S4 June 2015 Q4

EdexcelOld spec11 marksIncludes hypothesis testingQuality of Tests

4. A poultry farm produces eggs which are sold in boxes of 6. The farmer believes that the proportion, \(p\), of eggs that are cracked when they are packed in the boxes is approximately 5%. She decides to test the hypotheses

\[\mathrm{H}_0 : p = 0.05 \quad \text{against} \quad \mathrm{H}_1 : p \gt 0.05\]

To test these hypotheses she randomly selects a box of eggs and rejects \(\mathrm{H}_0\) if the box contains 2 or more eggs that are cracked. If the box contains 1 egg that is cracked, she randomly selects a second box of eggs and rejects \(\mathrm{H}_0\) if it contains at least 1 egg that is cracked. If the first or the second box contains no cracked eggs, \(\mathrm{H}_0\) is immediately accepted and no further boxes are sampled.

(a) Show that the power function of this test is \[1 - (1 - p)^6 - 6p(1 - p)^{11}\] (3)

(b) Calculate the size of this test. (2)

Given that \(p = 0.1\)

(c) find the expected number of eggs inspected each time this test is carried out, giving your answer correct to 3 significant figures, (3)

(d) calculate the probability of a Type II error. (2)

Given that \(p = 0.1\) is an unacceptably high value for the farmer,

(e) use your answer from part (d) to comment on the farmer’s test. (1)

Mark scheme (a)

Scheme	Marks
Power function \(= \mathrm{P}(\mathrm{H}_0 \text{ rejected}) = \mathrm{P}(X_1 \geqslant 2) + \mathrm{P}(X_1 = 1) \times \mathrm{P}(X_2 \geqslant 1)\) \(= 1 - (1 - p)^6 - 6p(1 - p)^5 + 6p(1 - p)^5 \times (1 - (1 - p)^6)\) \(= 1 - (1 - p)^6 - 6p(1 - p)^5 + 6p(1 - p)^5 - 6p(1 - p)^{11}\)	M1A1
\(= 1 - (1 - p)^6 - 6p(1 - p)^{11}\)	A1cso
	(3)

Notes

M1 for \(\mathrm{P}(X_1 \geqslant 2) + \mathrm{P}(X_1 = 1) \times \mathrm{P}(X_2 \geqslant 1)\) or \(1 - \left(\mathrm{P}(X_1 = 0) + \mathrm{P}(X_1 = 1) \times \mathrm{P}(X_2 = 0)\right)\) oe or a correct line of working

A1 a correct line of working before the final answer
A1 fully correct solution no errors.

Mark scheme (b)

Scheme	Marks
Size of test is value of power function when \(p = 0.05\) Size of test \(= 1 - 0.95^6 - 6 \times 0.05 \times 0.95^{11} = 0.094268\ldots\) (awrt 0.0943)	M1A1
	(2)

Notes

M1 attempt to subst 0.05 into (a)

Mark scheme (c)

Scheme	Marks
E[number of eggs inspected] \(= 12 \times \mathrm{P}(X_1 = 1) + 6 \times \mathrm{P}(X_1 \ne 1)\)	M1
\(= 12 \times 6 \times 0.1 \times 0.9^5 + 6 \times (1 - (6 \times 0.1 \times 0.9^5))\)	A1
\(= 8.1257\ldots\) (awrt 8.13)	A1
	(3)

Notes

M1 for \(12 \times \mathrm{P}(X_1 = 1) + 6 \times \mathrm{P}(X_1 \ne 1)\)

A1 \(12 \times 6 \times p \times 0.9(1 - p)^5 + 6 \times (1 - (6 \times p \times (1 - p)^5)\)

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(\text{Type II error} \mid p = 0.1) = 1 - (\text{value of power function when } p = 0.1)\)	M1
\(\mathrm{P}(\text{Type II error} \mid p = 0.1) = 1 - (1 - 0.9^6 - 6 \times 0.1 \times 0.9^{11}) = 0.7197\ldots\) (awrt 0.720)	A1
	(2)

Notes

M1 \(1 - (1 - (1 - p)^6 - 6 \times p \times (1 - p)^{11})\)

Mark scheme (e)

Scheme	Marks
Prob of Type II error, accepting \(p = 0.05\) when it is actually 0.1, unacceptably high, is large, therefore not a good test.	B1
	(1)
	(11 marks)

Notes

B1 idea that the Probability of a Type II error is too high or the power is too low so the test is not good/powerful or test needs changing