M1 | Correct integral with limits (may be implied subsequently). Allow use of "$+ c$" accompanied by valid attempt to evaluate it. |
| $= \left[-e^{-0.6x}\right]_0^t$ | A1 | Correctly integrated. |
| $= \left(-e^{-0.6t} - (-e^0)\right) = 1 - e^{-0.6t}$ | A1 | Limits used or $c$ evaluated correctly. Accept unsimplified form. If final answer given in terms of $\lambda$ then allow max M1A1A0. |
| **[3]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(T > 1) = 1 - F(1)$ | M1 | ft c's $F(t)$ |
| $= 1 - (1 - e^{-0.6}) = 0.5488$ | A1 | cao. Allow any exact form of the correct answer. |
| **[2]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(m) = \frac{1}{2}$ $\therefore 1 - e^{-0.6m} = \frac{1}{2}$ | M1 | Use of definition of median. Allow use of c's $F(t)$. |
| $\therefore e^{-0.6m} = \frac{1}{2}$ $\therefore -0.6m = -\ln 2$ $\therefore m = \frac{\ln 2}{0.6}$ | M1 | Convincing attempt to rearrange to "$m = \ldots$", to include use of logs. |
| $m = 1.155$ (days) | A1 | Cao obtained only from correct $F(t)$. Must be evaluated. Require 2 to 4 sf; condone 5. |
| **[3]** | | |4 The numbers of call-outs per day received by a fire station for a random sample of 255 weekdays were recorded as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of call-outs & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Frequency (days) & 145 & 79 & 22 & 6 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
The mean number of call-outs per day for these data is 0.6 . A Poisson model, using this sample mean of 0.6 , is fitted to the data, and gives the following expected frequencies (correct to 3 decimal places).
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of call-outs & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Expected frequency & 139.947 & 83.968 & 25.190 & 5.038 & 0.756 & 0.101 \\
\hline
\end{tabular}
\end{center}
(i) Using a $5 \%$ significance level, carry out a test to examine the goodness of fit of the model to the data.
The time $T$, measured in days, that elapses between successive call-outs can be modelled using the exponential distribution for which $\mathrm { f } ( t )$, the probability density function, is
$$\mathrm { f } ( t ) = \begin{cases} 0 & t < 0 , \\ \lambda \mathrm { e } ^ { - \lambda t } & t \geqslant 0 , \end{cases}$$
where $\lambda$ is a positive constant.\\
(ii) For the distribution above, it can be shown that $\mathrm { E } ( T ) = \frac { 1 } { \lambda }$. Given that the mean time between successive call-outs is $\frac { 5 } { 3 }$ days, write down the value of $\lambda$.\\
(iii) Find $\mathrm { F } ( t )$, the cumulative distribution function.\\
(iv) Find the probability that the time between successive call-outs is more than 1 day.\\
(v) Find the median time that elapses between successive call-outs.
\hfill \mbox{\textit{OCR MEI S3 2012 Q4 [18]}}