Question

The city of Dubai builds a new public parking garage in its Central business district downtown, hoping to attract more shoppers downtown. The city plans to pay for the structure through parking fees. For a random sample of 44 week- days, daily fees collected averaged $12, 600. The population standard deviation for daily fees collected is known to be $1500. Assume that parking fees are approximately normally distributed

1. Find a 90% confidence interval for the mean daily income this parking garage will generate and interpret your answer. Use z- distribution.

2. The consultant who advised the city on this project predicted that parking revenues would average $12,800 per day. Test the consultants claim at a level of confidence of 5% using a z distribution.

                                                                                                                                                             

Answer 1

a )

90% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.645
12600 ± Z (0.1/2 ) * 1500/√(44)
Lower Limit = 12600 - Z(0.1/2) 1500/√(44)
Lower Limit = 12228.01
Upper Limit = 12600 + Z(0.1/2) 1500/√(44)
Upper Limit = 12971.99
90% Confidence interval is ( 12228.01 , 12971.99 )

b)

H0: = 12800

Ha: 12800

Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 12600 - 12800 ) / ( 1500 / √( 44 ))
Z = -0.88

Test Criteria :-
Reject null hypothesis if | Z | > Z( α/2 )

Critical value Z(α/2) = Z( 0.05 /2 ) = 1.96
| Z | > Z( α/2 ) = 0.88 < 1.96

Result :- Fail to reject null hypothesis

We do not have sufficient evidence to support the consultants claim.