S3 June 2008 Q1

EdexcelOld spec8 marksConfidence Intervals

1. Some biologists were studying a large group of wading birds. A random sample of 36 were measured and the wing length, \(x\) mm of each wading bird was recorded. The results are summarised as follows

\[\sum x = 6046 \qquad \sum x^2 = 1\,016\,338\]

(a) Calculate unbiased estimates of the mean and the variance of the wing lengths of these birds. (3)

Given that the standard deviation of the wing lengths of this particular type of bird is actually 5.1 mm,

(b) find a 99% confidence interval for the mean wing length of the birds from this group. (5)

Mark scheme (a)

Scheme	Marks
\(\bar{x} = \left(\dfrac{6046}{36} =\right) 167.94\ldots\) awrt 168	B1
\(s^2 = \dfrac{1016338 - 36 \times \bar{x}^2}{35}\)	M1
\(= 27.0253\ldots.\) awrt 27.0 (Accept 27)	A1
	(3)

Notes

M1 for a correct expression for \(s^2\), follow through their mean, beware it is very “sensitive”
\(167.94 \to \dfrac{999.63..}{35} \to 28.56\ldots\)
\(167.9 \to \dfrac{1483.24..}{35} \to 42.37\ldots\)
\(168 \to \dfrac{274}{35} \to 7.82\)
These would all score M1A0

Use of 36 as the divisor (= 26.3… ) is M0A0

Mark scheme (b)

Scheme	Marks
99% Confidence Interval is: \(\bar{x} \pm 2.5758 \times \dfrac{5.1}{\sqrt{36}}\)	M1A1ft
2.5758	B1
\(= (165.755\ldots, 170.133\ldots)\) awrt (166, 170)	A1 A1
	(5)
	(8 marks)

Notes

M1 for substituting their values in \(\bar{x} \pm z \times \dfrac{5.1 \text{ or } s}{\sqrt{36}}\) where \(z\) is a recognizable value from tables

1^st A1 follow through their mean and their \(z\) (to 2dp) in \(\bar{x} \pm z \times \dfrac{5.1}{\sqrt{36}}\)

Beware: \(167.94 \pm 2.5758 \times \dfrac{5.1^2}{36} \to (166.07.., 169.8..)\) but scores B1M0A0A0A0

Correct answer only in (b) scores 0/5
2^nd & 3^rd A marks depend upon 2.5758 and M mark.