S3 June 2008 Q6

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

6. Ten cuttings were taken from each of 100 randomly selected garden plants. The numbers of cuttings that did not grow were recorded.

The results are as follows

No. of cuttings which did not grow	0	1	2	3	4	5	6	7	8, 9 or 10
Frequency	11	21	30	20	12	3	2	1	0

(a) Show that the probability of a randomly selected cutting, from this sample, not growing is 0.223 (2)

A gardener believes that a binomial distribution might provide a good model for the number of cuttings, out of 10, that do not grow.

He uses a binomial distribution, with the probability 0.2 of a cutting not growing. The calculated expected frequencies are as follows

No. of cuttings which did not grow	0	1	2	3	4	5 or more
Expected frequency	\(r\)	26.84	\(s\)	20.13	8.81	\(t\)

(b) Find the values of \(r\), \(s\) and \(t\). (4)

(c) State clearly the hypotheses required to test whether or not this binomial distribution is a suitable model for these data. (2)

The test statistic for the test is 4.17 and the number of degrees of freedom used is 4.

(d) Explain fully why there are 4 degrees of freedom. (2)

(e) Stating clearly the critical value used, carry out the test using a 5% level of significance. (3)

Mark scheme (a)

Scheme	Marks
\(p = \dfrac{0 \times 11 + 1 \times 21 + \ldots}{10 \times (11 + 21 + \ldots) \text{ or } 10 \times 100},\ = \dfrac{223}{1000} = 0.223\) (*) (Accept \(\dfrac{223}{1000}\))	M1, A1cso
	(2)

Notes

M1 Must show clearly how to get either 223 or 1000. As printed or better.
A1cso for showing how to get both 223 and 1000 and reaching \(p = 0.223\)

Mark scheme (b)

Scheme	Marks
\(r = (0.8)^{10} \times 100 = 10.7374\) awrt 10.74	M1A1
\(s = \dbinom{10}{2}(0.8)^8 \times (0.2)^2 \times 100 = 30.198\ldots\) awrt 30.2	A1
\(t = 100 - [r + s + 26.84 + 20.13 + 8.81] =\) awrt 3.28	A1cao
	(4)

Notes

M1 for any correct method (a correct expression) seen for \(r\) or \(s\).
1^st A1 for correct value for \(r\) awrt 10.74
2^nd A1 for \(s\) = awrt 30.2
3^rd A1 for \(t = 3.28\) only

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0\) : Binomial ([\(n = 10\)], \(p = 0.2\)) is a suitable model for these data	B1
\(\mathrm{H}_1\) : Binomial ([\(n = 10\)], \(p = 0.2\)) is NOT a suitable model for these data	B1
	(2)

Notes

B1 for each. The value of \(p\) must be mentioned at least once. Accept B(10, 0.2)
If hypotheses are correct but with no value of \(p\) then score B0B1
Minimum is \(X \sim \mathrm{B}(10, 0.2)\). If just B(10, 0.2) and not B(10, 0.2) award B1B0

Mark scheme (d)

Scheme	Marks
Since \(t \lt 5\), the last two groups are combined	M1
and \(\nu = 4 = 5 - 1\)	A1
	(2)

Notes

M1 for combining groups (must be stated or implied by a new table with combined cell seen)
A1 for the calculation 4 = 5 - 1

Mark scheme (e)

Scheme	Marks
Critical value \(\chi^2_4(5\%) = 9.488\)	B1
Not significant or do not reject null hypothesis	M1
The binomial distribution with \(p = 0.2\) is a suitable model for the number of cuttings that do not grow	A1
	(3)
	(13 marks)

Notes

M1 for a correct statement based on 4.17 and their cv (context not required) (may be implied)
Use of 4.17 as a critical value scores B0M0A0
A1 for a correct interpretation in context and \(p = 0.2\) and cuttings mentioned.