Question

1. A bus company wants to test the hypothesis that the proportions of the six types of passengers it carries on a certain route are EQUAL. The six types are: commuters, shoppers, college students, tourists, children, and mall employees. it carries on a certain route are EQUAL. The six types are: commuters, shoppers, college students, tourists, children, and mall employees. On the basis of the following sample of four hundred eighty bus riders, should the company accept this hypothesis at the .10 level of significance: (3 POINTS)

ROUND ALL CALCULATIONS TO 2 DECIMAL PLACES!    SHOW EVERY BIT OF YOUR WORK!!!                           

                           commuters – fifty-three

                           shoppers – seventy-five

                           college students – one hundred ten

                           tourists - eighty

                           children – one hundred

                            mall employees – sixty-two

Answer 1

Each expected frequency = 480 / 6 = 80

The Hypothesis

H0: p1 = p2 = p3 = p4 = p5 = p6

Ha: There is a difference in the proportions.

The Test Statistic:

	Observed	Expected	O-E	(O-E)2	(O-E)2/E
Commuters	53	80	-27	729	9.11
Shoppers	75	80	-5	25	0.31
College Students	110	80	30	900	11.25
Tourists	80	80	0	0	0.00
Children	100	80	20	400	5.00
Mall Emplyess	62	80	-18	324	4.05
Total	480	480		2378	29.73

Test = 29.73

The Critical value at = 0.10, df = n – 1 = 5, critical = 7.78

(4) The p value Foe Test = 29.73, for df = n – 1 = 5, p value = 0.000

The Decision Rule: If test is > critical, then Reject H0.

If p value is < , Then Reject H0.

The Decision:    Since test (29.73) is > critical (7.78), We Reject H0.

Since p value (0.0000) is < (0.10), We Reject H0.

The Conclusion: There is sufficient evidence at the 90% significance level to conclude that there is a difference in proportions of the 6 types of passengers. The company should not accept the hypothesis that they are equal.

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