S4 June 2012 Q2

EdexcelOld spec16 marksIncludes hypothesis testingt-Tests & Further Tests

2. A biologist investigating the shell size of turtles takes random samples of adult female and adult male turtles and records the length, \(x\) cm, of the shell. The results are summarised below.

	Number in sample	Sample mean \(\bar{x}\)	\(\sum x^2\)
Female	6	19.6	2308.01
Male	12	13.7	2262.57

You may assume that the samples come from independent normal distributions with the same variance.

The biologist claims that the mean shell length of adult female turtles is 5 cm longer than the mean shell length of adult male turtles.

(a) Test the biologist’s claim at the 5% level of significance. (10)

(b) Given that the true values for the variance of the population of adult male turtles and adult female turtles are both 0.9 cm²,

(i) show that when samples of size 6 and 12 are used with a 5% level of significance, the biologist’s claim will be accepted if \(4.07 \lt \bar{X}_F - \bar{X}_M \lt 5.93\) where \(\bar{X}_F\) and \(\bar{X}_M\) are the mean shell lengths of females and males respectively.

(ii) Hence find the probability of a type II error for this test if in fact the true mean shell length of adult female turtles is 6 cm more than the mean shell length of adult male turtles. (6)

Mark scheme (a)

Scheme	Marks
\(S_F^2 = \dfrac{1}{5}\{2308.01 - 6 \times 19.6^2\} = 0.61\)	B1
\(S_M^2 = \dfrac{1}{11}\{2262.57 - 12 \times 13.7^2\} = 0.93545..\)	B1
\(\mathrm{H}_0 : \mu_F = \mu_M + 5;\ \mathrm{H}_1 : \mu_F \neq \mu_M + 5\) both	B1
CR: \(t_{16}(0.025) \gt 2.120\) 2.12	B1
\(S_p^2 = \dfrac{5 \times 0.61 + 11 \times 0.93545\ldots}{16} = 0.83375\)	M1 A1
\(t = \dfrac{19.6 - 13.7 - 5}{\sqrt{0.83375\left(\frac{1}{6} + \frac{1}{12}\right)}} = 1.971\)	M1 A1ftA1
Since 1.971 is not in the critical region we accept \(\mathrm{H}_0\) and conclude that the mean shell length of female turtles does exceed the shell length of male turtles by 5cm.(or Biologists claim is correct)	A1 ft
	(10)

Notes

B1 – awrt 0.61
B1 – awrt 0.935
Both may be implied by correct \(t\) value or \(S_p\)
B1 allow rearrangements eg \(\mu_F - \mu_M = 5\). If \(M\) and \(F\) not used then they must make clear what each letter is.
B1 CV (if using one tail test allow 1.746)
M1 \(\dfrac{5 \times \text{their }0.61 + 11 \times \text{their }0.93545\ldots}{16}\)
A1 awrt 0.834
M1 \(\pm\left(\dfrac{19.6 - 13.7 - 5}{\sqrt{p\left(\frac{1}{6} + \frac{1}{12}\right)}}\right)\) where \(p\) is either their 0.61 or 0.94 or their \(S_p^2\) (awrt 0.834) (Allow 13.7 - 19.6 - 5)
A1 ft their \(S_p^2\)
A1 awrt 1.97

Mark scheme (b)

Scheme	Marks
(i) \(-1.96 \lt \dfrac{\bar{X}_F - \bar{X}_M - 5}{\sqrt{\left(\frac{0.9}{6} + \frac{0.9}{12}\right)}} \lt 1.96\)	B1 M1
\(4.07 \lt \bar{X}_F - \bar{X}_M \lt 5.93\)	A1cso
(ii) \(\mathrm{P}(\text{Type II error}) = \mathrm{P}(4.07 \lt \bar{X}_F - \bar{X}_M \lt 5.93 \mid \mathrm{N}(6, 0.225))\)	M1
\(= \mathrm{P}\left(\dfrac{4.07 - 6}{\sqrt{0.225}} \lt z \lt \dfrac{5.93 - 6}{\sqrt{0.225}}\right)\)	M1
\(= 0.44\) awrt 0.44	A1
	(6)
	(16 marks)

Notes

(i) B1 1.96
M1 must use z value

(ii) M1 writing or using N(6, 0.225)
M1 finding correct area and standardising (must use 6 but allow use of 0.9 and (0.9/18) for var)

(In the printed notes these two sets are labelled (b) and (c).)