AS June 2025 Q1

EdexcelAS paperCurrent spec10 marksIncludes hypothesis testingChi-Squared Tests

1. A researcher is investigating the relationship between a person’s age and their preferred method of shopping.

A random sample of 300 people is taken and the results are summarised in the table below.

		Preferred method of shopping
		Online	In-store	Total
Age	18 – 30	27	23	50
	31 – 50	31	59	90
	51 – 64	17	43	60
	65 and older	30	70	100
	Total	105	195	300

(a) Write down suitable hypotheses for a test to determine whether or not there is evidence of an association between age and preferred method of shopping. (1)

The researcher assumes that the null hypothesis is true and uses the data in the table to find expected values.

(b)

(i) Identify the cell which would have the smallest expected value.

(ii) Calculate the expected value for this cell. (2)

The value of \(\displaystyle\sum \frac{(O-E)^2}{E}\) for the other 7 cells is 5.060

(c) Test, at the 5% level of significance, whether or not there is evidence of an association between age and preferred method of shopping.
You should state the test statistic, degrees of freedom, critical value and conclusion clearly. (5)

(d) Explain whether or not your conclusion to part (c) would be the same if the test was carried out at a 1% level of significance. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between age and preferred method of shopping. \(\mathrm{H}_1\): There is an association between age and preferred method of shopping.	B1	3.4
	(1)

Notes

B1: For both hypotheses in terms of "association" or “independence"
Must mention age and shopping in at least one and be connected correctly to \(\mathrm{H}_0\) and \(\mathrm{H}_1\)
Use of link, relationship, correlation or connection is B0 here.
Hypotheses must be given in part (a).

Scheme	Marks	AO
(i) Online 18 – 30 or 27	B1	2.2a
(ii) [\((50 \times 105) / 300 =\)] 17.5	B1	1.1b
	(2)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{(\text{‘}27\text{’}-\text{‘}17.5\text{’})^2}{\text{‘}17.5\text{’}} + 5.060\)	M1	1.1b
\(= 10.217\ldots\) awrt 10.2	A1	1.1b
df [\(= (4 - 1)(2 - 1)\)] \(= 3\)	B1	1.1b
[\(10.2 \gt\)] \(\chi^2_{3,(0.05)} = 7.815\)	B1ft	3.1b
[Reject \(\mathrm{H}_0\)] There is evidence of an association between age and preferred method of shopping.	B1ft	2.2b
	(5)

Notes

M1: For use of \(\dfrac{(O-E)^2}{E}\) with their cell + 5.060 (this mark may be implied by awrt 10.2)
Watch out for \(\dfrac{(O-E)^2}{O}\) + 5.060 which is M0

A1: awrt 10.2 (ignore p-value if given)

B1: 3 cao

B1ft: Using the degrees of freedom to find the \(\chi^2\) CV for the appropriate model 7.815 or better
ft their df 1→3.841, 2→5.991, 4→9.488, 5→11.070, 6→12.592, 7→14.067, 8→15.507

B1ft: Correct conclusion or ft conclusion based on their values in (c), in context (age and shopping).
Must be consistent with their CV and test statistic. Independent of their hypotheses.
i.e. test statistic > CV → there is association… test statistic < CV → there is no association…
Allow relationship, link, connection BUT do not accept correlation or contradictory statements

Mark scheme (d)

Scheme	Marks	AO
\(\chi^2\) CV is now 11.345 [\(\gt 10.2\)]…	M1	1.1b
… \(\mathrm{H}_0\) is not rejected/conclusion is reversed	A1ft	2.2b
	(2)
	(10 marks)

Notes

M1: Obtaining new \(\chi^2\) CV 11.345 or awrt 11.3 Condone \(p\) = awrt 0.017 [\(\gt 0.01\)] for this mark
ft their df (all awrt 3sf) 1→6.64, 2→9.21, 4→13.3, 5→15.1, 6→16.8, 7→18.5, 8→20.1

A1ft: ft deduction which must be consistent with their 1% CV and their test statistic (need not be in context). Do not allow contradictory statements.