| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| 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 +0.3 This is a straightforward chi-squared goodness of fit test with clear structure: (i) routine binomial probability calculation using given parameters, (ii) standard chi-squared test procedure with provided expected values. While it's a Further Maths topic, the execution is mechanical with no conceptual challenges or novel problem-solving required—slightly easier than average A-level difficulty. |
| Spec | 5.02c Linear coding: effects on mean and variance5.06b Fit prescribed distribution: chi-squared test |
| Number of faulty chips | 0 | 1 | 2 | 3 | 4 |
| Number of samples | 2 | 12 | 27 | 49 | 10 |
| Number of faulty chips | 0 | 1 | 2 | 3 | 4 |
| Expected value | 2.56 | 15.36 | 34.56 | 34.56 | 12.96 |
| Answer | Marks |
|---|---|
| 9 | Show how value is found (allow M1 if 100 omitted): 100 4 C2 0⋅4 2 × 0⋅6 2 = 34⋅56 A.G. M1 A1 |
| Answer | Marks |
|---|---|
| State valid deduction (dep *A1; allow p ≠ 0⋅6): Prob. of faulty chips ≠ 0⋅6 (A.E.F.) B1 | (2) |
| (8) | [10] |
Question 9:
9 | Show how value is found (allow M1 if 100 omitted): 100 4 C2 0⋅4 2 × 0⋅6 2 = 34⋅56 A.G. M1 A1
State (at least) null hypothesis: H0: B(4, 0⋅6) fits data (A.E.F.) B1
Combine adjacent cells since exp. value < 5: O: 14 27 49 10
E: 17⋅92 34⋅56 34⋅56 12⋅96 *M1
Calculate value of χ2 (to 2 dp ; A1 dep *M1): χ2 = 9⋅22 M1 A1
Compare with consistent tabular value (to 2 dp): χ 3, 0.95 2 = 7⋅815 (cells combined)
χ 4, 0.95 2 = 9⋅488 (not combined) B1
χ2
Valid method for reaching conclusion: Reject H0 if > tabular value M1
Correct conclusion (A.E.F., requires correct values): 9⋅22 > 7⋅81[5] so distn. does not fit *A1
State valid deduction (dep *A1; allow p ≠ 0⋅6): Prob. of faulty chips ≠ 0⋅6 (A.E.F.) B1 | (2)
(8) | [10]It has been found that 60\% of the computer chips produced in a factory are faulty. As part of quality control, 100 samples of 4 chips are selected at random, and each chip is tested. The number of faulty chips in each sample is recorded, with the results given in the following table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Number of faulty chips & 0 & 1 & 2 & 3 & 4 \\
\hline
Number of samples & 2 & 12 & 27 & 49 & 10 \\
\hline
\end{tabular}
The expected values for a binomial distribution with parameters $n = 4$ and $p = 0.6$ are given in the following table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Number of faulty chips & 0 & 1 & 2 & 3 & 4 \\
\hline
Expected value & 2.56 & 15.36 & 34.56 & 34.56 & 12.96 \\
\hline
\end{tabular}
Show how the expected value 34.56 corresponding to 2 faulty chips is obtained. [2]
Carry out a goodness of fit test at the 5\% significance level, and state what can be deduced from the outcome of the test. [8]
\hfill \mbox{\textit{CAIE FP2 2009 Q9 [10]}}