S2 June 2010 Q6

EdexcelOld spec15 marksIncludes hypothesis testingBinomial Distribution

6. A company claims that a quarter of the bolts sent to them are faulty. To test this claim the number of faulty bolts in a random sample of 50 is recorded.

(a) Give two reasons why a binomial distribution may be a suitable model for the number of faulty bolts in the sample. (2)

(b) Using a 5% significance level, find the critical region for a two-tailed test of the hypothesis that the probability of a bolt being faulty is \(\dfrac{1}{4}\). The probability of rejection in either tail should be as close as possible to 0.025 (3)

In the sample of 50 the actual number of faulty bolts was 8.

(d) Comment on the company’s claim in the light of this value. Justify your answer. (2)

The machine making the bolts was reset and another sample of 50 bolts was taken. Only 5 were found to be faulty.

(e) Test at the 1% level of significance whether or not the probability of a faulty bolt has decreased. State your hypotheses clearly. (6)

Mark scheme (a)

Scheme	Marks
2 outcomes/faulty or not faulty/success or fail A constant probability Independence Fixed number of trials (fixed \(n\))	B1 B1
	(2)

Notes

B1 B1 one mark for each of any of the four statements. Give first B1 if only one correct statement given. No context needed.

Mark scheme (b)

Scheme	Marks
\(X \sim \mathrm{B}(50, 0.25)\) \(\mathrm{P}(X \leqslant 6) = 0.0194\) \(\mathrm{P}(X \leqslant 7) = 0.0453\) \(\mathrm{P}(X \geqslant 18) = 0.0551\) \(\mathrm{P}(X \geqslant 19) = 0.0287\)	M1
CR \(X \leqslant 6\) and \(X \geqslant 19\)	A1 A1
	(3)

Notes

M1 for writing or using B(50,0.25) also may be implied by both CR being correct. Condone use of P in critical region for the method mark.
A1 \((X) \leqslant 6\) o.e. [0,6] DO NOT accept \(\mathrm{P}(X \leqslant 6)\)
A1 \((X) \geqslant 19\) o.e. [19,50] DO NOT accept \(\mathrm{P}(X \geqslant 19)\)

Scheme	Marks
\(0.0194 + 0.0287 = 0.0481\)	M1A1
	(2)

Mark scheme (d)

Scheme	Marks
8(It) is not in the Critical region or 8(It) is not significant or 0.0916 > 0.025;	M1;
There is evidence that the probability of a faulty bolt is 0.25 or the company’s claim is correct.	A1ft
	(2)

Notes

M1 one of the given statements followed through from their CR.
A1 contextual comment followed through from their CR.
NB A correct contextual comment alone followed through from their CR.will get M1 A1

Mark scheme (e)

Scheme	Marks
\(\mathrm{H}_0 : p = 0.25 \quad \mathrm{H}_1 : p \lt 0.25\)	B1B1
\(\mathrm{P}(X \leqslant 5) = 0.0070\) or CR \(X \leqslant 5\)	M1A1
\(0.007 \lt 0.01\), 5 is in the critical region, reject H₀, significant.	M1
There is evidence that the probability of faulty bolts has decreased	A1ft
	(6)
	(15 marks)

Notes

B1 for H₀ must use \(p\) or \(\pi\) (pi)
B1 for H₁ must use \(p\) or \(\pi\) (pi)
M1 for finding or writing \(\mathrm{P}(X \leqslant 5)\) or attempting to find a critical region or a correct critical region
A1 awrt 0.007/CR \(X \leqslant 5\)

M1 correct statement using their Probability and 0.01 if one tail test or a correct statement using their Probability and 0.005 if two tail test.
The 0.01 or 0.005 needn’t be explicitly seen but implied by correct statement compatible with their H₁. If no H₁ given M0

A1 correct contextual statement follow through from their prob and H₁. Need faulty bolts and decreased.
NB A correct contextual statement alone followed through from their prob and H₁ get M1 A1