| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 19 |
| 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 standard chi-squared question with routine calculations (finding expected frequencies from areas, interpreting given test statistics, and stating hypotheses). Part (a) requires calculating areas of annuli and proportions, parts (b-c) are straightforward interpretation, parts (d-e) are standard contingency table setup, and parts (f-g) involve standard degrees of freedom reasoning and critical value comparison. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables5.06b Fit prescribed distribution: chi-squared test |
| Colour of region | Green | Red | Blue | Yellow | Total |
| Boys | 10 | 12 | 10 | 3 | 35 |
| Girls | 12 | 27 | 15 | 11 | 65 |
| Colour of region | Green | Red | Blue | Yellow |
| Boys | 7.7 | 13.65 | 8.75 | 4.9 |
| Girls | 14.3 | 25.35 | 16.25 | 9.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): U[0,10] is a suitable model; \(H_1\): U[0,10] is not a suitable model | B1 | |
| Values of \(D\): correct intervals | B1 | For correct values of \(D\) (can be implied by 40, 30, 20, 10) |
| Expected frequencies: 40, 30, 20, 10 | M1A1 | M1 for at least 2 expected frequencies or clear use of correct formula e.g. \(0.4N\). A1 for all correct \(E_i\) |
| \(\frac{(O_i - E_i)^2}{E_i}\): 8.1, 2.7, 1.25, 1.6 | M1 | For at least 2 correct calculations from 4th or 5th column |
| \(\chi^2 = 13.65\) | A1 | For test statistic of 13.65 (accept 13.7 to 3 sf). Awrt 13.7 only scores 2nd B1M1A1M1A1 |
| \(\nu = 3\), \(\chi^2_3(1\%) = 11.345\) | B1, B1 | |
| Reject \(H_0\); the uniform distribution over [0,10] is not a suitable model | A1 | For correct conclusion rejecting uniform model, provided test statistic \(> 11.345\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(\propto \pi r^2\) so \(r = 81, -49 = \mathbf{32}\) | M1, A1 | M1 for some attempt to use \(\pi r^2\) to find \(r\) |
| \(s = 100 - \text{"32"} - 49\) or \(100 - 81 = \mathbf{19}\) | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Not significant, Henry's model is suitable | M1, A1 | M1 for correct statement that it is not significant. A1 for correctly stating Henry's model is suitable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): The colour/region chosen for the points is independent of gender (or no association) | B1 | Independence or association mentioned at least once. Allow connection but not correlation |
| \(H_1\): The colour/region chosen for the points is dependent on gender (or association) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{39 \times 65}{100}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expected frequency for Yellow and Boys is \(4.9 < 5\) so cells must be pooled/combined. This gives a \(2 \times 3\) table so \(\nu = (2-1)\times(3-1) = 2\) | B1 | For recognising there is an \(E_i < 5\) and need for pooling/combining |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{cv} = 4.605\) | B1 | |
| Not significant, so data do not support Phoebe's belief | B1 | For correctly stating Phoebe's belief is not supported (depends on their cv being \(> 1.411\)) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: U[0,10] is a suitable model; $H_1$: U[0,10] is not a suitable model | B1 | |
| Values of $D$: correct intervals | B1 | For correct values of $D$ (can be implied by 40, 30, 20, 10) |
| Expected frequencies: 40, 30, 20, 10 | M1A1 | M1 for at least 2 expected frequencies or clear use of correct formula e.g. $0.4N$. A1 for all correct $E_i$ |
| $\frac{(O_i - E_i)^2}{E_i}$: 8.1, 2.7, 1.25, 1.6 | M1 | For at least 2 correct calculations from 4th or 5th column |
| $\chi^2 = 13.65$ | A1 | For test statistic of 13.65 (accept 13.7 to 3 sf). Awrt 13.7 only scores 2nd B1M1A1M1A1 |
| $\nu = 3$, $\chi^2_3(1\%) = 11.345$ | B1, B1 | |
| Reject $H_0$; the uniform distribution over [0,10] is not a suitable model | A1 | For correct conclusion rejecting uniform model, provided test statistic $> 11.345$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $\propto \pi r^2$ so $r = 81, -49 = \mathbf{32}$ | M1, A1 | M1 for some attempt to use $\pi r^2$ to find $r$ |
| $s = 100 - \text{"32"} - 49$ or $100 - 81 = \mathbf{19}$ | B1ft | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Not significant, Henry's model is suitable | M1, A1 | M1 for correct statement that it is not significant. A1 for correctly stating Henry's model is suitable |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: The colour/region chosen for the points is independent of gender (or no association) | B1 | Independence or association mentioned at least once. Allow connection but not correlation |
| $H_1$: The colour/region chosen for the points is dependent on gender (or association) | | |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{39 \times 65}{100}$ | B1 | |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expected frequency for Yellow and Boys is $4.9 < 5$ so cells must be pooled/combined. This gives a $2 \times 3$ table so $\nu = (2-1)\times(3-1) = 2$ | B1 | For recognising there is an $E_i < 5$ and need for pooling/combining |
## Part (g):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{cv} = 4.605$ | B1 | |
| Not significant, so data do not support Phoebe's belief | B1 | For correctly stating Phoebe's belief is not supported (depends on their cv being $> 1.411$) |
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6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{740f7555-3a9a-4526-9048-39908aa8f8dd-10_684_694_239_625}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The sketch in Figure 1 represents a target which consists of 4 regions formed from 4 concentric circles of radii $4 \mathrm {~cm} , 7 \mathrm {~cm} , 9 \mathrm {~cm}$ and 10 cm . The regions are coloured as labelled in Figure 1.\\
A random sample of 100 children each choose a point on the target and their results are summarised in the table below.
(b) Find the value of $r$ and the value of $s$.
Henry obtained a test statistic of 6.188 and no groups were pooled.\\
(c) State what conclusion Henry should make about his claim.
Phoebe believes that the children chose the region of the target according to colour. She believes that boys and girls would favour different colours and splits the original data by gender to obtain the following table.
\section*{Observed frequencies}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Colour of region & Green & Red & Blue & Yellow & Total \\
\hline
Boys & 10 & 12 & 10 & 3 & 35 \\
\hline
Girls & 12 & 27 & 15 & 11 & 65 \\
\hline
\end{tabular}
\end{center}
(d) State suitable hypotheses to test Phoebe's belief.
Phoebe calculated the following expected frequencies to carry out a suitable test.
\section*{Expected frequencies}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Colour of region & Green & Red & Blue & Yellow \\
\hline
Boys & 7.7 & 13.65 & 8.75 & 4.9 \\
\hline
Girls & 14.3 & 25.35 & 16.25 & 9.1 \\
\hline
\end{tabular}
\end{center}
(e) Show how the value of 25.35 was obtained.
Phoebe carried out the test using 2 degrees of freedom and a $10 \%$ level of significance. She obtained a test statistic of 1.411\\
(f) Explain clearly why Phoebe used 2 degrees of freedom.\\
(g) Stating your critical value clearly, determine whether or not these data support Phoebe's belief.
\hfill \mbox{\textit{Edexcel S3 2015 Q6 [19]}}