S3 June 2016 Q6

EdexcelOld spec17 marksIncludes hypothesis testingChi-Squared Tests

6. An airport manager carries out a survey of families and their luggage. Each family is allowed to check in a maximum of 4 suitcases. She observes 50 families at the check-in desk and counts the total number of suitcases each family checks in. The data are summarised in the table below.

Number of suitcases	0	1	2	3	4
Frequency	6	25	12	6	1

The manager claims that the data can be modelled by a binomial distribution with \(p = 0.3\)

(a) Test the manager’s claim at the 5% level of significance. State your hypotheses clearly.
Show your working clearly and give your expected frequencies to 2 decimal places. (8)

The manager also carries out a survey of the time taken by passengers to check in. She records the number of passengers that check in during each of 100 five-minute intervals.

The manager makes a new claim that these data can be modelled by a Poisson distribution. She calculates the expected frequencies given in the table below.

Number of passengers	0	1	2	3	4	5 or more
Observed frequency	5	40	31	18	6	0
Expected frequency	16.53	29.75	\(r\)	\(s\)	7.23	3.64

(b) Find the value of \(r\) and the value of \(s\) giving your answers to 2 decimal places. (3)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H_0}\): Binomial with \(p = 0.3\) is a good fit. \(\mathrm{H_1}\): Binomial with \(p = 0.3\) is not a good fit.

	0	1	2 or more
Observed	6	25	19
Expected	50x0.2401 =12.005 or 12.01 or 12.00	50x0.4116 =20.58	50x0.2646+50x0.0756+50x0.0081 =13.23+3.78+0.405 =17.415 or 17.41 or 17.42
\(\frac{(O - E)^2}{E}\)	3.003751	0.949291	0.144256
\(\frac{O^2}{E}\)	2.998751	30.36929	20.72926

M1A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 4.097\ldots\) or \(\displaystyle\sum \frac{O^2}{E} - N = 54.097\ldots - 50 = 4.097\ldots\) awrt 4.09-4.1(0)

dM1A1

\(\nu = 3 - 1 = 2\)

B1ft

\(\chi^2_2(5\%) = 5.991\ (\gt 4.1(0))\)

B1ft

Insufficient evidence to reject \(\mathrm{H_0}\) (Accept \(\mathrm{H_0}\))
Binomial with \(p = 0.3\) is a good fit.

(8)

Notes

B1 both including \(p = 0.3\)

M1 with some combined columns and at least one \(E\) correct to 2sf

A1 all correct to 2dp and total of expected values is 50.

dM1 either method

A1 awrt 4.09-4.1(0)

B1 ft their columns -1

B1 ft their

A1 cao

Scheme	Marks
\(\bar{x} = \dfrac{40 + 62 + 54 + 24}{100} = 1.8\)	B1 cao
\(r = 26.78\)	B1 cao
\(s = 16.07\)	B1 cao
	(3)

Mark scheme (c)

Scheme

Marks

\(\mathrm{H_0}\): Poisson is a good fit. \(\mathrm{H_1}\): Poisson is not a good fit.

	0	1	2	3	4 or more
Observed	5	40	31	18	6
Expected	16.53	29.75	26.78	16.07	10.87
\(\dfrac{(O - E)^2}{E}\)	\(\frac{11.53^2}{16.53} = 8.042\ldots\)	\(\frac{10.25^2}{29.75} = 3.532\ldots\)	\(\frac{4.22^2}{26.78} = 0.665\ldots\)	\(\frac{1.93^2}{16.07} = 0.232\ldots\)	\(\frac{4.87^2}{10.87} = 2.182\ldots\)
\(\dfrac{O^2}{E}\)	\(\frac{5^2}{16.53} = 1.512\ldots\)	\(\frac{40^2}{29.75} = 53.782\ldots\)	\(\frac{31^2}{26.78} = 35.885\ldots\)	\(\frac{18^2}{16.07} = 20.162\ldots\)	\(\frac{6^2}{10.87} = 3.312\ldots\)

\(\displaystyle\sum \frac{(O - E)^2}{E} = 14.65 - 14.66\) or \(\displaystyle\sum \frac{O^2}{E} - N = 114.65 - 100 = 14.65 - 14.66\)

M1A1

\(\nu = 5 - 1 - 1 = 3\)

B1 cao

\(\chi^2_3(1\%) = 11.345\ (\lt 14.65)\)

B1ft

Sufficient evidence to reject \(\mathrm{H_0}\)
Poisson is not a good fit.

A1 cao

(6)

(17 marks)

Notes

B1 no parameters included

M1 either method

B1 ft their \(\nu\)