CAIE FP2 2019 June — Question 9 10 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Other continuous
DifficultyStandard +0.3 This is a standard chi-squared goodness of fit test with straightforward calculations. Part (i) requires integrating a given pdf (using substitution u=4-x), which is routine A-level integration. Part (ii) follows the standard chi-squared test procedure: calculate test statistic, find degrees of freedom, compare to critical value. The only minor complication is combining cells to ensure expected frequencies ≥5, but this is a standard textbook technique. Overall, this is slightly easier than average as it's a methodical application of a standard statistical test with no novel problem-solving required.
Spec5.06c Fit other distributions: discrete and continuous
A random sample of 50 observations of the continuous random variable \(X\) was taken and the values are summarised in the following table.
Interval\(0 \leqslant x < 0.8\)\(0.8 \leqslant x < 1.6\)\(1.6 \leqslant x < 2.4\)\(2.4 \leqslant x < 3.2\)\(3.2 \leqslant x < 4\)
Observed frequency1816862
It is required to test the goodness of fit of the distribution with probability density function f given by $$f(x) = \begin{cases} \frac{3}{16}(4 - x)^{\frac{1}{2}} & 0 \leqslant x < 4, \\ 0 & \text{otherwise}. \end{cases}$$ The relevant expected frequencies, correct to 2 decimal places, are given in the following table.
Interval\(0 \leqslant x < 0.8\)\(0.8 \leqslant x < 1.6\)\(1.6 \leqslant x < 2.4\)\(2.4 \leqslant x < 3.2\)\(3.2 \leqslant x < 4\)
Expected frequency14.2212.5410.598.184.47
  1. Show how the expected frequency for \(1.6 \leqslant x < 2.4\) is obtained. [3]
  2. Carry out a goodness of fit test at the 5% significance level. [7]
Question 9:
AnswerMarks
9 (i)E = (3/16) ∫ 2.4 (4 – x)1/2 dx
3 1.6
= (3/16) [ – (2/3) (4 – x)3/2 ] 2.4
AnswerMarks Guidance
1.6M1 State or imply expression for required expected value E of X
3
AnswerMarks Guidance
= (2⋅43/2 – 1⋅63/2) / 8 = 1⋅694/8 or 0⋅2118A1 Find expected value E (may be implied in finding 50 E )
3 3
(M1 A1 requires adequate explicit working)
50 E = 10⋅59 AG
AnswerMarks Guidance
3A1 Hence verify corresponding expected frequency
3
AnswerMarks Guidance
9(ii)H : Distribution fits data (AEF)
0B1 State (at least) null hypothesis in full
O 18 16 8 8
i
E: 14⋅22 12⋅54 10⋅59 12⋅65
i
AnswerMarks Guidance
X2 = 1⋅005 + 0⋅955 + 0⋅633 + 1⋅709 = 4⋅30M1 Combine values consistent with all exp. values ≥ 5
M1 A1Find value of X2 from Σ (E – O)2 / E [or Σ O2/E – n ]
i i i i i
No. n of cells: 5 4 3
χ 2: 9⋅488 7⋅815 5⋅991 (to 3 s.f.)
AnswerMarks Guidance
n-1, 0.95B1 FT on number, n, of cells used to find X2
State or use consistent tabular value χ 2
n-1, 0.95
Accept H if X2 < tabular value (AEF)
AnswerMarks Guidance
0M1 State or imply valid method for conclusion
4⋅30 [± 0⋅01] < 7⋅81[5] so distribution fits [data]
AnswerMarks Guidance
or distribution is a suitable model (AEF)A1 Conclusion (requires both values approx. correct)
7
AnswerMarks Guidance
QuestionAnswer Marks 
Question 9:
--- 9 (i) ---
9 (i) | E = (3/16) ∫ 2.4 (4 – x)1/2 dx
3 1.6
= (3/16) [ – (2/3) (4 – x)3/2 ] 2.4
1.6 | M1 | State or imply expression for required expected value E of X
3
= (2⋅43/2 – 1⋅63/2) / 8 = 1⋅694/8 or 0⋅2118 | A1 | Find expected value E (may be implied in finding 50 E )
3 3
(M1 A1 requires adequate explicit working)
50 E = 10⋅59 AG
3 | A1 | Hence verify corresponding expected frequency
3
--- 9(ii) ---
9(ii) | H : Distribution fits data (AEF)
0 | B1 | State (at least) null hypothesis in full
O 18 16 8 8
i
E: 14⋅22 12⋅54 10⋅59 12⋅65
i
X2 = 1⋅005 + 0⋅955 + 0⋅633 + 1⋅709 = 4⋅30 | M1 | Combine values consistent with all exp. values ≥ 5
M1 A1 | Find value of X2 from Σ (E – O)2 / E [or Σ O2/E – n ]
i i i i i
No. n of cells: 5 4 3
χ 2: 9⋅488 7⋅815 5⋅991 (to 3 s.f.)
n-1, 0.95 | B1 | FT on number, n, of cells used to find X2
State or use consistent tabular value χ 2
n-1, 0.95
Accept H if X2 < tabular value (AEF)
0 | M1 | State or imply valid method for conclusion
4⋅30 [± 0⋅01] < 7⋅81[5] so distribution fits [data]
or distribution is a suitable model (AEF) | A1 | Conclusion (requires both values approx. correct)
7
Question | Answer | Marks | Guidance
A random sample of 50 observations of the continuous random variable $X$ was taken and the values are summarised in the following table.

\begin{tabular}{|l|c|c|c|c|c|}
\hline
Interval & $0 \leqslant x < 0.8$ & $0.8 \leqslant x < 1.6$ & $1.6 \leqslant x < 2.4$ & $2.4 \leqslant x < 3.2$ & $3.2 \leqslant x < 4$ \\
\hline
Observed frequency & 18 & 16 & 8 & 6 & 2 \\
\hline
\end{tabular}

It is required to test the goodness of fit of the distribution with probability density function f given by
$$f(x) = \begin{cases}
\frac{3}{16}(4 - x)^{\frac{1}{2}} & 0 \leqslant x < 4, \\
0 & \text{otherwise}.
\end{cases}$$

The relevant expected frequencies, correct to 2 decimal places, are given in the following table.

\begin{tabular}{|l|c|c|c|c|c|}
\hline
Interval & $0 \leqslant x < 0.8$ & $0.8 \leqslant x < 1.6$ & $1.6 \leqslant x < 2.4$ & $2.4 \leqslant x < 3.2$ & $3.2 \leqslant x < 4$ \\
\hline
Expected frequency & 14.22 & 12.54 & 10.59 & 8.18 & 4.47 \\
\hline
\end{tabular}

(i) Show how the expected frequency for $1.6 \leqslant x < 2.4$ is obtained. [3]

(ii) Carry out a goodness of fit test at the 5% significance level. [7]

\hfill \mbox{\textit{CAIE FP2 2019 Q9 [10]}}
This paper (11 questions)
View full paper
Q1 4 Q2 8 Q3 10 Q4 10 Q5 12 Q6 7 Q7 8 Q8 8 Q9 10 Q10 11 Q11 24