S2 June 2009 Q4

EdexcelOld spec8 marksIncludes hypothesis testingBinomial Distribution

4. Past records suggest that 30% of customers who buy baked beans from a large supermarket buy them in single tins. A new manager questions whether or not there has been a change in the proportion of customers who buy baked beans in single tins. A random sample of 20 customers who had bought baked beans was taken.

(a) Using a 10% level of significance, find the critical region for a two-tailed test to answer the manager’s question. You should state the probability of rejection in each tail which should be less than 0.05. (5)

(b) Write down the actual significance level of a test based on your critical region from part (a). (1)

The manager found that 11 customers from the sample of 20 had bought baked beans in single tins.

Mark scheme (a)

Scheme	Marks
\(X \sim \mathrm{B}(20, 0.3)\)	M1
\(\mathrm{P}(X \leqslant 2) = 0.0355\)	A1
\(\mathrm{P}(X \leqslant 9) = 0.9520\) so \(\mathrm{P}(X \geqslant 10) = 0.0480\)	A1
Therefore the critical region is \(\{X \leqslant 2\} \cup \{X \geqslant 10\}\)	A1A1
	(5)

Notes

M1 for B(20,0.3) seen or used

1^st A1 for 0.0355

2^nd A1 for 0.048

3^rd A1 for \((X) \leqslant 2\) or \((X) \lt 3\) or [0,2] They get A0 if they write \(\mathrm{P}(X \leqslant 2/\ X \lt 3)\)

4^th A1 \((X) \geqslant 10\) or \((X) \gt 9\) or [10,20] They get A0 if they write \(\mathrm{P}(X \geqslant 10/\ X \gt 9)\)
\(\mathbf{10} \leqslant X \leqslant 20\) etc is accepted (corrected from the printed mark scheme: printed \(10 \leqslant X \leqslant 2\))

To describe the critical regions they can use any letter or no letter at all. It does not have to be \(X\).

Scheme	Marks
\(0.0355 + 0.0480 = 0.0835\) awrt (0.083 or 0.084)	B1
	(1)

Mark scheme (c)

Scheme	Marks
11 is in the critical region	B1ft
there is evidence of a change/ increase in the proportion/number of customers buying single tins	B1ft
	(2)
	(8 marks)

Notes

1^st B1 for a correct statement about 11 and their critical region.

2^nd B1 for a correct comment in context consistent with their CR and the value 11

Alternative solution
1^st B0 \(P(X \geqslant 11) = 1 - 0.9829 = 0.0171\) since no comment about the critical region
2^nd B1 a correct contextual statement.