S4 June 2015 Q3

EdexcelOld spec14 marksIncludes hypothesis testingt-Tests & Further Tests

3. As part of their research two sports science students, Ali and Bea, select a random sample of 10 adult male swimmers and a random sample of 13 adult male athletes from local sports clubs. They measure the arm span, \(x\) cm, of each person selected.
The data are summarised in the table below

	\(n\)	\(s^2\)	\(\bar{x}\)
Swimmers	10	48	195
Athletes	13	161	186

The students know that the arm spans of adult male swimmers and of adult male athletes may each be assumed to be normally distributed.
They decide to share out the data analysis, with Ali investigating the means of the two distributions and Bea investigating the variances of the two distributions.

Ali assumes that the variances of the two distributions are equal. She calculates the pooled estimate of variance, \({s_p}^2\)

(a) Show that \({s_p}^2 = 112.6\) to 1 decimal place. (2)

Ali claims that there is no difference in the mean arm spans of adult male swimmers and of adult male athletes.

(b) Stating your hypotheses clearly, test this claim at the 10% level of significance. (5)

Bea believes that the variances of the arm spans of adult male swimmers and adult male athletes are not equal.

(c) Show that, at the 10% level of significance, the data support Bea’s belief. State your hypotheses and show your working clearly. (5)

Ali and Bea combine their work and present their results to their tutor, Clive.

(d) Explain why Clive is not happy with their research and state, with a reason, which of the tests in parts (b) and (c) is not valid. (2)

Mark scheme (a)

Scheme	Marks
\({s_p}^2 = \dfrac{12 \times 161 + 9 \times 48}{13 + 10 - 2} = \dfrac{2364}{21} = 112.571\ldots = 112.6\ (1\text{dp})\)	M1A1cso
	(2)

Notes

M1 for \(\dfrac{12 \times 161 + 9 \times 48}{13 + 10 - 2}\)

A1cso need to get awrt112.57 or \(\dfrac{2364}{21}\) then write 112.6

Mark scheme (b)

Scheme	Marks
To test \(\mathrm{H}_0 : \mu_s = \mu_a\) against \(\mathrm{H}_1 : \mu_s \ne \mu_a\) (o.e.)	B1
Test stat, \(t = \pm\dfrac{195 - 186}{\sqrt{112.57\ldots\left(\frac{1}{10} + \frac{1}{13}\right)}} = \pm 2.016\ldots\) (awrt2.02)	M1A1
Critical values, \(t_{21} = (\pm)1.721\)	B1
In critical region, therefore significant evidence to reject \(\mathrm{H}_0\) and accept \(\mathrm{H}_1\) Evidence of difference in mean arm span of adult male swimmers and adult male athletes or No evidence to support Ali’s claim.	A1
	(5)

Notes

M1 \(\dfrac{195 - 186}{\sqrt{112.6\left(\frac{1}{10} + \frac{1}{13}\right)}}\)

2nd B1 alternate method, \(p\) value of 0.0566 in place of critical value

Final A1 requires correct conclusion in context

Mark scheme (c)

Scheme	Marks
To test \(\mathrm{H}_0 : {\sigma_s}^2 = {\sigma_a}^2\) against \(\mathrm{H}_1 : {\sigma_s}^2 \ne {\sigma_a}^2\)	B1
Test stat, \(F_{12,9} = \dfrac{161}{48} = 3.354\ldots\) \(\left(\dfrac{1}{\mathrm{F}_{12,9}} = \dfrac{48}{161} = 0.2981\ldots\right)\)	M1A1
Critical value, \(F_{12,9} = 3.07\ (0.3257\ldots)\)	B1
In critical region, therefore significant evidence to reject \(\mathrm{H}_0\) and accept \(\mathrm{H}_1\) Evidence of difference in variance of arm span of adult male swimmers and adult male athletes or the data supports Bea’s belief	A1cso
	(5)

Notes

1st B1 allow \(\mathrm{H}_0 : \sigma_s = \sigma_a\) against \(\mathrm{H}_1 : \sigma_s \ne \sigma_a\)

M1 allow \(\dfrac{161^2}{48^2}\) if they write the formula down

Final A1 requires correct conclusion

Mark scheme (d)

Scheme	Marks
Should do test for variance first as equal variances is necessary assumption for \(t\) test for means	B1
but is not supported in (c), so result in (b) is invalid.	B1d
	(2)
	(14 marks)

Notes

1st B1 equal variances is necessary assumption (may be implied by saying not equal)

2nd B1d but not supported in (c)/(variances not equal) therefore (b) result invalid