Question

An inexperienced fair promoter has printed 10,000 tickets for the Eurelia Solstice Fair. There are 5000 blue tickets for adults, costing $10 each, 3000 green tickets for children, costing $5 each, and 2000 gold tickets for seniors, costing $4 each. A random sample of 200 Eurelians intending to go to the fair has been selected by Eurelifax, the major polling company in Eurelia. In the sample, there are 120 adults, 55 children and 25 seniors.

a) Test whether the proportions of the different types of ticket are significantly different from the proportions printed by the inexperienced promoter

b) Based on the sample of 200 Eurelians chosen by Eurelifax, form a 95% confidence interval for the mean revenue per ticket at the fair. If 8000 Eurelians go to the fair, what does the interval you have obtained translate into in terms of total ticket revenue for the fair?

c) How large a sample would be needed to form a 99% confidence interval for the mean ticket revenue if the desired margin of error is 5 cents?

Answer 1

There are 120 adults out of 200, whereas there should have been (5000/10100)*200 = 100 adults

There are 55 children out of 200, whereas there should have been (3000/10000)*200 = 60 children

There are 25 seniors out of 200, whereas there should have been (2000/10000)*200 = 40 seniors

(O - E) values are respectively 120 - 100 = 20, 55 - 60 = -5, 25 - 40 = -15

(O - E)^2/E values are respectively 20^2/100 = 4, (-5)^2/60 = 0.42, (-15)^2/40 = 5.63

χ2 statistic = 4 + 0.42 + 5.63 = 10.04

Critical χ2 value (for df = 2, α = 0.05) is 5.99

10.04 > 5.99, so we can conclude that the sample proportions are significantly different from those printed.

(1)

x	f	P(x)	x P(x)	x^2 P(x)
10	120	0.6	6	60
5	55	0.275	1.375	6.875
4	25	0.125	0.5	2
∑	200	1	7.875	68.875

Mean price per ticket = 7.88

Standard deviation = √(∑x^2 P(x) - Mean^2) = 2.6

Standard error = σ/√n = 2.6/√200 = 0.184

95% confidence interval is 7.88 - 1.96*0.184 < μ < 7.88 + 1.96*0.184, that is 7.52 < μ < 8.24

If 8000 Eurelians go to the fair, the revenue will be between 7.52*8000 = $60160 and 8.24*8000 = $65920

(2)

n = (z*σ/E)^2 = (2.58*2.6/0.05)^2 = 17999