S3 June 2016 Q5

EdexcelOld spec12 marksIncludes hypothesis testingt-Tests & Further Tests

5. A doctor claims there is a higher mean lung capacity in people who exercise regularly compared to people who do not exercise regularly. He measures the lung capacity, \(x\), of 35 people who exercise regularly and 42 people who do not exercise regularly. His results are summarised in the table below.

	\(n\)	\(\bar{x}\)	\(s^2\)
Exercise regularly	35	26.3	12.2
Do not exercise regularly	42	24.8	10.1

(a) Test, at the 5% level of significance, the doctor’s claim. State your hypotheses clearly. (6)

(b) State any assumptions you have made in testing the doctor’s claim. (2)

The doctor decides to add another person who exercises regularly to his data. He measures the person’s lung capacity and finds \(x = 31.7\)

(c) Find the unbiased estimate of the variance for the sample of 36 people who exercise regularly. Give your answer to 3 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(\mathrm{H_0}\): \(\mu_e = \mu_n\), \(\mathrm{H_1}\): \(\mu_e \gt \mu_n\)	B1
\(z = \dfrac{26.3 - 24.8}{\sqrt{\frac{12.2}{35} + \frac{10.1}{42}}} = \dfrac{1.5}{\sqrt{0.58904\ldots}} = \dfrac{1.5}{0.76749\ldots}\)	M1M1
\(z = 1.9544\ldots\) awrt 1.95	A1
Critical value is 1.6449	B1
Reject \(\mathrm{H_0}\). Doctor’s claim is supported.	A1
	(6)

Notes

Both hyps, one tailed only oe.
Accept \(\mu_1, \mu_2\) or \(\mu_A, \mu_B\) etc if there is some indication of which is which.

M1 for correct method for standard error

M1 for whole expression

A1 awrt 1.95

B1 1.6449 or \(p = 0.974\ldots\) (>0.95)

A1 must mention doctor and claim or description of claim that includes ‘mean lung capacity’ and ‘exercise’.

ALT (a)

Scheme	Marks
M1 for \(\sqrt{\dfrac{12.2}{35} + \dfrac{10.1}{42}}\)
M1 for \(1.6449 = \dfrac{c}{\sqrt{\frac{12.2}{35} + \frac{10.1}{42}}}\)
A1 for awrt \(c = 1.26\) seen
B1 1.5

Mark scheme (b)

Scheme	Marks
Either assume \(\bar{X}\) has a normal distribution (for both samples) or assume sample sizes are large enough to use CLT Assume individual results are independent Assume \(\sigma^2 = s^2\) for both populations or a single general population	B1 B1
	(2)

Mark scheme (c)

Scheme	Marks
\(\bar{x} = \left(\dfrac{35 \times 26.3 + 31.7}{36} = \dfrac{952.2}{36}\right) 26.45\)	B1
For \(n = 35\), \(\sum x^2 = 34 \times 12.2 + 35 \times 26.3^2\ (= 24623.95)\)	M1
For \(n = 36\), \(s^2 = \dfrac{25628.84 - 36 \times 26.45^2}{35} = 12.661\ldots\) awrt 12.7	dM1A1
	(4)
	(12 marks)

Notes

M1 Attempt \(\sum x^2 = 34 \times 12.2 + 35 \times 26.3^2\)
or \(\sum (x - \bar{x})^2 = 34 \times 12.2 + 35(26.45 - 26.3)^2\ (= 415.5875)\)

dM1 \(s^2 = \dfrac{\sum x^2 + 31.7^2 - 36 \times 26.45^2}{35}\) or \(s^2 = \dfrac{415.5875 + (31.7 - 26.45)^2}{35}\)

A1 awrt 12.7