| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring calculation of sample mean and standard deviation, then applying the t-distribution formula. The two-part structure (calculate CI, then interpret) is standard. While it involves multiple steps, each is routine for Further Maths students and requires no novel insight—slightly easier than average due to its textbook nature. |
| Spec | 5.05d Confidence intervals: using normal distribution |
**Question 2(i) and 2(ii)**
**(i)**
$\bar{x} = \dfrac{114}{20} = 5.7$, $\quad s^2 = \dfrac{2.382}{19} = 0.1254$ — B1B1
$t_{19} = 2.5395$ — B1
$98\%$ c.l.: $5.7 \pm \left(2.5395 \times \dfrac{\sqrt{0.1254}}{\sqrt{20}}\right)$ — M1 (Normal: B2B0M1A0)
$98\%$ C.I. is $\mathbf{(5.499,\ 5.901)}$ — A1A1
**[6]**
**(ii)**
5.5 is (just) within the confidence interval. Some evidence to suggest that the average pH is 5.5 in villages where rhododendrons grow well. — B1FT (Relate to CI), B1FT (Conclusion, FT on their confidence interval)
**[2]**2 The pH value, $X$, which is a measure of acidity, was measured for soil taken from a random sample of 20 villages in which rhododendrons grow well. The results are summarised below, where $\bar { x }$ denotes the sample mean. You may assume that the sample is selected from a normal population.
$$\Sigma x = 114 \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 2.382$$
(i) Calculate a $98 \%$ confidence interval for the mean pH value in villages where rhododendrons grow well, giving 3 decimal places in your answer.\\
(ii) Comment, justifying your answer, on a suggestion that the average pH value in villages where rhododendrons grow well is 5.5.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q2 [8]}}