| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Normal |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with quartiles already provided. Part (i) is routine verification using normal tables, part (ii) is a standard χ² test with given expected frequencies (all 12.5), and part (iii) is a basic confidence interval calculation. All techniques are standard S3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04f Find normal probabilities: Z transformation5.05d Confidence intervals: using normal distribution5.06b Fit prescribed distribution: chi-squared test |
| Measurement \(( x )\) | \(x \leqslant 8.084\) | \(8.084 < x \leqslant 8.592\) | \(8.592 < x \leqslant 9.100\) | \(x > 9.100\) |
| Frequency | 8 | 22 | 11 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\Phi\!\left(\frac{8.084-8.592}{0.7534}\right) = \Phi(-0.674) = 0.25\) | M1 | Standardise once, allow \(\sqrt{}\) confusions, ignore sign |
| A1 | Obtain 0.25 for one interval | |
| \(\Phi(0) - \Phi(\text{above}) = 0.25\) | A1 | For a second interval, justified e.g. using \(\Phi(0)=0.5\) |
| \(P(8.592 \leq X \leq 9.1)\) = same by symmetry | A1 4 | For a third, justified e.g. "by symmetry" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{x-8.592}{0.7534} = 0.674\) | M1A1 | A1 for art 0.674 |
| \(x = 8.592 \pm 0.674 \times 0.7534 = (8.084, 9.100)\) | A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): normal distribution fits data | B1 | *Not* N(8.592, 0.7534). Allow "it's normally distributed" |
| All E values \(50/4 = 12.5\) | B1 | |
| \(X^2 = \frac{4.5^2+9.5^2+1.5^2+3.5^2}{12.5} = 10\) | M1, A1 | [Yates: 8.56: A0] |
| \(10 > 7.8794\) | B1 | CV 7.8794 seen |
| Reject \(H_0\). Significant evidence that normal distribution is not a good fit. | M1, \(A1\sqrt{}\) 7 | Correct method, incl. formula for \(\chi^2\) and comparison, allow wrong \(\nu\); conclusion in context, not too assertive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8.592 \pm 2.576 \times \frac{0.7534}{\sqrt{49}}\) | M1 | Allow \(\sqrt{}\) errors, wrong \(\sigma\) or \(z\), allow 50 |
| A1 | Correct, including \(z=2.576\) or \(t_{49}=2.680\), *not* 50 | |
| \((8.315,\ 8.869)\) | A1 3 | In range [8.31, 8.32] and in range [8.86, 8.87], even from 50, or (8.306, 8.878) from \(t_{49}\) |
## Question 7:
### Part (i) α:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\Phi\!\left(\frac{8.084-8.592}{0.7534}\right) = \Phi(-0.674) = 0.25$ | M1 | Standardise once, allow $\sqrt{}$ confusions, ignore sign |
| | A1 | Obtain 0.25 for one interval |
| $\Phi(0) - \Phi(\text{above}) = 0.25$ | A1 | For a second interval, justified e.g. using $\Phi(0)=0.5$ |
| $P(8.592 \leq X \leq 9.1)$ = same by symmetry | A1 **4** | For a third, justified e.g. "by symmetry" |
### Part (i) β:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x-8.592}{0.7534} = 0.674$ | M1A1 | A1 for art 0.674 |
| $x = 8.592 \pm 0.674 \times 0.7534 = (8.084, 9.100)$ | A1A1 | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: normal distribution fits data | B1 | *Not* N(8.592, 0.7534). Allow "it's normally distributed" |
| All E values $50/4 = 12.5$ | B1 | |
| $X^2 = \frac{4.5^2+9.5^2+1.5^2+3.5^2}{12.5} = 10$ | M1, A1 | [Yates: 8.56: A0] |
| $10 > 7.8794$ | B1 | CV 7.8794 seen |
| Reject $H_0$. Significant evidence that normal distribution is not a good fit. | M1, $A1\sqrt{}$ **7** | Correct method, incl. formula for $\chi^2$ and comparison, allow wrong $\nu$; conclusion in context, not too assertive |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8.592 \pm 2.576 \times \frac{0.7534}{\sqrt{49}}$ | M1 | Allow $\sqrt{}$ errors, wrong $\sigma$ or $z$, allow 50 |
| | A1 | Correct, including $z=2.576$ or $t_{49}=2.680$, *not* 50 |
| $(8.315,\ 8.869)$ | A1 **3** | In range [8.31, 8.32] and in range [8.86, 8.87], even from 50, or (8.306, 8.878) from $t_{49}$ |7 In 1761, James Short took measurements of the parallax of the sun based on the transit of Venus. The mean and standard deviation of a random sample of 50 of these measurements are 8.592 and 0.7534 respectively, in suitable units.\\
(i) Show that if $X \sim \mathrm {~N} \left( 8.592,0.7534 ^ { 2 } \right)$, then
$$\mathrm { P } ( X \leqslant 8.084 ) = \mathrm { P } ( 8.084 < X \leqslant 8.592 ) = \mathrm { P } ( 8.592 < X \leqslant 9.100 ) = \mathrm { P } ( X > 9.100 ) = 0.25 \text {. }$$
The following table summarises the 50 measurements using these intervals.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Measurement $( x )$ & $x \leqslant 8.084$ & $8.084 < x \leqslant 8.592$ & $8.592 < x \leqslant 9.100$ & $x > 9.100$ \\
\hline
Frequency & 8 & 22 & 11 & 9 \\
\hline
\end{tabular}
\end{center}
(ii) Carry out a test, at the $\frac { 1 } { 2 } \%$ significance level, of whether a normal distribution fits the data.\\
(iii) Obtain a 99\% confidence interval for the mean of all similar parallax measurements.
\hfill \mbox{\textit{OCR S3 2009 Q7 [14]}}