S4 June 2014 (R) Q4

EdexcelOld spec12 marksIncludes hypothesis testingt-Tests & Further Tests

4. At the start of each academic year, a large college carries out a diagnostic test on a random sample of new students. Past experience has shown that the standard deviation of the scores on this test is 19.71

The admissions tutor claimed that the new students in 2013 would have more varied scores than usual. The scores for the students taking the test can be assumed to come from a normal distribution. A random sample of 10 new students was taken and the score \(x\), for each student was recorded. The data are summarised as \(\sum x = 619\) \(\sum x^2 = 42397\)

(a) Stating your hypotheses clearly, and using a 5% level of significance, test the admission tutor’s claim. (6)

The admissions tutor decides that in future he will use the same hypotheses but take a larger sample of size 30 and use a significance level of 1%.

(b) Use the tables to show that, to 3 decimal places, the critical region for \(S^2\) is \(S^2 \gt 664.281\) (3)

(c) Find the probability of a type II error using this test when the true value of the standard deviation is in fact 22.20 (3)

Mark scheme (a)

Scheme	Marks
\(s^2 = \dfrac{42397 - 10 \times \left(\dfrac{619}{10}\right)^2}{9} = 453.433\ldots\) = awrt 453	B1
\(\mathrm{H}_0 : \sigma = 19.71\) (or \(\sigma^2 = \ldots\)) \(\mathrm{H}_1 : \sigma \gt 19.71\) (or \(\sigma^2 \gt \ldots\))	B1
\(\dfrac{(n-1)s^2}{\sigma^2} \sim {\chi^2}_9\) test statistic = 10.5046… = awrt 10.5	M1A1
\({\chi^2}_9(0.05)\) cv = 16.919	B1
Not significant so insufficient evidence that the scores of the students are more varied than normal. Or Admission tutor’s claim is not supported	A1
	(6)

Notes

M1 for use of the correct test statistic

Mark scheme (b)

Scheme	Marks
\({\chi^2}_{29}(0.01)\) cv = 49.588	B1
Reject \(\mathrm{H}_0\) if \(\dfrac{29S^2}{19.71^2} \gt 49.588\)	M1
So critical region is \(S^2 \gt 664.281\ldots\) = awrt 664.281	A1cso
	(3)

Notes

M1 for use of a correct expression (LHS) only

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(\text{Type II error}) = \mathrm{P}(S^2 \lt 664.281\ldots \mid \sigma = 22.20)\) or \(\mathrm{P}\left({\chi_{29}}^2 \lt \dfrac{664.281\ldots \times 29}{22.20^2}\right)\)	M1 A1ft
\(= \mathrm{P}({\chi^2}_{29} \lt 39.088\ldots) = 0.90\) = awrt 0.90	A1
	(3)
	(12 marks)

Notes

M1 for a correct probability expression involving \(S^2\) or \({\chi^2}_{29}\). Ft their CR, may be implied by a correct answer

1st A1ft for a correct probability expression with \({\chi^2}_{29}\) but ft their CR, may be implied by a correct answer