| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.8 This is a standard chi-squared goodness of fit test with binomial distribution, requiring calculation of the sample proportion, expected frequencies, appropriate grouping of cells, correct degrees of freedom (accounting for estimated parameter), and hypothesis test execution. While methodical, it involves multiple computational steps and the subtlety of df adjustment, placing it moderately above average difficulty. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of defective mugs | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Number of samples | 11 | 43 | 35 | 8 | 2 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| mean \(= \frac{150}{100}\), \(p = \frac{1.5}{6} = \frac{1}{4}\) | M1 A1 | Find value of \(p\) for binomial distribution |
| \(H_0\): \(B(6, p)\) fits data (A.E.F.) | B1 | State null hypothesis |
| \(17.80\ 35.60\ 29.66\ 13.18\ 3.30\ 0.44\ 0.02\) | M1 A1 | Find expected binomial values (to 1 d.p.) |
| \(O\): \(11\ 43\ 35\ 11\) | Combine last four cells since exp. value \(< 5\) | |
| \(E\): \(17.80\ 35.60\ 29.66\ 16.94\) | *M1 | |
| \(\chi^2 = 2.60 + 1.54 + 0.96 + 2.08\) | M1 *A1 | Calculate \(\chi^2\) (to 2 d.p.; A1 dep *M1) |
| \(= 7.18\) (or \(7.14\) if 1 d.p.) | ||
| \(\chi^2_{2,\ 0.95} = 5.991\) (cells combined) | *B1 | State or use consistent tabular value |
| \([\chi^2_{3,\ 0.95} = 7.815,\ \chi^2_{4,\ 0.95} = 9.488]\) | ||
| \(\chi^2 > 5.99\) so distribution does not fit | B1 | Correct conclusion (A.E.F., dep *A1, *B1) |
## Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| mean $= \frac{150}{100}$, $p = \frac{1.5}{6} = \frac{1}{4}$ | M1 A1 | Find value of $p$ for binomial distribution |
| $H_0$: $B(6, p)$ fits data (A.E.F.) | B1 | State null hypothesis |
| $17.80\ 35.60\ 29.66\ 13.18\ 3.30\ 0.44\ 0.02$ | M1 A1 | Find expected binomial values (to 1 d.p.) |
| $O$: $11\ 43\ 35\ 11$ | | Combine last four cells since exp. value $< 5$ |
| $E$: $17.80\ 35.60\ 29.66\ 16.94$ | *M1 | |
| $\chi^2 = 2.60 + 1.54 + 0.96 + 2.08$ | M1 *A1 | Calculate $\chi^2$ (to 2 d.p.; A1 dep *M1) |
| $= 7.18$ (or $7.14$ if 1 d.p.) | | |
| $\chi^2_{2,\ 0.95} = 5.991$ (cells combined) | *B1 | State or use consistent tabular value |
| $[\chi^2_{3,\ 0.95} = 7.815,\ \chi^2_{4,\ 0.95} = 9.488]$ | | |
| $\chi^2 > 5.99$ so distribution does not fit | B1 | Correct conclusion (A.E.F., dep *A1, *B1) |
---8 A factory produces china mugs. Random samples of size 6 are selected at regular intervals, and the mugs are inspected for defects. During one week, 100 samples are selected and the numbers of defective mugs found are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of defective mugs & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Number of samples & 11 & 43 & 35 & 8 & 2 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
Fit a binomial distribution to the data and carry out a goodness of fit test at the 5\% significance level.
\hfill \mbox{\textit{CAIE FP2 2013 Q8 [10]}}