A2 June 2025 Q6
6. Xiang owns a takeaway restaurant and wants to employ a delivery company. Xiang uses two companies, \(A\) and \(B\), for one week each and records the delivery rate, \(x\) minutes per km, for each delivery.
Company \(A\) made 61 deliveries with \(\bar{x}_A = 3.7\) and \(s_A^{\,2} = 1.64\)
Company \(B\) made 51 deliveries with \(\sum x_B = 142.8\) and \(\sum x_B^{\,2} = 461.34\)
Xiang wants to carry out a two-sample \(t\)-test to determine whether there is a difference in the mean delivery rate for these companies.
Xiang noticed that a greater proportion of company \(A\)’s deliveries were in the town, with lots of traffic, whereas a greater proportion of company \(B\)’s deliveries were rural, with less traffic.
Xiang decides to carry out a further test. He selects a random sample of 10 customers who regularly place an order at the same time each week. He asks company \(A\) to deliver these orders one week and company \(B\) the next week. There were no roadworks or other incidents affecting travel time in either week.
The delivery rates are given in the table below.
| Customer | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Company \(A\) | 5.4 | 3.6 | 2.5 | 4.7 | 2.3 | 3.2 | 3.4 | 4.1 | 5.9 | 4.5 |
| Company \(B\) | 4.9 | 3.9 | 2.3 | 4.5 | 2.6 | 2.8 | 3.1 | 3.8 | 5.2 | 4.3 |
You should clearly state your hypotheses, test statistic and critical value. (6)
| Scheme | Marks | AO |
|---|---|---|
| \(\bar{x}_B = 2.8\) oe | B1 | 1.1b |
| \(s^2{}_B = \dfrac{461.34 - 51 \times 2.8^2}{50}\) ; = 1.23 oe | M1;A1 | 1.1b;1.1b |
| (3) |
Notes
B1: for 2.8 (may be seen in a calculation but must be seen somewhere) oe
M1: for a correct expression for \(s^2\) Can ft their mean or implied by correct answer.
A1: for 1.23 oe
| Scheme | Marks | AO |
|---|---|---|
| Delivery rates must be normally distributed with equal variances | B1 | 2.4 |
| (1) |
Notes
B1: for mention of “normal” and “equal variance” (or standard deviation)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \mu_A = \mu_B \qquad \mathrm{H}_1 : \mu_A \ne \mu_B\) oe | B1 | 2.5 |
| \(s_p^{\,2} = \dfrac{1.64 \times 60 + \text{“}1.23\text{”} \times 50}{110}\) ; \(= 1.4536\ldots\) awrt 1.45 | M1;A1 | 1.1b;1.1b |
| \(t_{110} = \pm\dfrac{3.7 - \text{“}2.8\text{”}}{\sqrt{\text{“}1.45\ldots\text{”}\left(\frac{1}{61} + \frac{1}{51}\right)}} = 3.93419\ldots\) awrt 3.93 or 3.94 | M1;A1 | 3.4;1.1b |
| 5% 2-tail cv \(= \pm 1.982\) | B1 | 1.1b |
| [significant result] there is evidence of a difference in mean delivery rates oe | A1 | 2.2b |
| (7) |
Notes
1st B1: for correct hypotheses, both in terms of \(\mu\). Can be \(\mu_1\) and \(\mu_2\) etc
1st M1: for correct attempt at \(s_p^{\,2}\) ft their value for \(B\) ;
1st A1: for awrt 1.45 (implied by 2nd A1) \(\left(\frac{1599}{1100}\right)\)
2nd M1: for correct ft expression for test statistic using 3.7 allow (\(\pm\)) standardisation ;
2nd A1: for awrt (\(\pm\)) 3.93 or awrt (\(\pm\)) 3.94
2nd B1: for (\(\pm\))1.982 (or better)
3rd A1: dep on all previous Ms and \(1 \lt |\text{cv}| \lt \) “3.93” for correct conclusion mentioning “mean delivery rate”. Do not allow contradictory statements. Incorrect comparisons A0.
| Scheme | Marks | AO |
|---|---|---|
| [Let \(D\) = delivery \(A\) – delivery \(B\)] \(\mathrm{H}_0 : \mu_D = 0 \qquad \mathrm{H}_1 : \mu_D \ne 0\) | B1 | 2.5 |
| \(d = 0.5,\ -0.3,\ 0.2,\ 0.2,\ -0.3,\ 0.4,\ 0.3,\ 0.3,\ 0.7,\ 0.2\) (may be opposite signs) | M1 | 2.1 |
| \(\left[\bar{x}_D = 0.22 \quad s_D = 0.31552\ldots\right]\ \ t_{[9]} = \pm\dfrac{\text{“}0.22\text{”} - 0}{\frac{\text{“}0.31552\ldots\text{”}}{\sqrt{10}}} = 2.2049\ldots\); awrt 2.20 | M1;A1 | 3.4;1.1b |
| Two-tail \(t_9(5\%)\) cv \(= \pm 2.262\) | B1 | 1.1b |
| [Not significant] insufficient evidence of a difference in delivery rates oe | A1 | 2.2b |
| (6) | ||
| (17 marks) |
Notes
1st B1: for both hypotheses in terms of \(\mu\) (or \(\mu_d\)) but not \(\mu_1\) and \(\mu_2\) etc
1st M1: for attempting differences (at least 6 correct values) implied by mean or \(s_D^{\,2}\) or \(s_D\)
2nd M: for a correct express’n for \(t\) (ignore df) with their values for \(\bar{x}_D\) and \(s_D\) (correct method)
1st A1: for awrt \(\pm 2.20\) or \(\pm 2.205\) (accept 2.2 if correct expression or \(\bar{x}_D\) and \(s_D\) seen)
2nd B1: for correct cv i.e. \(\pm 2.262\) or better
2nd A1: dep on all previous Ms and “2.20” \(\lt |\text{cv}| \lt 3\) for correct conclusion in context. Do not allow contradictory statements. “mean delivery rate” or incorrect comparison is A0.
SC: If they carry out a \(t\)-test for two independent samples then allow 2nd M1 for standardising with diff of their means and pooled variance and 2nd B1 for cv of \(\pm 2.101\)