| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| 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 straightforward chi-squared goodness of fit test with standard steps: estimating a parameter from sample mean, calculating expected frequencies, computing chi-squared statistic, and comparing to critical values. Part (i) is routine parameter estimation, part (ii) follows standard test procedure (though requires combining cells), and part (iii) is simple interpretation given the test statistic. The binomial context and calculations are typical S3 material with no novel insights required. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of underfilled bottles | 0 | 1 | 2 | 3 |
| Number of packs | 44 | 20 | 8 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= (20+16+9)/75 = 0.6\) | M1, A1 | |
| \(3p = 0.6\), \(p = 0.2\) AG | A1 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): \(B(3,p)\) fits the data; \(H_1\): \(B(3,p)\) does not fit the data | B1 | Or: \(X \sim B(3,p)\) or \(B(3,0.2)\). Not 'Data fits model' |
| Expected values: 38.4, 28.8, 7.2, 0.6 | M1, A1, A1 | Use \(B(3,0.2)\times 75\). At least 2 correct, All correct |
| Combine last two cells | B1 | |
| \(\chi^2 = 5.6^2/38.4 + 8.8^2/28.8 + 3.2^2/7.8 = 4.818\) | M1, \(\text{A1}\sqrt{}\), A1 | With one correct. At least 2 correct Ft values. Accept 4.82 cao |
| \(4.818 > 3.841\) | \(\text{B1}\sqrt{}\) | ft 4.818. SR1 if cells not combined: B1M1A1A1B0M1A1A0B1(5.991)M1 |
| Reject \(H_0\) and conclude there is insufficient evidence that \(B(3,p)\) fits the data | M1 10 | SR2: E-values rounded: B1M1A1A1A0B1M1A1A0(4.865)B1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2.74 < 3.841\), accept \(H_0\), conclude that \(B(6,p)\) fits the data | B1 1 | Accept with no reason if evidence of method in (ii) |
# Question 8:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= (20+16+9)/75 = 0.6$ | M1, A1 | |
| $3p = 0.6$, $p = 0.2$ AG | A1 **3** | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: $B(3,p)$ fits the data; $H_1$: $B(3,p)$ does not fit the data | B1 | Or: $X \sim B(3,p)$ or $B(3,0.2)$. Not 'Data fits model' |
| Expected values: 38.4, 28.8, 7.2, 0.6 | M1, A1, A1 | Use $B(3,0.2)\times 75$. At least 2 correct, All correct |
| Combine last two cells | B1 | |
| $\chi^2 = 5.6^2/38.4 + 8.8^2/28.8 + 3.2^2/7.8 = 4.818$ | M1, $\text{A1}\sqrt{}$, A1 | With one correct. At least 2 correct Ft values. Accept 4.82 cao |
| $4.818 > 3.841$ | $\text{B1}\sqrt{}$ | ft 4.818. SR1 if cells not combined: B1M1A1A1B0M1A1A0B1(5.991)M1 |
| Reject $H_0$ and conclude there is insufficient evidence that $B(3,p)$ fits the data | M1 **10** | SR2: E-values rounded: B1M1A1A1A0B1M1A1A0(4.865)B1M1 |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2.74 < 3.841$, accept $H_0$, conclude that $B(6,p)$ fits the data | B1 **1** | Accept with no reason if evidence of method in (ii) |8 A soft drinks factory produces lemonade which is sold in packs of 6 bottles. As part of the factory's quality control, random samples of 75 packs are examined at regular intervals. The number of underfilled bottles in a pack of 6 bottles is denoted by the random variable $X$. The results of one quality control check are shown in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number of underfilled bottles & 0 & 1 & 2 & 3 \\
\hline
Number of packs & 44 & 20 & 8 & 3 \\
\hline
\end{tabular}
\end{center}
A researcher assumes that $X \sim \mathrm {~B} ( 3 , p )$.\\
(i) By finding the sample mean, show that an estimate of $p$ is 0.2 .\\
(ii) Show that, at the $5 \%$ significance level, there is evidence that this binomial distribution does not fit the data.\\
(iii) Another researcher suggests that the goodness of fit test should be for $\mathrm { B } ( 6 , p )$. She finds that the corresponding value of $\chi ^ { 2 }$ is 2.74 , correct to 3 significant figures. Given that the number of degrees of freedom is the same as in part (ii), state the conclusion of the test at the same significance level.
\hfill \mbox{\textit{OCR S3 2009 Q8 [14]}}