S3 June 2014 Q5

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

5. A research station is doing some work on the germination of a new variety of genetically modified wheat.

They planted 120 rows containing 7 seeds in each row.

The number of seeds germinating in each row was recorded. The results are as follows

Number of seeds germinating in each row	0	1	2	3	4	5	6	7
Observed number of rows	2	6	11	19	25	32	16	9

(a) Write down two reasons why a binomial distribution may be a suitable model. (2)

(b) Show that the probability of a randomly selected seed from this sample germinating is 0.6 (2)

The research station used a binomial distribution with probability 0.6 of a seed germinating. The expected frequencies were calculated to 2 decimal places. The results are as follows

Number of seeds germinating in each row	0	1	2	3	4	5	6	7
Expected number of rows	0.20	2.06	\(s\)	23.22	\(t\)	31.35	15.68	3.36

(d) Stating your hypotheses clearly, test, at the 1% level of significance, whether or not the data can be modelled by a binomial distribution. (7)

Mark scheme (a)

Scheme	Marks
The seeds are independent / There are a fixed number of seeds in a row / There are only two outcomes to the seed germinating – either it germinates or it does not / The probability of a seed germinating is constant	B1 B1
	(2)

Notes

Any two and at least one must have context. 2 correct, no context B1B0. Do not award B0B1.

Mark scheme (b)

Scheme	Marks
\(\dfrac{(0 \times 2) + (1 \times 6) + (2 \times 11) + (3 \times 19) + (4 \times 25) + (5 \times 32) + (6 \times 16) + (7 \times 9)}{120 \times 7} = \dfrac{504}{840}\)	M1
\(= 0.6\) **	A1cso
	(2)

Notes

M1 require at least two correct terms in numerator and /(120x7) or /120 then /7

A1 cso as given answer

Mark scheme (c)

Scheme	Marks
\(p = 0.6 \quad q = 0.4\) \(s = 120 \times 21q^5p^2 = 120 \times 21 \times 0.4^5 \times 0.6^2 = 9.29\)	B1
\(t = 120 \times 35q^3p^4 = 120 \times 35 \times 0.4^3 \times 0.6^4 = 34.84\)	B1
	(2)

Notes

Cao for each B1

Mark scheme (d)

Scheme

Marks

\(\mathrm{H_0}\): A binomial distribution is a suitable model.
\(\mathrm{H_1}\): A binomial distribution is not a suitable model.

Observed number of rows	19	19	25	32	25
Expected number of rows	11.55	23.22	34.84	31.35	19.04
\(\dfrac{(O - E)^2}{E}\)	4.81	0.77	2.78	0.013	1.87
\(\dfrac{O^2}{E}\)	31.26	15.55	17.94	32.66	32.83

\(\nu = 5 - 2 = 3\)

B1ft

Critical value for \(\chi^2 = 11.345\)

B1ft

\(\displaystyle\sum \frac{(O - E)^2}{E} = 10.23\) or \(\displaystyle\sum \frac{O^2}{E} - N = 130.23 - 120 = 10.23\)

M1A1

10.23 < 11.345 therefore do not reject \(\mathrm{H_0}\)
A binomial is a suitable model.

(7)

(13 marks)

Notes

1st B1 for both hypotheses. B0 if they include 0.6 Condone \(X \sim \mathrm{B}(n, p)\) etc

1st M1 for using some combined columns (<8)

2nd B1ft follows from ‘their no of columns’ -2

3rd B1ft follows from the degrees of freedom

2nd M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\) with at least 2nd (3 seeds) and 4th (5 seeds) accurate to 2sf
Contradictory statements score M0 e.g. “significant” do not reject \(\mathrm{H_0}\)

1st A1 for awrt 10.2

2nd A1 dependent on 2nd M for a correct comment suggesting that binomial model is suitable. No follow through.
Condone mention of 0.6 here. Hypotheses wrong way round scores A0