| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Validity and assumptions questions |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with summary statistics requiring calculation of means and variances, stating assumptions, performing the test, and commenting on validity. While it involves multiple steps, all techniques are routine S3 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Numbers of men and women should have normal distributions | B1 | Context & 3 points: 2 of these, B1; 3, B2; 4, B3 |
| with equal variance | B1 | Summary data: 14.73, 49.06, 52.57, 16.24, 62.18, 66.07 |
| distributions should be independent | B1 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_M = \mu_W\); \(H_1: \mu_M \neq \mu_W\) | B1 | Both hypotheses correctly stated |
| \(3992 - \frac{221^2}{15} + 5538 - \frac{276^2}{17} \approx 1793\) | M1 | Attempt at this expression |
| Either 1793 or 30 | A1 | |
| \(1793/(14+16) = 59.766\) | A1 | Variance estimate in range [59.7, 59.8] (or \(\sqrt{} = 7.73\)) |
| \((\pm)\frac{221/15 - 276/17}{\sqrt{59.766(\frac{1}{15}+\frac{1}{17})}} = (-)0.548\) | M1 | Standardise, allow wrong (but not missing) \(1/n\) |
| \(A1\sqrt{}\) | Correct formula, allow \(s^2(\frac{1}{15}+\frac{1}{17})\) or \((\frac{s_1^2}{15}+\frac{s_2^2}{17})\) | |
| A1 | allow 14 & 16 in place of 15, 17; 0.548 or −0.548 | |
| Critical region: \( | t | \geq 2.042\) |
| Do not reject \(H_0\). Insufficient evidence of a difference in mean number of days | M1 | Correct method and comparison type, must be \(t\), allow 1-tail; conclusion in context, not too assertive |
| \(A1\sqrt{}\) 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. Samples not independent so test invalid | B1 1 | Any relevant valid comment, e.g. "not representative" |
## Question 5:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Numbers of men and women should have normal distributions | B1 | Context & 3 points: 2 of these, B1; 3, B2; 4, B3 |
| with equal variance | B1 | Summary data: 14.73, 49.06, 52.57, 16.24, 62.18, 66.07 |
| distributions should be independent | B1 **3** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_M = \mu_W$; $H_1: \mu_M \neq \mu_W$ | B1 | Both hypotheses correctly stated |
| $3992 - \frac{221^2}{15} + 5538 - \frac{276^2}{17} \approx 1793$ | M1 | Attempt at this expression |
| Either 1793 or 30 | A1 | |
| $1793/(14+16) = 59.766$ | A1 | Variance estimate in range [59.7, 59.8] (or $\sqrt{} = 7.73$) |
| $(\pm)\frac{221/15 - 276/17}{\sqrt{59.766(\frac{1}{15}+\frac{1}{17})}} = (-)0.548$ | M1 | Standardise, allow wrong (but not missing) $1/n$ |
| | $A1\sqrt{}$ | Correct formula, allow $s^2(\frac{1}{15}+\frac{1}{17})$ or $(\frac{s_1^2}{15}+\frac{s_2^2}{17})$ |
| | A1 | allow 14 & 16 in place of 15, 17; 0.548 or −0.548 |
| Critical region: $|t| \geq 2.042$ | B1 | 2.042 seen |
| Do not reject $H_0$. Insufficient evidence of a difference in mean number of days | M1 | Correct method and comparison type, must be $t$, allow 1-tail; conclusion in context, not too assertive |
| | $A1\sqrt{}$ **10** | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. Samples not independent so test invalid | B1 **1** | Any relevant valid comment, e.g. "not representative" |
---5 Each person in a random sample of 15 men and 17 women from a university campus was asked how many days in a month they took exercise. The numbers of days for men and women, $x _ { M }$ and $x _ { W }$ respectively, are summarised by
$$\Sigma x _ { M } = 221 , \quad \Sigma x _ { M } ^ { 2 } = 3992 , \quad \Sigma x _ { W } = 276 , \quad \Sigma x _ { W } ^ { 2 } = 5538 .$$
(i) State conditions for the validity of a suitable test of the difference in the mean numbers of days for men and women on the campus.\\
(ii) Given that these conditions hold, carry out the test at the $5 \%$ significance level.\\
(iii) If in fact the random sample was drawn entirely from the university Mathematics Department, state with a reason whether the validity of the test is in doubt.
\hfill \mbox{\textit{OCR S3 2009 Q5 [14]}}