AS June 2018 Q4

EdexcelAS paperCurrent spec7 marksIncludes hypothesis testingChi-Squared Tests

4. Abram carried out a survey of two treatments for a plant fungus. The contingency table below shows the results of a survey of a random sample of 125 plants with the fungus.

		Treatment
		No action	Plant sprayed once	Plant sprayed every day
Outcome	Plant died within a month	15	16	25
	Plant survived for 1 – 6 months	8	25	10
	Plant survived beyond 6 months	7	14	5

Abram calculates expected frequencies to carry out a suitable test. Seven of these are given in the partly-completed table below.

		Treatment
		No action	Plant sprayed once	Plant sprayed every day
Outcome	Plant died within a month			17.92
	Plant survived for 1 – 6 months	10.32	18.92	13.76
	Plant survived beyond 6 months	6.24	11.44	8.32

The value of \(\displaystyle\sum \frac{(O-E)^2}{E}\) for the 7 given values is 8.29

Test at the 2.5% level of significance, whether or not there is an association between the treatment of the plants and their survival. State your hypotheses and conclusion clearly. (7)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\): There is no association between the treatment of the plants and their survival/outcome.
\(\mathrm{H}_1\): There is an association between the treatment of the plants and their survival/outcome

3.4

	No action	Plant sprayed once	Plant sprayed every day
Plant died within a month	13.44	24.64	17.92
Plant survived for 1 – 6 months	10.32	18.92	13.76
Plant survived beyond 6 months	6.24	11.44	8.32

M1
A1

1.1b
1.1b

\(\displaystyle\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(15 - \text{“}13.44\text{”})^2}{\text{“}13.44\text{”}} + \frac{(16 - \text{“}24.64\text{”})^2}{\text{“}24.64\text{”}} + 8.29\)

1.1b

awrt 11.5

1.1b

Degrees of freedom \((3-1)(3-1) = 4\)
\(\chi^2_{4,0.025} = 11.143\)

3.1b

Reject \(\mathrm{H}_0\) There is an association between the treatment of the plants and their survival/outcome

dA1ft

2.2b

(7 marks)

Notes

B1: For correct hypotheses at least one in context. Allow independent and not independent. Do not accept correlation.

M1: For attempt at \(\dfrac{(\text{Row Total})(\text{Column Total})}{(\text{Grand Total})}\) to find expected frequencies. ( they may put numbers in table)

A1: awrt 13.44 and 24.64 This may be implied by a correct value of \(\chi^2\)

M1: For applying \(\displaystyle\sum \frac{(O-E)^2}{E}\) ft their expected values. If no method shown at least 1 of the two missing \(\chi^2\) contributions must be correct – you may need to check this (correct ones are \(0.181\ldots\) and \(3.0296\ldots\) allow 2sf) (condone missing 8.29)

A1: awrt 11.5

M1: For using degrees of freedom to set up \(\chi^2\) model critical value, implied by CV 11.143 or better

dA1ft: dependent on the 2^nd and 3^rd M marks. Correct conclusion ft their \(\displaystyle\sum \frac{(O-E)^2}{E}\) there is an association between the treatment of the plants and their survival/outcome: - do not allow contradicting statements. Do not award if hypotheses are the wrong way round or there are no hypotheses.