| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Degrees of freedom determination |
| Difficulty | Moderate -0.3 This is a straightforward chi-squared goodness of fit test with minimal complexity. Part (a) is trivial arithmetic (4×43 - sum), part (b) tests basic understanding of degrees of freedom (k-1), and part (c) is a standard hypothesis test requiring only calculation of the test statistic and comparison to a table value. All steps are routine applications of the chi-squared test procedure with no problem-solving or novel insight required. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number | 1 | 2 | 3 | 4 |
| Frequency | 47 | 34 | 36 | \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 4 \times 43 - 47 - 34 - 36 = 55\) | B1* | Using uniform model: \(x = \frac{43 - 0.25 \times (47 + 34 + 36)}{0.25} = 55\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\nu = 4 - 1 = 3\) since the only constraint is that the totals agree | B1 | \(4 - 1 = 3\) (may be in words) and explanation of what the constraint is |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): The die is unbiased; \(H_1\): The die is biased | B1 | Both hypotheses correct, e.g. "The data fits a discrete uniform distribution" |
| Test Statistic \(= \frac{(47-43)^2}{43} + \frac{(34-43)^2}{43} + \frac{(36-43)^2}{43} + \frac{(55-43)^2}{43}\) | M1 | Attempting \(\sum \frac{(O-E)^2}{E}\) or \(\sum \frac{O^2}{E} - N\); may be implied by awrt 6.74 or \(p\) value 0.0805 |
| \(= 6.744\ldots\) | A1 | awrt 6.74 or \(\frac{290}{43}\) oe; may be implied by \(p\) value 0.0805 |
| \(\chi^2_{(3, 0.05)} = 7.815\) | B1 | awrt 7.82 (Calc 7.8147…) |
| Not in critical region since \(7.815 > 6.74\ldots\), therefore insufficient evidence to reject \(H_0\); consistent with the die being unbiased | A1 | Drawing correct inference in context; need the word "die" or "tetrahedral" |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4 \times 43 - 47 - 34 - 36 = 55$ | B1* | Using uniform model: $x = \frac{43 - 0.25 \times (47 + 34 + 36)}{0.25} = 55$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\nu = 4 - 1 = 3$ since the only constraint is that the totals agree | B1 | $4 - 1 = 3$ (may be in words) and explanation of what the constraint is |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: The die is unbiased; $H_1$: The die is biased | B1 | Both hypotheses correct, e.g. "The data fits a discrete uniform distribution" |
| Test Statistic $= \frac{(47-43)^2}{43} + \frac{(34-43)^2}{43} + \frac{(36-43)^2}{43} + \frac{(55-43)^2}{43}$ | M1 | Attempting $\sum \frac{(O-E)^2}{E}$ or $\sum \frac{O^2}{E} - N$; may be implied by awrt 6.74 or $p$ value 0.0805 |
| $= 6.744\ldots$ | A1 | awrt 6.74 or $\frac{290}{43}$ oe; may be implied by $p$ value 0.0805 |
| $\chi^2_{(3, 0.05)} = 7.815$ | B1 | awrt 7.82 (Calc 7.8147…) |
| Not in critical region since $7.815 > 6.74\ldots$, therefore insufficient evidence to reject $H_0$; consistent with the **die** being unbiased | A1 | Drawing correct inference in context; need the word "die" or "tetrahedral" |
---\begin{enumerate}
\item Kelly throws a tetrahedral die $n$ times and records the number on which it lands for each throw.
\end{enumerate}
She calculates the expected frequency for each number to be 43 if the die was unbiased.\\
The table below shows three of the frequencies Kelly records but the fourth one is missing.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number & 1 & 2 & 3 & 4 \\
\hline
Frequency & 47 & 34 & 36 & $x$ \\
\hline
\end{tabular}
\end{center}
(a) Show that $x = 55$
Kelly wishes to test, at the $5 \%$ level of significance, whether or not there is evidence that the tetrahedral die is unbiased.\\
(b) Explain why there are 3 degrees of freedom for this test.\\
(c) Stating your hypotheses clearly and the critical value used, carry out the test.
\hfill \mbox{\textit{Edexcel FS1 2021 Q1 [7]}}