| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with standard components: (i) finding mean/variance of an exponential distribution using standard formulas, (ii) verifying an expected frequency calculation using the given pdf, and (iii) performing a routine hypothesis test. All steps are mechanical applications of learned techniques with no novel problem-solving required, making it slightly easier than average for Further Maths Statistics. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Range | \(0 \leq x < 1\) | \(1 \leq x < 2\) | \(2 \leq x < 3\) | \(3 \leq x < 4\) | \(x \geq 4\) |
| Observed | 24 | 22 | 10 | 3 | 1 |
| Expected | 33.040 | 14.846 | 6.671 | 2.997 | 2.446 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (i) | (cid:102) |
| Answer | Marks |
|---|---|
| (cid:32)1.5625 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Attempt (cid:179)xf(cid:11)x(cid:12) dx |
| Answer | Marks |
|---|---|
| 16 | BC |
| Answer | Marks |
|---|---|
| (ii) | 2 |
| Answer | Marks |
|---|---|
| frequency is 0.247432(cid:117)60(cid:32)14.846 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| i2.2a | e |
| Answer | Marks |
|---|---|
| correctly obtain given answer AG | Requires clear use of notation |
| Answer | Marks |
|---|---|
| (iii) | H : data consistent with distribution |
| Answer | Marks |
|---|---|
| consistent with distribution | c |
| Answer | Marks |
|---|---|
| [7] | 2.5 |
| Answer | Marks |
|---|---|
| 2.2b | Or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| FT on numerical errors only | BC | |
| O | E | (cid:11)O(cid:16)E(cid:12)2 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(i) | 4 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(ii) | 2 | 20 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(iii) | 4 | 20 |
Question 8:
8 | (i) | (cid:102)
(cid:80)(cid:32)(cid:179) 0.8xe(cid:16)0.8xdx(cid:32)1.25
0
E (cid:11) X2(cid:12) (cid:32)(cid:179) (cid:102) 0.8x2e(cid:16)0.8xdx(cid:62)(cid:32)3.125(cid:64)
0
Var(cid:11)X(cid:12)(cid:32)3.125(cid:16)1.252
(cid:32)1.5625 | M1
A1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | Attempt (cid:179)xf(cid:11)x(cid:12) dx
Obtain 1.25 or exact equivalent
Attempt (cid:179)x2f(cid:11)x(cid:12) dx(cid:16)(cid:80)2
n
25
Obtain or exact equivalent
16 | BC
or awrt 1.56
BC
(ii) | 2
P(cid:11)1(cid:100)x(cid:31)2(cid:12)(cid:32)(cid:179) 0.8e(cid:16)0.8xdx
1
(cid:32)0.247432 (6 s.f.)
There are 60 specimens, so the expected
frequency is 0.247432(cid:117)60(cid:32)14.846 | M1
E1
A1
E1
[4] | 1.1
2.1
1.1
i2.2a | e
mCorrect pdf
Integrate between 1 and 2
Correct answer, allow 3 s.f.
Multiply probability by 60 and
correctly obtain given answer AG | Requires clear use of notation
BC
(iii) | H : data consistent with distribution
0
H : data not consistent
1 p
Combine cells to get
O E (cid:11)O(cid:16)E(cid:12)S 2 /E
24 33.040 2.4734
22 14.846 3.4474
10 6.6707 1.6613
4 5.4431 0.3826
(cid:11)O(cid:16)E(cid:12)2
(cid:166) (cid:32)7.965
E
(cid:70)2(cid:11)0.95(cid:12)(cid:32)7.815 and 7.965(cid:33)7.815
3
Reject H . Evidence that the data is not
0
consistent with distribution | c
e
B1
M1
M1
A1
A1
B1
A1FT
[7] | 2.5
1.1a
1.1
1.1
3.4
1.1
2.2b | Or equivalent
Combine last two cells
(cid:11)O(cid:16)E(cid:12)2
Calculate for at least one cell
E
(cid:11)O(cid:16)E(cid:12)2
At least two values correct
E
(cid:70)2 in range [7.96, 7.97]
Comparison with 7.815
State not consistent with distribution
FT on numerical errors only | BC
O | E | (cid:11)O(cid:16)E(cid:12)2
/E
--- 8(i) ---
8(i) | 4 | 0 | 0 | 0 | 4
--- 8(ii) ---
8(ii) | 2 | 20 | 04 | i
--- 8(iii) ---
8(iii) | 4 | 20 | 1 | 78 A continuous random variable $X$ has probability density function given by
$$\mathrm { f } ( x ) = \left\{ \begin{array} { c c }
0.8 \mathrm { e } ^ { - 0.8 x } & x \geq 0 \\
0 & x < 0
\end{array} \right.$$
(i) Find the mean and variance of $X$.
The lifetime of a certain organism is thought to have the same distribution as $X$. The lifetimes in days of a random sample of 60 specimens of the organism were found. The observed frequencies, together with the expected frequencies correct to 3 decimal places, are given in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Range & $0 \leq x < 1$ & $1 \leq x < 2$ & $2 \leq x < 3$ & $3 \leq x < 4$ & $x \geq 4$ \\
\hline
Observed & 24 & 22 & 10 & 3 & 1 \\
\hline
Expected & 33.040 & 14.846 & 6.671 & 2.997 & 2.446 \\
\hline
\end{tabular}
\end{center}
(ii) Show how the expected frequency for $1 \leq x < 2$ is obtained.\\
(iii) Carry out a goodness of fit test at the $5 \%$ significance level.
\hfill \mbox{\textit{OCR Further Statistics Q8 [15]}}