| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Uniform |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with uniform distribution. Part (a) requires simple proportion calculations (e.g., 45/360 × 100), and part (b) is a standard hypothesis test following a routine procedure. The uniform distribution makes expected frequencies trivial to compute, and the test execution is mechanical with no conceptual challenges beyond S3 syllabus basics. |
| Spec | 5.06a Chi-squared: contingency tables5.06b Fit prescribed distribution: chi-squared test |
| Angle (degrees) | \(0 - 45\) | \(45 - 90\) | \(90 - 180\) | \(180 - 315\) | \(315 - 360\) |
| Frequency | 18 | 16 | 18 | 29 | 19 |
| Expected frequency | 12.5 | \(a\) | \(b\) | \(c\) | 12.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{45}{360} \times 100 = 12.5\) | M1 | For one correct calculation |
| \(b = \frac{90}{360} \times 100 = 25\) | ||
| \(c = \frac{135}{360} \times 100 = 37.5\) | A1 | 12.5, 25, 37.5 only |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Continuous uniform distribution is a suitable model | B1 | Both; accept in terms of Linda's claim; accept \(U[0,360)\) |
| \(H_1\): Continuous uniform distribution is not a suitable model | ||
| \(O\) | \(E\) | \(\frac{(O-E)^2}{E}\) |
| 18 | 12.5 | 2.42 |
| 16 | 12.5 | 0.98 |
| 18 | 25 | 1.96 |
| 29 | 37.5 | 1.927 |
| 19 | 12.5 | 3.38 |
| 100 | 100 | 10.67 |
| M1 for attempting \(\frac{(O-E)^2}{E}\) or \(\frac{O^2}{E}\) with at least 3 correct expressions or values | M1A1 | A1 for all values correct to 2dp or as fractions |
| \(\sum \frac{(O-E)^2}{E} = 10.67\) or \(\sum \frac{O^2}{E} - 100 = 10.67\) | A1 | awrt 10.7 |
| \(\nu = 5-1=4,\ \chi^2_4(5\%) = 9.488\) | B1,B1ft | 4 can be implied by awrt 9.488 seen or \(p =\) awrt 0.03 |
| \((10.7 > 9.488)\) Reject \(H_0\) | M1 | Correct for their test stat and cv or their \(p\) to 5%; can be implied by correct conclusion from their test stat and cv |
| Linda's claim is not supported | A1ft | Correct comment suggesting continuous uniform model is not suitable or Linda's claim is not correct; Linda's claim must be described in full if Linda is not mentioned; follow through from their test stat and cv, but hypotheses must be correct |
## Question 2:
### Part (a):
$a = \frac{45}{360} \times 100 = 12.5$ | M1 | For one correct calculation
$b = \frac{90}{360} \times 100 = 25$ | |
$c = \frac{135}{360} \times 100 = 37.5$ | A1 | 12.5, 25, 37.5 only
### Part (b):
$H_0$: Continuous uniform distribution is a suitable model | B1 | Both; accept in terms of Linda's claim; accept $U[0,360)$
$H_1$: Continuous uniform distribution is not a suitable model | |
| $O$ | $E$ | $\frac{(O-E)^2}{E}$ | $\frac{O^2}{E}$ |
|-----|-----|-----|-----|
| 18 | 12.5 | 2.42 | 25.92 |
| 16 | 12.5 | 0.98 | 20.48 |
| 18 | 25 | 1.96 | 12.96 |
| 29 | 37.5 | 1.927 | 22.43 |
| 19 | 12.5 | 3.38 | 28.88 |
| 100 | 100 | 10.67 | 110.67 |
M1 for attempting $\frac{(O-E)^2}{E}$ or $\frac{O^2}{E}$ with at least 3 correct expressions or values | M1A1 | A1 for all values correct to 2dp or as fractions
$\sum \frac{(O-E)^2}{E} = 10.67$ or $\sum \frac{O^2}{E} - 100 = 10.67$ | A1 | awrt 10.7
$\nu = 5-1=4,\ \chi^2_4(5\%) = 9.488$ | B1,B1ft | 4 can be implied by awrt 9.488 seen or $p =$ awrt 0.03
$(10.7 > 9.488)$ Reject $H_0$ | M1 | Correct for their test stat and cv or their $p$ to 5%; can be implied by correct conclusion from their test stat and cv
Linda's claim is not supported | A1ft | Correct comment suggesting continuous uniform model is not suitable or Linda's claim is not correct; Linda's claim must be described in full if Linda is not mentioned; follow through from their test stat and cv, but **hypotheses must be correct**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{585de4b0-906e-40c4-9045-966d68505eff-04_430_438_260_753}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The pointer shown in Figure 1 is spun so that it comes to rest between 0 and 360 degrees.\\
Linda claims that it is equally likely to come to rest at any point between 0 and 360 degrees. She spins the pointer 100 times and her results are summarised in the table below. She calculates expected frequencies for some of the possible outcomes and these are also given in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Angle (degrees) & $0 - 45$ & $45 - 90$ & $90 - 180$ & $180 - 315$ & $315 - 360$ \\
\hline
Frequency & 18 & 16 & 18 & 29 & 19 \\
\hline
Expected frequency & 12.5 & $a$ & $b$ & $c$ & 12.5 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the values of the missing expected frequencies $a , b$ and $c$.
\item Stating your hypotheses clearly and using a $5 \%$ level of significance, test whether or not Linda's claim is supported by these data.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q2 [10]}}