| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a standard two-sample hypothesis test with clearly stated data and a straightforward alternative hypothesis (H₁: μ₁ - μ₂ > 1.5). The large sample sizes (n=80, n=65) make the CLT application in part (b) routine. While it requires proper hypothesis formulation and test statistic calculation, it's a textbook S3 procedure with no novel insight needed—slightly easier than average due to its mechanical nature. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_g - \mu_s = 1.5\) | B1 | \(g\) = in a group, \(s\) = on their own. Accept \(\mu_1, \mu_2\) or \(\mu_A, \mu_B\) etc if there is indication of which is which |
| \(H_1: \mu_g - \mu_s > 1.5\) | B1 | |
| \(\text{s.e.} = \sqrt{\frac{2.1^2}{80} + \frac{1.4^2}{65}} = \sqrt{0.08527...} = [0.292]\) | M1 | Attempt at s.e. with 3 out of 4 values correct. Condone switching 2.1 and 1.4 |
| \(z = \frac{8.7 - 6.6 - 1.5}{\sqrt{\frac{2.1^2}{80} + \frac{1.4^2}{65}}}\) | dM1 | Dependent on 1st M1 for correct numerator (must have \(-1.5\)) and ft their s.e. |
| \(= 2.0546...\) awrt \(2.05(5)\) | A1 | |
| cv \(1\%\) one tailed \(= 2.3263\) | B1 | For \(\pm 2.3263\) or better |
| Not significant, accept \(H_0\) | dM1 | Dependent on 1st M1 for correct statement based on normal cv and test statistic |
| Insufficient evidence that using plan as part of a group leads to weight loss of more than 1.5 kg more than using plan on one's own, or researcher's belief not supported | A1ft | Must mention "plan" and "group or individual" and "1.5" or "researcher" and "belief or claim" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Since sample is large, Central Limit Theorem (CLT) applies | B1 | For mentioning "large samples" and "CLT" |
| No need to assume normal distribution | dB1 | Dependent on 1st B1 for stating no need to assume normality (since CLT assures it) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_g - \mu_s = 1.5$ | B1 | $g$ = in a group, $s$ = on their own. Accept $\mu_1, \mu_2$ or $\mu_A, \mu_B$ etc if there is indication of which is which |
| $H_1: \mu_g - \mu_s > 1.5$ | B1 | |
| $\text{s.e.} = \sqrt{\frac{2.1^2}{80} + \frac{1.4^2}{65}} = \sqrt{0.08527...} = [0.292]$ | M1 | Attempt at s.e. with 3 out of 4 values correct. Condone switching 2.1 and 1.4 |
| $z = \frac{8.7 - 6.6 - 1.5}{\sqrt{\frac{2.1^2}{80} + \frac{1.4^2}{65}}}$ | dM1 | Dependent on 1st M1 for correct numerator (must have $-1.5$) and ft their s.e. |
| $= 2.0546...$ awrt $2.05(5)$ | A1 | |
| cv $1\%$ one tailed $= 2.3263$ | B1 | For $\pm 2.3263$ or better |
| Not significant, accept $H_0$ | dM1 | Dependent on 1st M1 for correct statement based on normal cv and test statistic |
| Insufficient evidence that using plan as part of a group leads to weight loss of more than 1.5 kg more than using plan on one's own, or researcher's belief not supported | A1ft | Must mention "plan" and "group or individual" and "1.5" or "researcher" and "belief or claim" |
**NB:** Use of cv for difference in means $D$ will have $D = 1.5 + 2.3263 \times \text{s.e.} = \text{awrt } 2.18$ and requires sight of $d = 2.1$ with a comment for the 3rd M1
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Since sample is large, Central Limit Theorem (CLT) applies | B1 | For mentioning "large samples" and "CLT" |
| No need to assume normal distribution | dB1 | Dependent on 1st B1 for stating no need to assume normality (since CLT assures it) |
---2. A researcher believes that the mean weight loss of those people using a slimming plan as part of a group is more than 1.5 kg a year greater than the mean weight loss of those using the plan on their own. The mean weight loss of a random sample of 80 people using the plan as part of a group is 8.7 kg with a standard deviation of 2.1 kg . The mean weight loss of a random sample of 65 people using the plan on their own is 6.6 kg with a standard deviation of 1.4 kg .
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test the researcher's claim. Use a $1 \%$ level of significance.
\item For the test in part (a), state whether or not it is necessary to assume that the weight loss of a person using this plan has a normal distribution. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2015 Q2 [10]}}