| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| 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 expected frequencies using normal tables, a straightforward chi-squared test, a confidence interval calculation, and interpretation. All techniques are textbook applications with no novel insight required, making it slightly easier than average for S3 level. |
| Spec | 5.05d Confidence intervals: using normal distribution5.06c Fit other distributions: discrete and continuous |
| Values | \(y < 20\) | \(20 \leqslant y < 25\) | \(25 \leqslant y < 30\) | \(y \geqslant 30\) |
| Frequency | 3 | 27 | 12 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(20 \leq y < 25) = \Phi(0) - \Phi(-5/\sqrt{(20)})\) | M1 | |
| Multiply by 50 to give 18.41 AG. 18.41 for 25 ≤ y < 30 and 6.59 for y < 20, y ≥ 30 | A1 | |
| A1 | ||
| A1 | 4 marks | |
| (ii) \(H_0\): \(N(25, 20)\) fits data. \(\chi^2 = 3.59^2/6.59 + 8.59^2/18.41 + 6.41^2/18.41 + 1.41^2/6.59\) | B1 | OR: \(Y \sim N(25, 20)\) |
| \(= 8.497\) | M1√ | ft values from (i); art 8.5 |
| 8.497 > 7.815. Accept that \(N(25, 20)\) is not a good fit | M1 | |
| A1 | 5 marks | |
| (iii) Use \(24.91 \pm z\sqrt{(20/50)}\); \(z = 2.326\) | M1 | With \(\sqrt{(20/50)}\) |
| \((23.44, 26.38)\) | B1 | |
| A1 | 3 marks; art (23.4, 26.4) Must be interval | |
| (iv) No- Sample size large enough to apply CLT. Sample mean will be (approximately) normally distributed whatever the distribution of \(Y\) | B1 | Refer to large sample size |
| B1 | 2 marks; Refer to normality of sample mean |
(i) $P(20 \leq y < 25) = \Phi(0) - \Phi(-5/\sqrt{(20)})$ | M1 |
Multiply by 50 to give 18.41 AG. 18.41 for 25 ≤ y < 30 and 6.59 for y < 20, y ≥ 30 | A1 |
| A1 |
| A1 | 4 marks |
(ii) $H_0$: $N(25, 20)$ fits data. $\chi^2 = 3.59^2/6.59 + 8.59^2/18.41 + 6.41^2/18.41 + 1.41^2/6.59$ | B1 | OR: $Y \sim N(25, 20)$
$= 8.497$ | M1√ | ft values from (i); art 8.5
8.497 > 7.815. Accept that $N(25, 20)$ is not a good fit | M1 |
| A1 | 5 marks |
(iii) Use $24.91 \pm z\sqrt{(20/50)}$; $z = 2.326$ | M1 | With $\sqrt{(20/50)}$
$(23.44, 26.38)$ | B1 |
| A1 | 3 marks; art (23.4, 26.4) Must be interval
(iv) No- Sample size large enough to apply CLT. Sample mean will be (approximately) normally distributed whatever the distribution of $Y$ | B1 | Refer to large sample size
| B1 | 2 marks; Refer to normality of sample mean8 The continuous random variable $Y$ has a distribution with mean $\mu$ and variance 20. A random sample of 50 observations of $Y$ is selected and these observations are summarised in the following grouped frequency table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Values & $y < 20$ & $20 \leqslant y < 25$ & $25 \leqslant y < 30$ & $y \geqslant 30$ \\
\hline
Frequency & 3 & 27 & 12 & 8 \\
\hline
\end{tabular}
\end{center}
(i) Assuming that $Y \sim \mathrm {~N} ( 25,20 )$, show that the expected frequency for the interval $20 \leqslant y < 25$ is 18.41, correct to 2 decimal places, and obtain the remaining expected frequencies.\\
(ii) Test, at the $5 \%$ significance level, whether the distribution $\mathrm { N } ( 25,20 )$ fits the data.\\
(iii) Given that the sample mean is 24.91 , find a $98 \%$ confidence interval for $\mu$.\\
(iv) Does the outcome of the test in part (ii) affect the validity of the confidence interval found in part (iii)? Justify your answer.
\hfill \mbox{\textit{OCR S3 2007 Q8 [14]}}