Date | May 2014 | Marks available | 11 | Reference code | 14M.1.hl.TZ0.7 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Determine and Find | Question number | 7 | Adapted from | N/A |
Question
The weights, in grams, of 10 apples were measured with the following results:
\(212.2\) \(216.9\) \(209.0\) \(215.5\) \(215.9\) \(213.5\) \(208.9\) \(213.8\) \(216.4\) \(209.9\)
You may assume that this is a random sample from a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\).
(a) Giving all your answers correct to four significant figures,
(i) determine unbiased estimates for \(\mu \) and \({\sigma ^2}\);
(ii) find a \(95\%\) confidence interval for \(\mu \).
Another confidence interval for \(\mu \), \([211.5, 214.9]\), was calculated using the above data.
(b) Find the confidence level of this interval.
Markscheme
(a) (i) \(\bar x = {\text{213.2}}\) A1
\(s = 3.0728 \ldots \) (A1)
\({s^2} = 9.442\) A1
(ii) \([211.0, 215.4]\) A1A1
Note: Accept \(211\) in place of \(211.0\).
Note: Apart from the above note, accept any answers which round to the correct 4 significant figure answers.
[5 marks]
(b) use of the fact that the width of the interval is \(2t \times \frac{s}{{\sqrt n }}\) (A1)
so that \(3.4 = 2t \times \frac{{3.0728 \ldots }}{{\sqrt {10} }}\) M1
\(t = 1.749\) A1
degrees of freedom \( = 9\) (A1)
\({\text{P}}(T > 1.749) = 0.0571\) (M1)
confidence level \( = 1 - 2 \times 0.0571 = 0.886{\text{ }}(88.6\% )\) A1
Note: Award the \({\text{DF}} = 9\) (A1) mark if the following line has \(0.00337\) on the RHS.
Note: Accept any answer which rounds to \(88.6\%\).
[6 marks]