Question

 

An inexperienced fair promoter has printed 10,000 tickets for the year Eurelia Solstice fair. There are 5,000 blue tickets for adults, costing $10 each, 3000 green tickets for children, costing $5 each, and 2,000 gold tickets for seniors, costing $4 each. A random sample of 200 Eurelians intending to go to the fair have has been selected by EURELIFAX, the major polling company in Eurelia. in the sample there are a 120 adults, 55 children, and 25 series seniors. test whether the proportion of the different types of tickets are significantly different from the proportions printed by the inexperienced promoter.

1. 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 terms of total ticket revenue for the fair?

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

Answer 1

Ho: The sampled tickets are in proportion to those printed

Ha: The sampled tickets are not in proportion to those printed

Observed(O)	Expected (E)	(O - E)^2 /E
120	100	4.00
55	60	0.42
25	40	5.63
	χ2 =	10.04
	Df = 2
	p- value =	0.0066

Since the p- value < 0.05, we reject Ho      

Conclusion: There is sufficient evidence that the sampled tickets are not in proportion to those printed

(1) 95% confidence interval:
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
Sums =	200	1	7.875	68.875
Mean revenue per ticket = ∑x P(x) = 7.875
Variance of the revenue per ticket = ∑x^2 P(x) - Mean^2 = 68.875 - 7.875^2 = 6.86
Standard deviation of the revenue per ticket = 2.62

n = 200              

x-bar = 7.875             

s = 2.62              

% = 95              

Standard Error, SE = σ/√n =    2.62 /√200 = 0.185261977       

z- score = 1.959963985             

Width of the confidence interval = z * SE =    1.95996398454005 * 0.185261976670875 = 0.363106802

Lower Limit of the confidence interval = x-bar - width =     7.875 - 0.363106801979616 = 7.511893198

Upper Limit of the confidence interval = x-bar + width =     7.875 + 0.363106801979616 = 8.238106802

The 95% confidence interval for revenue = [$7.51, $8.24]     

If 8000 people go to the fair, the net revenue will lie in the range [$7.51 * 8000, $8.24 * 8000] = [$60080, $65920]

(2) Sample size:        

z- score for 99% confidence is z = 2.5758       

N = (z * σ/E)^2 = (2.5758 * 2.62/0.05)^2 = 18218