| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| 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 for a binomial distribution with straightforward calculations. Part (a) requires simple weighted mean calculation, part (b) uses binomial probability formula (routine for S3), part (c) is a textbook chi-squared test with cell combination, and part (d) requires basic interpretation. All steps are procedural with no novel insight required, making it slightly easier than average for S3 level. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of tasks failed by an applicant | 0 | 1 | 2 | 3 | 4 | 5 | 6 or more |
| Frequency | 2 | 21 | 45 | 42 | 12 | 3 | 0 |
| Number of tasks failed by an applicant | 0 | 1 | 2 | 3 | 4 | 5 | 6 or more |
| Expected frequency | 7.21 | 24.71 | 37.06 | \(r\) | 17.02 | 5.83 | \(s\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\hat{p} = \frac{0(2)+1(21)+2(45)+3(42)+4(12)+5(3)}{8(2+21+45+42+12+3)}\) or \(8(125)\) \(\left\{=\frac{300}{1000}\right\} = 0.3\) | M1 | Must show clearly how to get either 300 or 1000 |
| Answer is given. See notes. | A1 cso | Showing how to get both 300 and 1000 and reaching \(p=0.3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = 125 \times {}^8C_3(0.3)^3(0.7)^5 \{= 31.76523...\}\) (formula) or \(r = 125 \times (0.8059-0.5518)\{=31.7625\}\) (tables) | M1 | For any correct method (or correct expression) for finding either \(r\) or \(s\) |
| \(s = 125-(7.21+24.71+37.06+\text{their } r+17.02+5.83)\{=1.40477...\text{ or }1.4075\}\) or \(s=125\times(1-0.9887)\{=1.4125\}\) | ||
| \(r = 31.76523\) or \(31.7625\) or \(31.7575\) | A1 | \(r=\) awrt 31.77 or \(r=\) awrt 31.76 |
| \(s = 1.40477\) or \(1.4075\) or \(1.4125\) | A1 | \(s=1.4\) or awrt 1.40 or \(s=\) awrt 1.41 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Pooling 5 failed tasks and \(\geqslant 6\) ONLY | 1st M1 | For an attempt to pool 5 failed tasks and \(\geqslant 6\) failed tasks ONLY. Give 1st M0 for pooling 0 failed tasks and 1 failed task |
| At least 2 correct expressions/values for test statistic (to awrt 2 d.p. or truncated 2 d.p.) | 2nd M1 | |
| \(X^2 = \sum\frac{(O-E)^2}{E}\) or \(\sum\frac{O^2}{E}-125 :=\) awrt 13.3 | dM1 | Dependent on 2nd M1. For applying either \(\sum\frac{(O-E)^2}{E}\) or \(\sum\frac{O^2}{E}-125\) |
| awrt 13.3 | 1st A1 | See notes |
| \(\nu = 6-1-1=4\) | 1st B1 ft | For their evaluated \(n-1-1\), i.e. realising they must subtract 2 from their \(n\) |
| \(\chi^2_4(0.05)=9.488 \Rightarrow\) CR: \(X^2 \geqslant 9.488\) | 2nd B1 | A correct ft for their \(\chi^2_k(0.05)\), where \(k=n-1-1\) from their \(n\) |
| \(H_0\): Binomial distribution is a good (or suitable) model (or fit). \(H_1\): Binomial distribution is not a suitable model. | 3rd B1 | Must have both hypotheses and mention Binomial at least once. Inclusion of 0.3 for \(p\) in hypotheses is B0 but condone in conclusion |
| [In the CR/significant/Reject \(H_0\)] Binomial distribution is not a suitable model. | Final A1 | Dependent on 2nd and 3rd Method marks only. A correct conclusion (context not required) which is based on their \(X^2\)-value and their \(\chi^2\)-critical value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Following from a correct conclusion in part (c), a comment conveying either: \(p\) is not constant or employer's belief is not justified | B1 |
# Question 6:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{p} = \frac{0(2)+1(21)+2(45)+3(42)+4(12)+5(3)}{8(2+21+45+42+12+3)}$ or $8(125)$ $\left\{=\frac{300}{1000}\right\} = 0.3$ | M1 | Must show clearly how to get either 300 or 1000 |
| Answer is given. See notes. | A1 cso | Showing how to get both 300 and 1000 and reaching $p=0.3$ |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = 125 \times {}^8C_3(0.3)^3(0.7)^5 \{= 31.76523...\}$ (formula) or $r = 125 \times (0.8059-0.5518)\{=31.7625\}$ (tables) | M1 | For any correct method (or correct expression) for finding either $r$ or $s$ |
| $s = 125-(7.21+24.71+37.06+\text{their } r+17.02+5.83)\{=1.40477...\text{ or }1.4075\}$ or $s=125\times(1-0.9887)\{=1.4125\}$ | | |
| $r = 31.76523$ or $31.7625$ or $31.7575$ | A1 | $r=$ awrt 31.77 or $r=$ awrt 31.76 |
| $s = 1.40477$ or $1.4075$ or $1.4125$ | A1 | $s=1.4$ or awrt 1.40 or $s=$ awrt 1.41 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Pooling 5 failed tasks and $\geqslant 6$ ONLY | 1st M1 | For an attempt to pool 5 failed tasks and $\geqslant 6$ failed tasks ONLY. Give 1st M0 for pooling 0 failed tasks and 1 failed task |
| At least 2 correct expressions/values for test statistic (to awrt 2 d.p. or truncated 2 d.p.) | 2nd M1 | |
| $X^2 = \sum\frac{(O-E)^2}{E}$ or $\sum\frac{O^2}{E}-125 :=$ awrt 13.3 | dM1 | Dependent on 2nd M1. For applying either $\sum\frac{(O-E)^2}{E}$ or $\sum\frac{O^2}{E}-125$ |
| awrt 13.3 | 1st A1 | See notes |
| $\nu = 6-1-1=4$ | 1st B1 ft | For their evaluated $n-1-1$, i.e. realising they must subtract 2 from their $n$ |
| $\chi^2_4(0.05)=9.488 \Rightarrow$ CR: $X^2 \geqslant 9.488$ | 2nd B1 | A correct ft for their $\chi^2_k(0.05)$, where $k=n-1-1$ from their $n$ |
| $H_0$: Binomial distribution is a good (or suitable) model (or fit). $H_1$: Binomial distribution is not a suitable model. | 3rd B1 | Must have both hypotheses and mention Binomial at least once. Inclusion of 0.3 for $p$ in hypotheses is B0 but condone in conclusion |
| [In the CR/significant/Reject $H_0$] Binomial distribution is not a suitable model. | Final A1 | Dependent on 2nd and 3rd Method marks only. A correct conclusion (context not required) which is based on **their** $X^2$-value and **their** $\chi^2$-critical value |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Following from a correct conclusion in part (c), a comment conveying **either**: $p$ is not constant **or** employer's belief is not justified | B1 | |
---6. Eight tasks were given to each of 125 randomly selected job applicants. The number of tasks failed by each applicant is recorded.
The results are as follows
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Number of tasks failed by an applicant & 0 & 1 & 2 & 3 & 4 & 5 & 6 or more \\
\hline
Frequency & 2 & 21 & 45 & 42 & 12 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that the probability of a randomly selected task, from this sample, being failed is 0.3
An employer believes that a binomial distribution might provide a good model for the number of tasks, out of 8, that an applicant fails.
He uses a binomial distribution, with the estimated probability 0.3 of a task being failed. The calculated expected frequencies are as follows
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Number of tasks failed by an applicant & 0 & 1 & 2 & 3 & 4 & 5 & 6 or more \\
\hline
Expected frequency & 7.21 & 24.71 & 37.06 & $r$ & 17.02 & 5.83 & $s$ \\
\hline
\end{tabular}
\end{center}
\item Find the value of $r$ and the value of $s$ giving your answers to 2 decimal places.
\item Test, at the $5 \%$ level of significance, whether or not a binomial distribution is a suitable model for these data. State your hypotheses and show your working clearly.
The employer believes that all applicants have the same probability of failing each task.
\item Use your result from part(c) to comment on this belief.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q6 [14]}}