| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2021 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Known variance confidence interval |
| Difficulty | Standard +0.3 This is a standard S3 confidence interval and hypothesis testing question with known variance. Part (a) is routine CI calculation, part (b) is a standard two-sample z-test, and part (c) requires understanding of critical values but is still algorithmic. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(\bar{x} \pm z \times \frac{18}{\sqrt{25}}\); \(z = 2.3263\) (or better) | M1; B1 | M1 for use of correct expression with 18, 25 and \(1 < z < 3\) (Ignore \(\bar{x}\) for this mark); B1 for \(z = 2.3263\) or better (calc: 2.3263787...) |
| \(= (44.0253..., 60.7746...)\) awrt (44.0, 60.8) | A1, A1 | 1st A1 for awrt 44.0 (ans only of 44.02...or awrt 44.03 scores M1B1 implied); 2nd A1 for awrt 60.8 (ans only of 60.77... or awrt 60.77 scores M1B1 implied) |
| Subtotal: (4 marks) | ||
| \(H_0: \mu_A = \mu_B\); \(H_1: \mu_B > \mu_A\) | B1 | 1st B1 for both hypotheses in terms of \(\mu\)s (If using \(\mu_1\) etc they must define which is which) |
| \(z = (\pm) \frac{57.8 - 52.4}{18\sqrt{\frac{1}{25} + \frac{1}{30}}}\) | M1dM1 | 1st M1 for correct denominator \((18 \text{ needn't be outside square root})\) [4.87(44...)]. 2nd dM1 for correct expression for test statistic |
| \(= (\pm) 1.1078...\) awrt (± 1.11) | A1 | 1st A1 for awrt (±) 1.11 |
| 5% one-tail critical value is 1.6449 (or p-value = 0.13396... i.e. awrt 0.134) | B1 | 2nd B1 for critical value 1.6449 or better (If B0 in (a) for 2.33 allow 1.64 or 1.645 here). [Allow p-value of awrt 0.134 and condone awrt 0.866 if compared with 0.95] |
| (not sig') so insufficient evidence (in these data) to support newspaper's claim | A1 | 2nd A1 for contextual conclusion. It comparing their "1.11" with 1.64 (or their cv) must be not significant and mention "claim" or "score in town A" and "score in town B" |
| Subtotal: (6 marks) | ||
| Require \(\frac{\bar{x} - \mu}{\frac{18}{\sqrt{n}}} > z\) where \(z = -1.6449\) (o.e.) | M1 | 1st M1 for correct starting inequality with any \(z\) such that \(\ |
| \(\mu < 52.4 + 1.64(49) \times \frac{18}{5}\) or \(\mu < 57.8 + 1.64(49) \times \frac{18}{\sqrt{30}}\) | A1 | 1st A1 for either correct inequality for \(\mu\), allow \(z = 1.64\) or better |
| i.e. \(\mu < 58.3216...\) and \(\mu < 63.2056...\) | M1 | 2nd M1 for both cases of \(\bar{x} \pm z \frac{18}{\sqrt{n}}\) (z > 1) can allow "=" or inequality, may be in CI |
| So \(\mu = \mathbf{58.3}\) | A1 | 2nd A1 (dep on both Ms) for sight of both awrt 58.3 and awrt 63.2 and selecting awrt 58.3 |
| Subtotal: (4 marks) |
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $\bar{x} \pm z \times \frac{18}{\sqrt{25}}$; $z = 2.3263$ (or better) | M1; B1 | M1 for use of correct expression with 18, 25 and $1 < z < 3$ (Ignore $\bar{x}$ for this mark); B1 for $z = 2.3263$ or better (calc: 2.3263787...) |
| $= (44.0253..., 60.7746...)$ awrt **(44.0, 60.8)** | A1, A1 | 1st A1 for awrt 44.0 (ans only of 44.02...or awrt 44.03 scores M1B1 implied); 2nd A1 for awrt 60.8 (ans only of 60.77... or awrt 60.77 scores M1B1 implied) |
| **Subtotal: (4 marks)** | | |
| $H_0: \mu_A = \mu_B$; $H_1: \mu_B > \mu_A$ | B1 | 1st B1 for both hypotheses in terms of $\mu$s (If using $\mu_1$ etc they must define which is which) |
| $z = (\pm) \frac{57.8 - 52.4}{18\sqrt{\frac{1}{25} + \frac{1}{30}}}$ | M1dM1 | 1st M1 for correct denominator $(18 \text{ needn't be outside square root})$ [4.87(44...)]. 2nd dM1 for correct expression for test statistic |
| $= (\pm) 1.1078...$ awrt **(± 1.11)** | A1 | 1st A1 for awrt (±) 1.11 |
| 5% one-tail critical value is 1.6449 (or p-value = 0.13396... i.e. awrt 0.134) | B1 | 2nd B1 for critical value 1.6449 or better (If B0 in (a) for 2.33 allow 1.64 or 1.645 here). [Allow p-value of awrt 0.134 and condone awrt 0.866 if compared with 0.95] |
| (not sig') so insufficient evidence (in these data) to support newspaper's claim | A1 | 2nd A1 for contextual conclusion. It comparing their "1.11" with 1.64 (or their cv) must be not significant and mention "claim" or "score in town A" and "score in town B" |
| **Subtotal: (6 marks)** | | |
| Require $\frac{\bar{x} - \mu}{\frac{18}{\sqrt{n}}} > z$ where $z = -1.6449$ (o.e.) | M1 | 1st M1 for correct starting inequality with any $z$ such that $\|z\| > 1$ (Allow $\geq$) |
| $\mu < 52.4 + 1.64(49) \times \frac{18}{5}$ or $\mu < 57.8 + 1.64(49) \times \frac{18}{\sqrt{30}}$ | A1 | 1st A1 for either correct inequality for $\mu$, allow $z = 1.64$ or better |
| i.e. $\mu < 58.3216...$ and $\mu < 63.2056...$ | M1 | 2nd M1 for both cases of $\bar{x} \pm z \frac{18}{\sqrt{n}}$ (z > 1) can allow "=" or inequality, may be in CI |
| So $\mu = \mathbf{58.3}$ | A1 | 2nd A1 (dep on both Ms) for sight of both awrt 58.3 and awrt 63.2 and selecting awrt 58.3 |
| **Subtotal: (4 marks)** | | |
**Total: [14 marks]**
---4. The scores in a national test of seven-year-old children are normally distributed with a standard deviation of 18\\
A random sample of 25 seven-year-old children from town $A$ had a mean score of 52.4
\begin{enumerate}[label=(\alph*)]
\item Calculate a 98\% confidence interval for the mean score of the seven-year-old children from town $A$.\\
(4)
An independent random sample of 30 seven-year-old children from town $B$ had a mean score of 57.8\\
A local newspaper claimed that the mean score of seven-year-old children from town $B$ was greater than the mean score of seven-year-old children from town $A$.
\item Stating your hypotheses clearly, use a $5 \%$ significance level to test the newspaper's claim. You should show your working clearly.
The mean score for the national test of seven-year-old children is $\mu$.
Considering the two samples of seven-year-old children separately, at the $5 \%$ level of significance, there is insufficient evidence that the mean score for town $A$ is less than $\mu$, and insufficient evidence that the mean score for town $B$ is less than $\mu$.
\item Find the largest possible value for $\mu$.
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel S3 2021 Q4 [14]}}