S2 June 2013 (R) Q6

EdexcelOld spec14 marksIncludes hypothesis testingPoisson Distribution

6. Frugal bakery claims that their packs of 10 muffins contain on average 80 raisins per pack. A Poisson distribution is used to describe the number of raisins per muffin.

A muffin is selected at random to test whether or not the mean number of raisins per muffin has changed.

(a) Find the critical region for a two-tailed test using a 10% level of significance. The probability of rejection in each tail should be less than 0.05 (4)

(b) Find the actual significance level of this test. (2)

The bakery has a special promotion claiming that their muffins now contain even more raisins.

A random sample of 10 muffins is selected and is found to contain a total of 95 raisins.

(c) Use a suitable approximation to test the bakery’s claim. You should state your hypotheses clearly and use a 5% level of significance. (8)

Mark scheme (a)

Scheme	Marks
[\(X\) = the number of raisins in a mini-muffin]
\(X \sim \mathrm{Po}(8)\)	B1
e.g. \(\mathrm{P}(X \leqslant 3) = 0.0424\), \(\mathrm{P}(X \leqslant 13) = 0.9658\) so \(\mathrm{P}(X \geqslant 14) = 0.0342\)	M1
So Critical Region is \(X \leqslant 3\) or \(X \geqslant 14\)	A1 A1
	(4)

Notes

B1 for Po(8) seen or implied by use

M1 for clear evidence of use of Po(8), may be implied by a correct CR (allow written as a probability statement) or a probability seen in part(b). If they give 3 and 14

1st A1 for \(X \leqslant 3\) or \(0 \leqslant X \leqslant 3\) or 0,1,2,3 or [0,3] Allow any letter

2nd A1 for \(X \geqslant 14\) or \([14, \infty)\) condone \([14, \infty]\) Allow any letter

These A marks must be for statements with \(X\) only – not in prob statements

Mark scheme (b)

Scheme	Marks
\(0.0424 + 0.0342\)	M1
\(= \underline{\mathbf{0.0766}}\) (or better)	A1
	(2)

Notes

M1 for showing they are adding together the two probabilities that correspond to their CR or allow M1 A1for correct answer

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \lambda = 8\) (or \(\mu = 80\)) \(\qquad \mathrm{H}_1 : \lambda \gt 8\) (or \(\mu \gt 80\))	B1
[\(R\) = no. of raisins in 10 muffins. \(R \sim \mathrm{Po}(80)\).] Use \(Y \sim \mathrm{N}(80, 80)\)	M1A1
\(\mathrm{P}(R \geqslant 95) \simeq \mathrm{P}(Y \geqslant 94.5)\)	M1
\(= \mathrm{P}\left(Z \gt \dfrac{94.5 - 80}{\sqrt{80}}\right)\)	M1
\(= \mathrm{P}(Z \gt 1.62\ldots) = 1 - 0.9474 =\) awrt 0.053	A1
Probability is greater than 0.05 so not significant (accept \(\mathrm{H}_0\))	M1
Insufficient evidence to support the bakery’s claim Or insufficient evidence of an increase in the (mean) number of raisins per muffin	A1cso
	(8)
	(14 marks)

Notes

B1 for both hypotheses. Must be in terms of \(\lambda\) or \(\mu\), 8 or 80 can be swapped

1st M1 for normal approx

1st A1 E(Y) = 80 and Var(Y) = 80 (or correct st. dev seen somewhere)

2nd M1 for use of a continuity correction 94.5 or 95.5

3rd M1 Standardising using their mean and their sd, If they have not written down a mean and sd then these need to be correct here to award the mark. They must also use 94.5, 95.5 or 95 and find the correct area ie using \(1 - \mathrm{P}(Z \leqslant\) “their 1.62”)

2nd A1 for awrt 0.053 or awrt 0.947

4th M1 for a correct statement based on their probability and 0.05

3rd A1 cso for a correct contextualised statement and a fully correct solution with no errors seen. Need either bakery’s claim or Raisins and muffin

NB If Found \(\mathrm{P}(X = 95)\) they can get B1 M1 A1 M0M0A0M0A0