| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | State meaning of Type I error |
| Difficulty | Easy -1.2 This is a straightforward recall question worth 1 mark asking for the definition of Type I error in context. It requires only stating that a Type I error means rejecting the null hypothesis when it's actually true, applied to the specific scenario. No calculation or problem-solving is involved—just recalling a standard statistical definition. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between country and air quality; \(H_1\): There is an association between country and air quality | B1 (2.5) | States both hypotheses using correct language. Variables need to be stated in at least the null hypothesis |
| Expected values: \(A_1 = 95.504\), \(A_2 = 92.496\), \(B_1 = 158.496\), \(B_2 = 153.504\) | M1 (3.3) | Translate situation into expected contingency table for \(\chi^2\) model |
| \(\sum\frac{(\ | O-E\ | -0.5)^2}{E} = \frac{(\ |
| \(= 2.1849\) | A1 (1.1b) | AWRT 2.2 |
| \(\chi^2\) cv for 1 df \(= 2.706\) | B1 (1.1b) | AWRT 2.7, or corresponding probability of test statistic AWRT 0.14 |
| \(2.1849 < 2.706\) | R1 (3.5a) | Evaluates \(\chi^2\) test statistic by correctly comparing with critical value or probability with 0.1 |
| Accept \(H_0\) | E1F (2.2b) | FT their calculated value of \(\sum\frac{(\ |
| No evidence to suggest/support that there is an association between country and air quality | E1F (3.2a) | Conclusion must not be definite. FT their incorrect rejection of \(H_0\) provided consistent with their comparison |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| To conclude that there is an association between country and air quality when there is not | E1 (3.2a) | Interprets Type I error in context |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between country and air quality; $H_1$: There is an association between country and air quality | B1 (2.5) | States both hypotheses using correct language. Variables need to be stated in at least the null hypothesis |
| Expected values: $A_1 = 95.504$, $A_2 = 92.496$, $B_1 = 158.496$, $B_2 = 153.504$ | M1 (3.3) | Translate situation into expected contingency table for $\chi^2$ model |
| $\sum\frac{(\|O-E\|-0.5)^2}{E} = \frac{(\|87-95.504\|-0.5)^2}{95.504} + \frac{(\|167-158.496\|-0.5)^2}{158.496} + \frac{(\|101-92.496\|-0.5)^2}{92.496} + \frac{(\|145-153.504\|-0.5)^2}{153.504}$ | M1 (3.4) | Uses $\chi^2$ model to calculate test statistic. Condone missing modulus sign. Condone use of $\sum\frac{(O-E)^2}{E}$ |
| $= 2.1849$ | A1 (1.1b) | AWRT 2.2 |
| $\chi^2$ cv for 1 df $= 2.706$ | B1 (1.1b) | AWRT 2.7, or corresponding probability of test statistic AWRT 0.14 |
| $2.1849 < 2.706$ | R1 (3.5a) | Evaluates $\chi^2$ test statistic by correctly comparing with critical value or probability with 0.1 |
| Accept $H_0$ | E1F (2.2b) | FT their calculated value of $\sum\frac{(\|O-E\|-0.5)^2}{E}$ |
| No evidence to suggest/support that there is an association between country and air quality | E1F (3.2a) | Conclusion must not be definite. FT their incorrect rejection of $H_0$ provided consistent with their comparison |
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## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| To conclude that there is an association between country and air quality when there is not | E1 (3.2a) | Interprets Type I error in context |
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7
\begin{enumerate}[label=(\alph*)]
\item Test the scientist's claim, using the 10\% level of significance.\\
7
\item For the context of the test carried out in part (a), state the meaning of a Type I error. [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2022 Q7 [9]}}