| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Challenging +1.2 This is a standard chi-squared goodness of fit test with binomial distribution requiring estimation of parameter p from data, calculation of expected frequencies, possible pooling of cells, and hypothesis test execution. While it involves multiple computational steps and understanding of degrees of freedom adjustment, it follows a well-established procedure taught directly in Further Maths statistics with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.06b Fit prescribed distribution: chi-squared test |
| Number of perfect glasses | 0 | 1 | 2 | 3 | 4 |
| Number of packs | 1 | 3 | 10 | 17 | 19 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find value of \(p\) for binomial distribution: mean \(= 150/50 = 3\), \(p = \frac{3}{4}\) | M1 A1 | |
| Find expected binomial values (to 2 d.p.): \(0.20\ \ 2.34\ \ 10.55\ \ 21.09\ \ 15.82\) | M1 A1 | |
| Combine adjacent cells since exp. value \(< 5\): \(O\): \(14\ \ \ \ 17\ \ \ \ 19\) | ||
| \(E\): \(13.09\ \ \ \ 21.09\ \ \ \ 15.82\) | *M1 | |
| Calculate value of \(\chi^2\) (to 2 d.p.; A1 dep *M1): \(\chi^2 = 1.50\) | M1 *A1 | |
| State/use consistent tabular value (to 2 d.p.): \(\chi^2_{1,0.9} = 2.706\) (cells combined) | *B1 | [\(\chi^2_{2,0.9} = 4.605\), \(\chi^2_{3,0.9} = 6.251\)] |
| Correct conclusion: \(1.50 < 2.71\) so distribution fits | A1 | (A.E.F., dep *A1, *B1) Total: [9] |
## Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find value of $p$ for binomial distribution: mean $= 150/50 = 3$, $p = \frac{3}{4}$ | M1 A1 | |
| Find expected binomial values (to 2 d.p.): $0.20\ \ 2.34\ \ 10.55\ \ 21.09\ \ 15.82$ | M1 A1 | |
| Combine adjacent cells since exp. value $< 5$: $O$: $14\ \ \ \ 17\ \ \ \ 19$ | | |
| $E$: $13.09\ \ \ \ 21.09\ \ \ \ 15.82$ | *M1 | |
| Calculate value of $\chi^2$ (to 2 d.p.; A1 dep *M1): $\chi^2 = 1.50$ | M1 *A1 | |
| State/use consistent tabular value (to 2 d.p.): $\chi^2_{1,0.9} = 2.706$ (cells combined) | *B1 | [$\chi^2_{2,0.9} = 4.605$, $\chi^2_{3,0.9} = 6.251$] |
| Correct conclusion: $1.50 < 2.71$ so distribution fits | A1 | (A.E.F., dep *A1, *B1) Total: [9] |8 Drinking glasses are sold in packs of 4. The manufacturer conducts a survey to assess the quality of the glasses. The results from a sample of 50 randomly chosen packs are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of perfect glasses & 0 & 1 & 2 & 3 & 4 \\
\hline
Number of packs & 1 & 3 & 10 & 17 & 19 \\
\hline
\end{tabular}
\end{center}
Fit a binomial distribution to the data and carry out a goodness of fit test at the $10 \%$ significance level.
\hfill \mbox{\textit{CAIE FP2 2012 Q8 [9]}}