The city of Laguna Beach operates two public parking lots. The Ocean Drive parki

Question

The city of Laguna Beach operates two public parking lots. The Ocean Drive parking lot can accommodate up to 125 cars and the Rio Rancho parking lot can accommodate up to 130 cars. City planners are considering increasing the size of the lots and changing the fee structure. To begin, the Planning Office would like some information on the number of cars in the lots at various times of the day. A junior planner officer is assigned the task of visiting the two lots at random times of the day and evening and counting the number of cars in the lots. The study lasted over a period of one month. Below is the number of cars in the lots for 25 visits of the Ocean Drive lot and 28 visits of the Rio Rancho lot. Assume the population standard deviations are equal.

State the decision rule for .05 significance level: H0:?1= ?2H1:?1??2. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.)

The city of Laguna Beach operates two public parking lots. The Ocean Drive parking lot can accommodate up to 125 cars and the Rio Rancho parking lot can accommodate up to 130 cars. City planners are considering increasing the size of the lots and changing the fee structure. To begin, the Planning Office would like some information on the number of cars in the lots at various times of the day. A junior planner officer is assigned the task of visiting the two lots at random times of the day and evening and counting the number of cars in the lots. The study lasted over a period of one month. Below is the number of cars in the lots for 25 visits of the Ocean Drive lot and 28 visits of the Rio Rancho lot. Assume the population standard deviations are equal.

Explanation / Answer

Ocean Drive:
n1=25
Mean 86.24
Standard deviation = 23.433

Rio Rancho:

n2=28
Mean = 88.4643
Standard deviation = 28.3869

H0: ? (1) = ? (2)
HA: ? (1) ? ? (2)

Sample 1 size 25
Sample 2 size 28
Sample 1 mean 86.24
Sample 2 mean 88.4643
Sample 1 S.D. 23.433
Sample 2 S.D. 28.3869
Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)
Pooled variance s = [(24)(549.105489)+(27)(805.816092))] / (51) =
Pooled variance s^2 = 685.011102
Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)
Standard error of difference in means = (0.275162)(26.172717)
= 7.201745 (denominator of t)
t = (mean1-mean2) / SE = (86.24 - 88.4643) /7.201745
t = -0.308856

Degrees of freedom = 51
Using a .05 significance level, the t-critical value is (two-tailed) :2.0076

Computed t is smaller than critical t, so we do not reject the null hypothesis.
There is no difference in the mean number of cars in the two lots.


Apply the t-test assuming equal variances in the population.

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