| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with a binomial distribution (n=3, p=1/6). Students must calculate expected frequencies, compute the test statistic, and compare to critical values. While it requires multiple steps, it's a standard textbook application with no conceptual challenges or novel insights—slightly easier than average due to its routine nature. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of sixes | 0 | 1 | 2 | 3 |
| Frequency | 625 | 384 | 81 | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Let the random variable \(X\) be the number of 6s thrown from 3 dice. If the dice are unbiased then \(X \sim B(3, \frac{1}{6})\) | B1 | si (implied by at least 3 correct expected frequencies) or equivalent |
| \(H_0\): The data can be modelled by the Binomial distribution \(B(3, \frac{1}{6})\). | B1 | |
| \(H_1\): The data cannot be modelled by the Binomial distribution \(B(3, \frac{1}{6})\). | ||
| Number of sixes | 0 | 1 |
| Observed | 625 | 384 |
| Expected | 636.574 | 381.944 |
| Use of \(\chi^2\) stat \(= \sum\frac{(O-E)^2}{E}\) or \(\sum\frac{O^2}{E} - N\) | M1 | Must see at least 2 terms added |
| \(= \frac{(625 - 636.574)^2}{636.574} + \frac{(384 - 381.944)^2}{381.944} + \frac{(81 - 76.389)^2}{76.389} + \frac{(10 - 5.093)^2}{5.093}\) | m1 | \(\frac{625^2}{636.574} + \frac{384^2}{381.944} + \frac{81^2}{76.389} + \frac{10^2}{5.093} - 1100\) |
| \(= 5.23\) | A1 | Accept anything which rounds to 5.2 |
| DF = 3 5% CV = 7.815 | B1 B1 | Accept other test levels. 1% CV = 11.345 10% CV = 6.251 |
| Since 5.23 < 7.815 we cannot reject \(H_0\). There is insufficient evidence at the 5% level to conclude that the set of dice are not fair. | B1 E1 | FT their \(\chi^2\) Only award E1 if all five previous B1 awarded E0 for categorical statements |
| Total [11] |
| Let the random variable $X$ be the number of 6s thrown from 3 dice. If the dice are unbiased then $X \sim B(3, \frac{1}{6})$ | B1 | si (implied by at least 3 correct expected frequencies) or equivalent
| $H_0$: The data can be modelled by the Binomial distribution $B(3, \frac{1}{6})$. | B1 |
| $H_1$: The data cannot be modelled by the Binomial distribution $B(3, \frac{1}{6})$. | |
| | Number of sixes | 0 | 1 | 2 | 3 |
| | --- | --- | --- | --- | --- |
| | Observed | 625 | 384 | 81 | 10 |
| | Expected | 636.574 | 381.944 | 76.389 | 5.093 | M1 A1 | At least one correct. All correct.
| Use of $\chi^2$ stat $= \sum\frac{(O-E)^2}{E}$ or $\sum\frac{O^2}{E} - N$ | M1 | Must see at least 2 terms added
| $= \frac{(625 - 636.574)^2}{636.574} + \frac{(384 - 381.944)^2}{381.944} + \frac{(81 - 76.389)^2}{76.389} + \frac{(10 - 5.093)^2}{5.093}$ | m1 | $\frac{625^2}{636.574} + \frac{384^2}{381.944} + \frac{81^2}{76.389} + \frac{10^2}{5.093} - 1100$
| $= 5.23$ | A1 | Accept anything which rounds to 5.2
| DF = 3 5% CV = 7.815 | B1 B1 | Accept other test levels. 1% CV = 11.345 10% CV = 6.251
| Since 5.23 < 7.815 we cannot reject $H_0$. There is insufficient evidence at the 5% level to conclude that the set of dice are not fair. | B1 E1 | FT their $\chi^2$ Only award E1 if all five previous B1 awarded E0 for categorical statements
| **Total [11]** |
---5. John has a game that involves throwing a set of three identical, cubical dice with faces numbered 1 to 6 . He wishes to investigate whether these dice are fair in terms of the number of sixes obtained when they are thrown. John throws the set of three dice 1100 times and records the number of sixes obtained for each throw. The results are shown in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number of sixes & 0 & 1 & 2 & 3 \\
\hline
Frequency & 625 & 384 & 81 & 10 \\
\hline
\end{tabular}
\end{center}
Using these results, conduct a goodness of fit test and draw an appropriate conclusion.\\
\hfill \mbox{\textit{WJEC Further Unit 2 2022 Q5 [11]}}