Question

For the following scenarios calculate the appropriate test statistic and make a decision based on the provided decision rule. (Steps 5 & 6 of the six step hypothesis test.) Be sure to explain what your decision means in the context of the problem.

a. The US Department of Transportation estimates that 18% of Americans carpool. Does that imply that 1% of all cars will have two or more occupants? A sample of 600 cars travelling southbound on the New Jersey Turnpike yesterday revealed that 128 had two or more occupants. Can we conclude that more than 18% of cars on the New Jersey Turnpike have two or more occupants? Note that H0: π ≤ .15 and H1: π > .15.

i. Test Statistic:

ii. What is your conclusion if the critical value for a right tail test is 1.96?

b. Vehicles heading west on Front Street may turn right, turn left, or go straight ahead at Elm Street. The city traffic engineer believes that 70% of the vehicles will continue straight through the intersection. Of the remaining 30%, equal proportions will turn right and left. Three hundred vehicles were observed, with the following results. Can we conclude that the traffic engineer is correct? Note that H0: the engineer is correct, the actual distribution of turn directions at this intersection follows his believed distribution and H1: the engineer is incorrect, the actual distribution of turn directions at this intersection does not followed the believed distribution.

	Straight	Right Turn	Left Turn
Frequency	218	56	46

i. Test Statistic:

ii. What is your conclusion if the critical value for a right tail test is 7.824?

(Please focus on the numbers given above in the question)

Answer 1

(a)

(i)

H0: π ≤ .15

H1: π > .15.

n = 600

= 128/600 = 0.2133

= 0.025

Test Statistic is given by:

(ii)

Given: Critical Value of Z = 1.96

Since calculated value of Z = 4.345 is greatr than critical value of Z= 1.96, thedifference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that more than 18% of cars on the New Jersey Turnpike have two or more occupants.

(b)

(i)

Test Statistic () is calculated as follows:

Observed (O)	Expected (E)	(O - E)2/E
218	320 X 0.70 = 224	0.161
56	320 X 0.15 = 48	1.333
46	320 X 0.15 = 48	0.083
Total = =		1.577

So,

Test Statistic () = 1.677

(ii)

Given:
Critical Value of = 7.824

Since calculatedvalue of = 1.577 is less than the critical value of = 7.824, the difference is not significant. Fail to rejectnull hypothesis.

Conclusion:

The data support the claim that the engineer is correct, the actual distribution of turn directions at this intersection follows his believed distribution.