| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| 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 with a binomial distribution. Part (a) is routine mean calculation, part (b) requires finding missing expected frequencies (straightforward using binomial probabilities or the constraint that frequencies sum to 100), and part (c) is a textbook application of the chi-squared test. The question follows a familiar S3 template with no novel insights required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.06b Fit prescribed distribution: chi-squared test |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |||
| 6 | 16 | 20 | 23 | 17 | 10 | 8 |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |||
| 1.87 | 10.54 | 24.82 | \(a\) | 22.01 | 8.29 | \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Mean} = \frac{1\times16 + 2\times20 + \ldots + 6\times8}{100} = 2.91\) | M1A1 | M1 for at least 2 correct terms in numerator and 100 in denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = \frac{2.91}{6} = 0.485\) | B1 | 0.485 implied by at least 1 correct answer |
| \(a = 100 \times C^6_3 \times 0.485^3 \times 0.515^3 = 31.17\) | M1A1 | Clear use of Binomial and \(\times 100\) required for method |
| \(b = 100 \times 0.485^6 = 1.3(0)\) | A1 | Accept awrt 2dp for final answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Binomial is a good fit; \(H_1\): Binomial is not a good fit | B1 | Parameters in hypotheses award B0 |
| Combined table with \(O\): 22, 20, 23, 17, 18 and \(E\): 12.41, 24.82, 31.17, 22.01, 9.59 | M1 | M1 for combining 0 and 1 or 5 and 6 or both; require at least 1 value in combined correct |
| \(\sum \frac{(O-E)^2}{E} = \frac{(22-12.41)^2}{12.41} + \frac{(20-24.82)^2}{24.82} + \ldots + \frac{(18-9.59)^2}{9.59} = 18.998...\) awrt 19.0 | M1A1 | M1 for at least 2 correct expressions or values |
| \(\nu = 5-2 = 3\) degrees of freedom | B1 | |
| \(\chi^2_3(5\%) = 7.815\) | B1ft | |
| \(18.998 > 7.815\) so reject \(H_0\) | M1 | |
| Binomial is not a good fit (and is not a good model for the number of defective items in samples of size 6) | A1 | Correct comment suggesting Binomial not suitable; no ft; condone parameters |
# Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Mean} = \frac{1\times16 + 2\times20 + \ldots + 6\times8}{100} = 2.91$ | M1A1 | M1 for at least 2 correct terms in numerator and 100 in denominator |
---
# Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = \frac{2.91}{6} = 0.485$ | B1 | 0.485 implied by at least 1 correct answer |
| $a = 100 \times C^6_3 \times 0.485^3 \times 0.515^3 = 31.17$ | M1A1 | Clear use of Binomial and $\times 100$ required for method |
| $b = 100 \times 0.485^6 = 1.3(0)$ | A1 | Accept awrt 2dp for final answers |
---
# Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Binomial is a good fit; $H_1$: Binomial is not a good fit | B1 | Parameters in hypotheses award B0 |
| Combined table with $O$: 22, 20, 23, 17, 18 and $E$: 12.41, 24.82, 31.17, 22.01, 9.59 | M1 | M1 for combining 0 and 1 or 5 and 6 or both; require at least 1 value in combined correct |
| $\sum \frac{(O-E)^2}{E} = \frac{(22-12.41)^2}{12.41} + \frac{(20-24.82)^2}{24.82} + \ldots + \frac{(18-9.59)^2}{9.59} = 18.998...$ awrt 19.0 | M1A1 | M1 for at least 2 correct expressions or values |
| $\nu = 5-2 = 3$ degrees of freedom | B1 | |
| $\chi^2_3(5\%) = 7.815$ | B1ft | |
| $18.998 > 7.815$ so reject $H_0$ | M1 | |
| Binomial is not a good fit (and is not a good model for the number of defective items in samples of size 6) | A1 | Correct comment suggesting Binomial not suitable; no ft; condone parameters |
---6. A total of 100 random samples of 6 items are selected from a production line in a factory and the number of defective items in each sample is recorded. The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of \\
defective \\
items \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\begin{tabular}{ l }
Number of \\
samples \\
\end{tabular} & 6 & 16 & 20 & 23 & 17 & 10 & 8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that the mean number of defective items per sample is 2.91
A factory manager suggests that the data can be modelled by a binomial distribution with $n = 6$. He uses the mean from the sample above and calculates expected frequencies as shown in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of \\
defective \\
items \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\begin{tabular}{ l }
Expected \\
frequency \\
\end{tabular} & 1.87 & 10.54 & 24.82 & $a$ & 22.01 & 8.29 & $b$ \\
\hline
\end{tabular}
\end{center}
\item Calculate the value of $a$ and the value of $b$ giving your answers to 2 decimal places.
\item Test, at the $5 \%$ level, whether or not the binomial distribution is a suitable model for the number of defective items in samples of 6 items. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2012 Q6 [14]}}