| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 12 |
| 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 a fully specified binomial distribution. Students must calculate binomial probabilities (routine), combine cells appropriately (standard procedure), compute the test statistic, and compare to critical values. While it requires multiple steps, each is algorithmic with no novel insight needed. Slightly easier than average due to some expected frequencies being pre-calculated. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number Correct | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Number of Couples | 26 | 56 | 28 | 8 | 2 | 0 | 0 |
| Number Correct | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Expected Frequency | 9.83 | 1.84 | 0.18 | 0.01 |
| Answer | Marks |
|---|---|
| \(\times 120\) to give exp. freqs. 31.46, 47.19, 29.49 | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| combining groups \(\geq 3\) | B1 M1 | |
| \(O\) | \(E\) | \((O-E)\) |
| 26 | 31.46 | \(\neg 5.46\) |
| 56 | 47.19 | 8.81 |
| 28 | 29.49 | \(\neg 1.49\) |
| 10 | 11.86 | \(\neg 1.86\) |
| \(\therefore \sum \frac{(O-E)^2}{E} = 2.959\) | M1 A2 |
| Answer | Marks |
|---|---|
| \(B(6, \frac{1}{5})\) is a suitable model | M1 A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore\) suggests the group are not telepathic | B1 | (12) |
(a) $P(0) = (\frac{4}{5})^0 = 0.2621$
$P(1) = 6(\frac{1}{5})(\frac{4}{5})^5 = 0.3932$
$P(2) = \frac{6 \times 5}{2}(\frac{1}{5})^2(\frac{4}{5})^4 = 0.2458$
$\times 120$ to give exp. freqs. 31.46, 47.19, 29.49 | M1 A2 |
(b) $H_0: B(6, \frac{1}{5})$ is a suitable model
$H_1: B(6, \frac{1}{5})$ is not a suitable model
combining groups $\geq 3$ | B1 M1 |
| $O$ | $E$ | $(O-E)$ | $\frac{(O-E)^2}{E}$ |
|---|---|---|---|
| 26 | 31.46 | $\neg 5.46$ | 0.9476 |
| 56 | 47.19 | 8.81 | 1.6448 |
| 28 | 29.49 | $\neg 1.49$ | 0.0753 |
| 10 | 11.86 | $\neg 1.86$ | 0.2917 |
$\therefore \sum \frac{(O-E)^2}{E} = 2.959$ | M1 A2 |
$v = 4 - 1 = 3$, $\chi^2_{\text{crit}}(5\%) = 7.815$
$2.959 < 7.815 \therefore$ do not reject $H_0$
$B(6, \frac{1}{5})$ is a suitable model | M1 A1 A1 |
(c) $B(6, \frac{1}{5})$ is the dist. expected with guessing
$\therefore$ suggests the group are not telepathic | B1 | (12)
---4. A paranormal investigator invites couples who believe they have a telepathic connection to participate in a trial. With each couple one person looks at a card with one of five shapes on it and the other person says which of the shapes they think it is. This is repeated six times and the number of correct answers recorded. The results from 120 couples are given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Number Correct & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Number of Couples & 26 & 56 & 28 & 8 & 2 & 0 & 0 \\
\hline
\end{tabular}
\end{center}
The investigator wishes to see if this data fits a binomial distribution with parameters $n = 6$ and $p = \frac { 1 } { 5 }$ and calculates to 2 decimal places the expected frequencies given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Number Correct & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Expected Frequency & & & & 9.83 & 1.84 & 0.18 & 0.01 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the other expected frequencies.
\item Stating your hypotheses clearly, test at the $5 \%$ level of significance whether or not the distribution is an appropriate model.
\item Comment on your findings.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q4 [12]}}