| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Comment on claim using CI |
| Difficulty | Standard +0.3 This question tests understanding of the relationship between confidence intervals and hypothesis tests, requiring students to recognize that a value inside a 90% CI would not be rejected at 10% significance. Part (b) is a standard CI calculation with given sample size and mean. Both parts are routine applications of S3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 0.5 \quad H_1: \mu \neq 0.5\) | B1 | For both hypotheses in terms of \(\mu\) |
| Significance level \(= 10\%\) | dB1 | Accept 5% if they have a one-tail test as \(H_1\) |
| \(0.5\) is in the interval so not significant, accept \(H_0\), can accept \(\mu = 0.5\) | B1 | For correct comment leading to accepting \(H_0\). Ignore any further calculations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.6449 \times \frac{\sigma}{\sqrt{100}} = 0.0247\) | M1 | For \(z \cdot \frac{\sigma}{\sqrt{100}} = k\), using \(n=100\) and where \( |
| B1 | For 1.6449 or better (could be \(1.6449\sigma = k\) or \(1.6449\sigma^2 = k\)) | |
| \(\sigma = 0.15016\) or \(\frac{10 \times 0.0247}{1.6449}\) (awrt \(0.15\)) | A1 | For correct expression for \(\sigma\), e.g. awrt 0.15 |
| \(0.479 \pm 1.96 \times \frac{\text{"}\sigma\text{"}}{\sqrt{150}}\) | M1 | For \(\bar{x} \pm z \times \frac{\sigma}{\sqrt{150}}\) for any \(z > 1\) and ft their \(\sigma\), allow \(\bar{x} \in (0.4633, 0.5127)\) |
| B1 | For 1.96 or better (could be \(1.96\sigma\) or \(1.96\sigma^2\)) | |
| awrt \(\mathbf{(0.455, 0.503)}\) | A1 | For awrt 0.455 and awrt 0.503 |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 0.5 \quad H_1: \mu \neq 0.5$ | B1 | For both hypotheses in terms of $\mu$ |
| Significance level $= 10\%$ | dB1 | Accept 5% if they have a one-tail test as $H_1$ |
| $0.5$ is in the interval so not significant, accept $H_0$, can accept $\mu = 0.5$ | B1 | For correct comment leading to accepting $H_0$. Ignore any further calculations |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.6449 \times \frac{\sigma}{\sqrt{100}} = 0.0247$ | M1 | For $z \cdot \frac{\sigma}{\sqrt{100}} = k$, using $n=100$ and where $|z| > 1.5$ and $0.02 < k < 0.03$ |
| | B1 | For 1.6449 or better (could be $1.6449\sigma = k$ or $1.6449\sigma^2 = k$) |
| $\sigma = 0.15016$ or $\frac{10 \times 0.0247}{1.6449}$ (awrt $0.15$) | A1 | For correct expression for $\sigma$, e.g. awrt 0.15 |
| $0.479 \pm 1.96 \times \frac{\text{"}\sigma\text{"}}{\sqrt{150}}$ | M1 | For $\bar{x} \pm z \times \frac{\sigma}{\sqrt{150}}$ for any $z > 1$ and ft their $\sigma$, allow $\bar{x} \in (0.4633, 0.5127)$ |
| | B1 | For 1.96 or better (could be $1.96\sigma$ or $1.96\sigma^2$) |
| awrt $\mathbf{(0.455, 0.503)}$ | A1 | For awrt 0.455 and awrt 0.503 |
---\begin{enumerate}
\item The weights of bags of rice, $X \mathrm {~kg}$, have a normal distribution with unknown mean $\mu \mathrm { kg }$ and known standard deviation $\sigma \mathrm { kg }$. A random sample of 100 bags of rice gave a $90 \%$ confidence interval for $\mu$ of $( 0.4633,0.5127 )$.\\
(a) Without carrying out any further calculations, use this confidence interval to test whether or not $\mu = 0.5$
\end{enumerate}
State your hypotheses clearly and write down the significance level you have used.
A second random sample, of 150 of these bags of rice, had a mean weight of 0.479 kg .\\
(b) Calculate a $95 \%$ confidence interval for $\mu$ based on this second sample.
\hfill \mbox{\textit{Edexcel S3 2015 Q4 [9]}}