| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | January |
| Marks | 15 |
| 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 t-test with straightforward calculations of sample means and variances from given summations, followed by routine hypothesis testing. While it requires multiple steps and understanding of the test procedure, it follows a well-practiced template with no novel insights needed. The context is clear, and part (c) tests understanding of CLT, making it slightly easier than average for S3 level. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{3610}{50} \Rightarrow \bar{x} = 72.2\) | B1 | \(\bar{x} = 72.2\) |
| \(s_x^2 = \frac{260955.6 - 50(72.2)^2}{50-1} = 6.4\) | M1 A1 | A correct method for finding an unbiased estimate of the variance e.g. \(\frac{\sum x^2 - n(\bar{x})^2}{n-1}\); 6.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y} = \frac{2585}{50} \Rightarrow \bar{y} = 51.7\) | B1 | \(\bar{y} = 51.7\) |
| \(s_y^2 = \frac{133757.2 - 50(51.7)^2}{50-1} = 2.3\) | A1 | 2.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_x - \mu_y = 20\); \(H_1: \mu_x - \mu_y > 20\) | B1 | Both hypotheses correct. Allow equivalent hypotheses. Must be in terms of \(\mu\) |
| \(z = \frac{72.2 - 51.7 - 20}{\sqrt{\frac{6.4}{50} + \frac{2.3}{50}}}\) | M1 M1 | For correct standard error (follow through from (a)); an attempt at \(\frac{a-b-20}{\sqrt{\frac{c}{50}+\frac{d}{50}}}\) with at least 2 of \(a,b,c,d\) correct |
| \(= 1.1986...\) awrt 1.20 | A1 | awrt 1.20. Allow 1.2 if no incorrect working shown |
| One-tailed c.v. \(Z = 1.6449\) or CR: \(Z \geq 1.6449\) | B1 | 1.6449 or better |
| Not in CR / Do not reject \(H_0\) | M1 | A correct statement - need not be contextual but do not allow contradicting non-contextual comments |
| No significant evidence to support Tammy's belief | A1 | Correct contextual statement. Allow "the difference in mean weights is not greater than 20 kg" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Since the sample is large the CLT applies. No need to assume the weights are normally distributed. | M1 A1 | A suitable comment that mentions large and CLT; correct answer, context not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assumed that \(s^2 = \sigma^2\) | B1 | For the assumption that sample variance = population variance |
## Question 5:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{3610}{50} \Rightarrow \bar{x} = 72.2$ | B1 | $\bar{x} = 72.2$ |
| $s_x^2 = \frac{260955.6 - 50(72.2)^2}{50-1} = 6.4$ | M1 A1 | A correct method for finding an unbiased estimate of the variance e.g. $\frac{\sum x^2 - n(\bar{x})^2}{n-1}$; 6.4 |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y} = \frac{2585}{50} \Rightarrow \bar{y} = 51.7$ | B1 | $\bar{y} = 51.7$ |
| $s_y^2 = \frac{133757.2 - 50(51.7)^2}{50-1} = 2.3$ | A1 | 2.3 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_x - \mu_y = 20$; $H_1: \mu_x - \mu_y > 20$ | B1 | Both hypotheses correct. Allow equivalent hypotheses. Must be in terms of $\mu$ |
| $z = \frac{72.2 - 51.7 - 20}{\sqrt{\frac{6.4}{50} + \frac{2.3}{50}}}$ | M1 M1 | For correct standard error (follow through from (a)); an attempt at $\frac{a-b-20}{\sqrt{\frac{c}{50}+\frac{d}{50}}}$ with at least 2 of $a,b,c,d$ correct |
| $= 1.1986...$ awrt 1.20 | A1 | awrt 1.20. Allow 1.2 if no incorrect working shown |
| One-tailed c.v. $Z = 1.6449$ or CR: $Z \geq 1.6449$ | B1 | 1.6449 or better |
| Not in CR / Do not reject $H_0$ | M1 | A correct statement - need not be contextual but do not allow contradicting non-contextual comments |
| No significant evidence to support **Tammy's belief** | A1 | Correct contextual statement. Allow "the difference in mean weights is **not greater than 20 kg**" |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Since the sample is **large** the **CLT** applies. No need to assume the weights are normally distributed. | M1 A1 | A suitable comment that mentions large and CLT; correct answer, context not required |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assumed that $s^2 = \sigma^2$ | B1 | For the assumption that sample variance = population variance |
---\begin{enumerate}
\item A dog breeder claims that the mean weight of male Great Dane dogs is 20 kg more than the mean weight of female Great Dane dogs.
\end{enumerate}
Tammy believes that the mean weight of male Great Dane dogs is more than 20 kg more than the mean weight of female Great Dane dogs.
She takes random samples of 50 male and 50 female Great Dane dogs and records their weights.
The results are summarised below, where $x$ denotes the weight, in kg , of a male Great Dane dog and $y$ denotes the weight, in kg, of a female Great Dane dog.
$$\sum x = 3610 \quad \sum x ^ { 2 } = 260955.6 \quad \sum y = 2585 \quad \sum y ^ { 2 } = 133757.2$$
(a) Find unbiased estimates for the mean and variance of the weights of\\
(i) the male Great Dane dogs,\\
(ii) the female Great Dane dogs.\\
(b) Stating your hypotheses clearly, carry out a suitable test to assess Tammy's belief. Use a $5 \%$ level of significance and state your critical value.\\
(c) For the test in part (b), state whether or not it is necessary to assume that the weights of the Great Dane dogs are normally distributed. Give a reason for your answer.\\
(d) State an assumption you have made in carrying out the test in part (b).
\hfill \mbox{\textit{Edexcel S3 2022 Q5 [15]}}