## Question 4(a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected frequency $= \frac{42}{80} \times 29 = 15.225$ | B1 | For 15.225 |
| Contribution $= \frac{(12-15.225)^2}{15.225}$ | M1 | For valid attempt at $\frac{(O-E)^2}{E}$ |
| $= 0.6831$ AG | A1 [3] | Leading to correct answer. **NB Answer given** |

## Question 4(a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no association between sex and attitude to Mathematics; $H_1$: some association between sex and attitude to Mathematics | B1 | For correct hypotheses in context (with context seen in at least one hypothesis). NB if $H_0$, $H_1$ reversed do not award first B1 or final A1. Allow hypotheses expressed in terms of independence, and in context |
| Test statistic $X^2 = 5.3236$ | B1 | Allow 5.324 or 5.32 |
| Refer to $\chi^2_2$ | B1 | Allow "2 degrees of freedom" or $\nu = 2$ seen |
| Critical value at 5% level $= 5.991$ | B1 | No further marks from here if wrong or omitted |
| $(5.3236 < 5.991$ so result is) not significant | M1 | For not significant oe. FT their test statistic. Allow 'Accept $H_0$' or 'Reject $H_1$' |
| There is insufficient evidence **to suggest** that there is association between sex and attitude to Mathematics | A1 [6] | For **non-assertive** conclusion in context. FT their test statistic. Do not allow "relationship" or "correlation" for "association" |

## Question 4(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 373/60 = 6.217$ | B1 | Allow 6.22 |
| $s = \sqrt{\frac{2498-(373)^2/60}{59}} = \sqrt{\frac{179.183}{59}} = \sqrt{3.0370} = 1.743$ | M1 | For correctly structured calculation (divisor $= 59$) for the sample standard deviation or variance |
| | A1 | Allow answers which round to 1.74 |
| $H_0: \mu = 5.64$; $H_1: \mu > 5.64$ where $\mu$ denotes the **mean** radioactivity level in (the population of) limpets | B1 | For both hypotheses correct |
| | B1 | For definition of $\mu$ in context. Do not allow other symbols unless clearly defined as population mean |
| Test statistic $= \frac{6.217-5.64}{1.743/\sqrt{60}} = \frac{0.5767}{0.2250} = 2.563$ | M1* | Structure of test statistic using their sd and mean. Must include correct use of $\sqrt{60}$. Do not condone numerator reversed |
| $= 2.563$ | A1 | Allow answers between 2.56 and 2.57 inclusive |
| Upper 5% level 1-tailed critical value of $z = 1.645$ | B1 | For 1.645. No further marks from here if wrong |
| $2.563 > 1.645$ The result is… significant. There is sufficient evidence to reject $H_0$ | depM1* | For sensible comparison leading to a conclusion (even if incorrect). FT their test statistic |
| There is sufficient evidence **to suggest** that the **mean** level of radioactivity has increased | A1 A1 [11] | Correct conclusion. FT their test statistic. For correct **non-assertive** conclusion in words in context |

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