| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with common variance assumption. The question provides all necessary statistics, requires stating the normality assumption, and executing a routine hypothesis test procedure. The calculations are straightforward (pooled variance, test statistic, comparison with critical value), making it slightly easier than average for an S3 question. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Each popn (of yields) should be N dist. | B1 | Allow X and Y. Allow 'data'. NOT increase. NOT it. NOT sample. NOT just 'Normally distributed'. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu_x = \mu_y\), \(H_1: \mu_x > \mu_y\) | B1 | If in words, must have population. |
| \(S_p^2 = \frac{9 \times 87.96 + 9 \times 31.96}{18}\) | M1 | |
| \(= 59.96\) | A1 | |
| \(\frac{211.2 - 200.8}{\sqrt{59.96 \times \left(\frac{1}{10}+\frac{1}{10}\right)}}\) | M1, A1 | Allow \(\frac{1}{9} + \frac{1}{9}\) and/or incorrect mean for M1. Allow \(\frac{87.96}{10} + \frac{31.96}{10}\) for M1A1 |
| \(= 3.00\) | A1 | Allow 3 |
| \(CV = 2.552\) | B1 | \(p = 0.00382\) B1 |
| "\(3.00\)" \(>\) "\(2.552\)", reject \(H_0\) | M1 | Follow through both TS, CV for this mark. \(p < 0.01\) reject \(H_0\) M1 |
| There is evidence that the phosphorus treatment has increased the yield. | A1 | Follow through TS, but not CV. Contextualised, not over-assertive. Correct \(p\) and satisfactory conc. A1. |
| [9] |
## Question 8:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Each popn (of yields) should be N dist. | B1 | Allow X and Y. Allow 'data'. NOT increase. NOT it. NOT sample. NOT just 'Normally distributed'. |
| **[1]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_x = \mu_y$, $H_1: \mu_x > \mu_y$ | B1 | If in words, must have population. |
| $S_p^2 = \frac{9 \times 87.96 + 9 \times 31.96}{18}$ | M1 | |
| $= 59.96$ | A1 | |
| $\frac{211.2 - 200.8}{\sqrt{59.96 \times \left(\frac{1}{10}+\frac{1}{10}\right)}}$ | M1, A1 | Allow $\frac{1}{9} + \frac{1}{9}$ and/or incorrect mean for M1. Allow $\frac{87.96}{10} + \frac{31.96}{10}$ for M1A1 |
| $= 3.00$ | A1 | Allow 3 |
| $CV = 2.552$ | B1 | $p = 0.00382$ B1 |
| "$3.00$" $>$ "$2.552$", reject $H_0$ | M1 | Follow through both TS, CV for this mark. $p < 0.01$ reject $H_0$ M1 |
| There is evidence that the phosphorus treatment has increased the yield. | A1 | Follow through TS, but not CV. Contextualised, not over-assertive. Correct $p$ and satisfactory conc. A1. |
| **[9]** | | |
---8 A random sample of 20 plots of land, each of equal area, was used to test whether the addition of phosphorus would increase the yield of corn. 10 plots were treated with phosphorus and 10 plots were untreated. The yields of corn, in litres, on a treated plot and on an untreated plot are denoted by $X$ and $Y$ respectively. You are given that
$$\sum x = 2112 , \quad \sum y = 2008$$
You are also given that an unbiased estimate for the variance of treated plots is 87.96 and an unbiased estimate for the variance of untreated plots is 31.96 , both correct to 4 significant figures.\\
(i) You may assume that the population variance estimates are sufficiently similar for the assumption of common variance to be made. What other assumption needs to be made for a $t$-test to be valid?\\
(ii) Carry out a suitable $t$-test at the $1 \%$ significance level, to test whether the use of phosphorus increases the yield of corn.
\hfill \mbox{\textit{OCR S3 2014 Q8 [10]}}