S4 June 2015 Q5

EdexcelOld spec9 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

5. A researcher is investigating the accuracy of IQ tests. One company offers IQ tests that it claims will give any individual’s IQ with a standard deviation of 5

The researcher takes these tests 9 times with the following results

123, 118, 127, 120, 134, 120, 118, 135, 121

(a) Find the sample mean, \(\bar{x}\), and the sample variance, \(s^2\), of these scores. (2)

Given that any individual’s IQ scores on these tests are independent and have a normal distribution,

(b) use the hypotheses \[\mathrm{H}_0 : \sigma^2 = 25 \quad \text{against} \quad \mathrm{H}_1 : \sigma^2 \gt 25\] to test the company’s claim at the 5% significance level. (4)

Gurdip works for the company and has taken these IQ tests 12 times. Gurdip claims that the sample variance of these 12 scores is \(s^2 = 8.17\)

(c) Use this value of \(s^2\) to calculate a 95% confidence interval for the variance of Gurdip’s IQ test scores.
[You may use \(\mathrm{P}(\chi^2_{11} \gt 3.816) = 0.975\) and \(\mathrm{P}(\chi^2_{11} \gt 21.920) = 0.025\)] (2)

(d) Assuming that \(\sigma^2 = 25\), comment on Gurdip’s claim. (1)

Mark scheme (a)

Scheme	Marks
\(\bar{x} = \dfrac{\sum x}{n} = \dfrac{1116}{9} = 124\)	B1
\(s^2 = \dfrac{9}{8}\left(\dfrac{138728}{9} - 124^2\right) = 43\) Or \(s^2 = \dfrac{1}{8}\left(138728 - \dfrac{1116^2}{9}\right) = 43\)	B1
	(2)

Notes

B1 124
B1 43

Mark scheme (b)

Scheme	Marks
Test stat \(\chi^2 = \dfrac{8 \times 43}{25} = 13.76\)	M1A1
Critical value \(\chi^2 = 15.507\)	B1
Therefore not in critical region, insufficient evidence to reject \(\mathrm{H}_0\) There is evidence at the 5% level that the company’s claim is supported	B1d
	(4)

Notes

M1 \(\dfrac{8 \times \text{their } 43}{25}\)

A1 awrt 13.8

B1 15.507

B1 dep on previous M1 being awarded. Allow the standard deviation of the IQ scores is 5 oe. Must have IQ

Mark scheme (c)

Scheme	Marks
CI given by \(\dfrac{11 \times 8.17}{21.920} \lt \sigma^2 \lt \dfrac{11 \times 8.17}{3.816}\)	M1
Therefore \(4.0999\ldots \lt \sigma^2 \lt 23.55\ldots\) awrt 4.10 and 23.6	A1
	(2)

Notes

M1 \(\dfrac{11 \times 8.17}{3.816 \text{ or } 21.92}\)

A1 both correct

Mark scheme (d)

Scheme	Marks
\(\sigma^2 = 25\) is not in CI which suggests Gurdip’s(his) claim may not be true.	B1ft
	(1)
	(9 marks)

Notes

B1ft their interval from part(c). Gurdip’s claim may not be true
NB, no interval in (c) then B0