| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Normal |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with normal distribution. Part (a) requires routine calculation of a normal probability using given parameters (z-scores and tables). Part (b) follows the standard chi-squared test procedure: combine cells for expected frequencies <5, calculate test statistic, find critical value, and conclude. The question is slightly easier than average because expected frequencies are provided, eliminating potential calculation errors, and the procedure is entirely algorithmic with no novel insight required. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| Height, \(x \mathrm {~cm}\) | \(x < 100\) | \(100 \leqslant x < 110\) | \(110 \leqslant x < 120\) | \(120 \leqslant x < 130\) | \(130 \leqslant x < 140\) | \(x \geqslant 140\) |
| Observed frequency | 0 | 3 | 15 | 23 | 11 | 8 |
| Expected frequency | 1.12 | 5.22 | 13.97 | 19.38 | 13.97 | 6.34 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(130 \leq x < 140) = P\left(\frac{130-125}{12} \leq Z < \frac{140-125}{12}\right)\) | M1 | Must see 0.4167 and 1.25, may be implied by 0.8944 and 0.6616. Accept 0.417. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.8944 - 0.6616 = 0.2328\); Multiply by 60 leads to 13.97 | A1 | AG. Accept 0.6615. Need to see 0.2328 or 0.2329 or 13.968 or \((0.8944-0.6616)\times 60\) |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | 3 | 15 |
| 1.12 | 5.22 | 13.97 |
| M1 | Combine first two values. | |
| Test stat \(= 1.7588 + 0.0761 + 0.6743 + 0.6309 + 0.4352\) | M1 | |
| \(3.58\) | A1 | Accept \(3.57 - 3.58\). SC: 3.88, if values not combined, scores M1A0. |
| \(H_0\): distribution fits data; \(N(125, 12^2)\) is a good model for the data | B1 | Must mention distribution and data. |
| 4 degrees of freedom, so tabular value \(= 9.488\); \(3.58 < 9.488\); Accept \(H_0\) | M1 | Compare their value with 9.488 (9.49) and correct FT conclusion. Or, if values not combined, compare their value with 11.07 and correct FT conclusion for M1A0. |
| There is sufficient evidence that the normal distribution fits the data / There is sufficient evidence that the claim is supported by the data | A1 | Correct conclusion, in context, following correct work. Level of uncertainty in language is used. |
## Question 2(a):
$P(130 \leq x < 140) = P\left(\frac{130-125}{12} \leq Z < \frac{140-125}{12}\right)$ | M1 | Must see 0.4167 and 1.25, may be implied by 0.8944 and 0.6616. Accept 0.417.
$P(0.4167 < Z < 1.25)$
$0.8944 - 0.6616 = 0.2328$; Multiply by 60 leads to 13.97 | A1 | AG. Accept 0.6615. Need to see 0.2328 or 0.2329 or 13.968 or $(0.8944-0.6616)\times 60$
---
## Question 2(b):
| 0 | 3 | 15 | 23 | 11 | 8 |
| 1.12 | 5.22 | 13.97 | 19.38 | 13.97 | 6.34 |
| M1 | Combine first two values.
Test stat $= 1.7588 + 0.0761 + 0.6743 + 0.6309 + 0.4352$ | M1 |
$3.58$ | A1 | Accept $3.57 - 3.58$. SC: 3.88, if values not combined, scores M1A0.
$H_0$: distribution fits data; $N(125, 12^2)$ is a good model for the data | B1 | Must mention distribution and data.
4 degrees of freedom, so tabular value $= 9.488$; $3.58 < 9.488$; Accept $H_0$ | M1 | Compare their value with 9.488 (9.49) and correct FT conclusion. Or, if values not combined, compare their value with 11.07 and correct FT conclusion for M1A0.
There is sufficient evidence that the normal distribution fits the data / There is sufficient evidence that the claim is supported by the data | A1 | Correct conclusion, in context, following correct work. Level of uncertainty in language is used.
---2 It is claimed that the heights of a particular age group of boys follow a normal distribution with mean 125 cm and standard deviation 12 cm . Observations for a randomly chosen group of 60 boys in this age group are summarised in the following table. The table also gives the expected frequencies, correct to 2 decimal places, based on the normal distribution with mean 125 cm and standard deviation 12 cm .
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
Height, $x \mathrm {~cm}$ & $x < 100$ & $100 \leqslant x < 110$ & $110 \leqslant x < 120$ & $120 \leqslant x < 130$ & $130 \leqslant x < 140$ & $x \geqslant 140$ \\
\hline
Observed frequency & 0 & 3 & 15 & 23 & 11 & 8 \\
\hline
Expected frequency & 1.12 & 5.22 & 13.97 & 19.38 & 13.97 & 6.34 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show how the expected frequency for $130 \leqslant x < 140$ is obtained.
\item Carry out a goodness of fit test, at the $5 \%$ significance level, to determine whether the claim is supported by the data.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q2 [8]}}