| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Normal |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with Normal distribution, requiring routine calculations of mean/SD from grouped data, finding expected frequencies using Normal tables, computing the test statistic, and interpreting results. While it involves multiple steps and some statistical reasoning in part (iv), all techniques are standard S3 material with no novel problem-solving required. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.06b Fit prescribed distribution: chi-squared test |
| Mass \(( m \mathrm {~kg} )\) | \(m < 1.6\) | \(1.6 \leqslant m < 1.8\) | \(1.8 \leqslant m < 2.0\) | \(2.0 \leqslant m < 2.2\) | \(2.2 \leqslant m < 2.4\) | \(2.4 \leqslant m < 2.6\) | \(2.6 \leqslant m\) |
| Frequency | 2 | 8 | 30 | 42 | 11 | 5 | 2 |
| Mass \(( m \mathrm {~kg} )\) | \(m < 1.6\) | \(1.6 \leqslant m < 1.8\) | \(1.8 \leqslant m < 2.0\) | \(2.0 \leqslant m < 2.2\) | \(2.2 \leqslant m < 2.4\) | \(2.4 \leqslant m < 2.6\) | \(2.6 \leqslant m\) | ||
| 2.17 | 10.92 | \(f\) | 33.85 | 19.22 | 5.13 | 0.68 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using mid-intervals 1.5, 1.7, etc | M1 | |
| \(\bar{x} = \frac{205}{100} = 2.05\) | A1 | Mean |
| \(s = \sqrt{\frac{425.16 - 100 \times 2.05^2}{99}} = 0.2227(01\ldots)\) | E1 | s.d. Answer given; must show convincingly. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f = 100 \times P(1.8 \leq M < 2.0)\) | M1 | Probability \(\times\) 100 |
| \(= 100 \times P(-1.1226 \leq z < -0.2245)\) | A1 | Correct Normal probabilities. ft c's mean |
| \(= 100 \times ((1 - 0.5888) - (1 - 0.8691))\) | Must show convincingly using Normal distribution | |
| \(= 100 \times (0.4112 - 0.1309) = 28.03\) | A1 | ft c's mean. Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): The Normal model fits the data. \(H_1\): The Normal model does not fit the data | B1, B1 | Ignore any reference to parameters |
| Merge first 2 and last 2 cells | M1 | |
| \(X^2 = 0.7294 + 0.1384 + 1.9623 + 3.5155 + 0.2437 = 6.589(3)\) | M1, A1 | Calculation of \(X^2\); c.a.o. |
| Refer to \(\chi^2_2\) | M1 | Allow correct df (= cells \(- 3\)) from wrongly grouped table and ft. Otherwise no ft if wrong. \(P(X^2 > 6.589) = 0.0371\) |
| Upper 5% point is 5.991 | A1 | No ft from here if wrong |
| Significant | A1 | ft only c's test statistic |
| Evidence suggests that the model does not fit the data | A1 | ft only c's test statistic. Conclusion in context. Total: 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The model overestimates in the \(2.2 - 2.4\) class | E1 | |
| Underestimates in the \(2 - 2.2\) class | E1 | |
| At lower significance levels the test would not have been significant | E1 | Total: 3 |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Using mid-intervals 1.5, 1.7, etc | M1 | |
| $\bar{x} = \frac{205}{100} = 2.05$ | A1 | Mean |
| $s = \sqrt{\frac{425.16 - 100 \times 2.05^2}{99}} = 0.2227(01\ldots)$ | E1 | s.d. Answer given; must show convincingly. Total: 3 |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f = 100 \times P(1.8 \leq M < 2.0)$ | M1 | Probability $\times$ 100 |
| $= 100 \times P(-1.1226 \leq z < -0.2245)$ | A1 | Correct Normal probabilities. ft c's mean |
| $= 100 \times ((1 - 0.5888) - (1 - 0.8691))$ | | Must show convincingly using Normal distribution |
| $= 100 \times (0.4112 - 0.1309) = 28.03$ | A1 | ft c's mean. Total: 3 |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: The Normal model fits the data. $H_1$: The Normal model does not fit the data | B1, B1 | Ignore any reference to parameters |
| Merge first 2 and last 2 cells | M1 | |
| $X^2 = 0.7294 + 0.1384 + 1.9623 + 3.5155 + 0.2437 = 6.589(3)$ | M1, A1 | Calculation of $X^2$; c.a.o. |
| Refer to $\chi^2_2$ | M1 | Allow correct df (= cells $- 3$) from wrongly grouped table and ft. Otherwise no ft if wrong. $P(X^2 > 6.589) = 0.0371$ |
| Upper 5% point is 5.991 | A1 | No ft from here if wrong |
| Significant | A1 | ft only c's test statistic |
| Evidence suggests that the model does not fit the data | A1 | ft only c's test statistic. Conclusion in context. Total: 9 |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| The model overestimates in the $2.2 - 2.4$ class | E1 | |
| Underestimates in the $2 - 2.2$ class | E1 | |
| At lower significance levels the test would not have been significant | E1 | Total: 3 |
---3 The masses, in kilograms, of a random sample of 100 chickens on sale in a large supermarket were recorded as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Mass $( m \mathrm {~kg} )$ & $m < 1.6$ & $1.6 \leqslant m < 1.8$ & $1.8 \leqslant m < 2.0$ & $2.0 \leqslant m < 2.2$ & $2.2 \leqslant m < 2.4$ & $2.4 \leqslant m < 2.6$ & $2.6 \leqslant m$ \\
\hline
Frequency & 2 & 8 & 30 & 42 & 11 & 5 & 2 \\
\hline
\end{tabular}
\end{center}
(i) Assuming that the first and last classes are the same width as the other classes, calculate an estimate of the sample mean and show that the corresponding estimate of the sample standard deviation is 0.2227 kg .
A Normal distribution using the mean and standard deviation found in part (i) is to be fitted to these data. The expected frequencies for the classes are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Mass $( m \mathrm {~kg} )$ & $m < 1.6$ & $1.6 \leqslant m < 1.8$ & $1.8 \leqslant m < 2.0$ & $2.0 \leqslant m < 2.2$ & $2.2 \leqslant m < 2.4$ & $2.4 \leqslant m < 2.6$ & $2.6 \leqslant m$ \\
\hline
\begin{tabular}{ l }
Expected \\
frequency \\
\end{tabular} & 2.17 & 10.92 & $f$ & 33.85 & 19.22 & 5.13 & 0.68 \\
\hline
\end{tabular}
\end{center}
(ii) Use the Normal distribution to find $f$.\\
(iii) Carry out a goodness of fit test of this Normal model using a significance level of 5\%.\\
(iv) Discuss the outcome of the test with reference to the contributions to the test statistic and to the possibility of other significance levels.
\hfill \mbox{\textit{OCR MEI S3 2011 Q3 [18]}}