| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| 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 standard chi-squared goodness of fit test with binomial distribution from S3. Part (a) is simple calculation of sample proportion, parts (b-e) follow routine procedures: calculating expected frequencies using given binomial probabilities, stating hypotheses, explaining degrees of freedom (cells - parameters - 1), and comparing test statistic to critical value. All steps are textbook applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8,9 or 10 | |||
| Frequency | 11 | 21 | 30 | 20 | 12 | 3 | 2 | 1 | 0 |
| 0 | 1 | 2 | 3 | 4 | 5 or more | ||
| Expected frequency | \(r\) | 26.84 | \(s\) | 20.13 | 8.81 | \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = \frac{0 \times 11 + 1 \times 21 + ...}{10 \times (11 + 21 + ...)} \text{ or } \frac{10 \times 100}{1000} = \frac{223}{1000} = 0.223\) (*) | M1, A1cso | (Accept \(\frac{223}{1000}\)) |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = (0.8)^{10} \times 100 = 10.7374\) | M1A1 | awrt 10.74 |
| \(s = \binom{10}{2}(0.8)^8 \times (0.2)^2 \times 100 = 30.198...\) | A1 | awrt 30.2 |
| \(t = 100 - [r + s + 26.84 + 20.13 + 8.81] =\) | A1cao | awrt 3.28 |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Binomial (\([n=10], p = 0.2\)) is a suitable model for these data | B1 | |
| \(H_1\): Binomial (\([n=10], p = 0.2\)) is NOT a suitable model for these data | B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(t < 5\), the last two groups are combined | M1 | |
| and \(\nu = 4 = 5 - 1\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Critical value \(\chi_4^2(5\%) = 9.488\) | B1 | |
| Not significant or do not reject null hypothesis | M1 | |
| The binomial distribution with \(p = 0.2\) is a suitable model for the number of cuttings that do not grow | A1 | (3 marks) |
**Part (a)**
$p = \frac{0 \times 11 + 1 \times 21 + ...}{10 \times (11 + 21 + ...)} \text{ or } \frac{10 \times 100}{1000} = \frac{223}{1000} = 0.223$ (*) | M1, A1cso | (Accept $\frac{223}{1000}$)
| | | (2 marks)
**Part (b)**
$r = (0.8)^{10} \times 100 = 10.7374$ | M1A1 | awrt 10.74
$s = \binom{10}{2}(0.8)^8 \times (0.2)^2 \times 100 = 30.198...$ | A1 | awrt 30.2
$t = 100 - [r + s + 26.84 + 20.13 + 8.81] =$ | A1cao | awrt 3.28
| | | (4 marks)
**Part (c)**
$H_0$: Binomial ($[n=10], p = 0.2$) is a suitable model for these data | B1 |
$H_1$: Binomial ($[n=10], p = 0.2$) is NOT a suitable model for these data | B1 | (2 marks)
**Part (d)**
Since $t < 5$, the last two groups are combined | M1 |
and $\nu = 4 = 5 - 1$ | A1 | (2 marks)
**Part (e)**
Critical value $\chi_4^2(5\%) = 9.488$ | B1 |
Not significant or do not reject null hypothesis | M1 |
The binomial distribution with $p = 0.2$ is a suitable model for the number of cuttings that do not grow | A1 | (3 marks)
**Total: 13 marks**
**Guidance Notes:**
(a) M1 Must show clearly how to get either 223 or 1000. As printed or better.
A1cso for showing how to get both 223 and 1000 and reaching $p = 0.223$
(b) M1 for any correct method (a correct expression) seen for $r$ or $s$.
1st A1 for correct value for $r$ awrt 10.74
2nd A1 for $s =$ awrt 30.2
3rd A1 for $t = 3.28$ only
(c) B1 for each. The value of $p$ must be mentioned at least once. Accept B(10, 0.2)
If hypotheses are correct but with no value of $p$ then score B0B1
Minimum is $X$-B(10, 0.2). If just B(10, 0.2) and not B(10, 0.2) award B1B0
(d) M1 for combining groups (must be stated or implied by a new table with combined cell seen)
A1 for the calculation $4 = 5 - 1$
(e) M1 for a correct statement based on 4.17 and their cv(context not required) (may be implied)
Use of 4.17 as a critical value scores B0M0A0
A1 for a correct interpretation in context and $p = 0.2$ and cuttings mentioned.
---\begin{enumerate}
\item Ten cuttings were taken from each of 100 randomly selected garden plants. The numbers of cuttings that did not grow were recorded.
\end{enumerate}
The results are as follows
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
No. of cuttings \\
which did \\
not grow \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8,9 or 10 \\
\hline
Frequency & 11 & 21 & 30 & 20 & 12 & 3 & 2 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
(a) Show that the probability of a randomly selected cutting, from this sample, not growing is 0.223
A gardener believes that a binomial distribution might provide a good model for the number of cuttings, out of 10 , that do not grow.
He uses a binomial distribution, with the probability 0.2 of a cutting not growing. The calculated expected frequencies are as follows
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
No. of cuttings which did \\
not grow \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Expected frequency & $r$ & 26.84 & $s$ & 20.13 & 8.81 & $t$ \\
\hline
\end{tabular}
\end{center}
(b) Find the values of $r , s$ and $t$.\\
(c) State clearly the hypotheses required to test whether or not this binomial distribution is a suitable model for these data.
The test statistic for the test is 4.17 and the number of degrees of freedom used is 4 .\\
(d) Explain fully why there are 4 degrees of freedom.\\
(e) Stating clearly the critical value used, carry out the test using a $5 \%$ level of significance.
\hfill \mbox{\textit{Edexcel S3 2008 Q6 [13]}}