Question

A city in Ontario 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, 700. The population standard deviation for daily fees collected is known to be $1450. Assume that parking fees are approximately normally distributed

1. Find a 95% 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 $12,600 per day. Test the consultants claim at a level of confidence of 10% using a z distribution.

Answer 1

a)

sample mean, xbar = 12700
sample standard deviation, σ = 1450
sample size, n = 44

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 1450/sqrt(44)
ME = 428.45

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (12700 - 1.96 * 1450/sqrt(44) , 12700 + 1.96 * 1450/sqrt(44))
CI = (12271.55 , 13128.45)

We rae 95% confident that the mean daily income this parking garage will generate is between (12271.55 , 13128.45)

b)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 12600
Alternative Hypothesis, Ha: μ ≠ 12600

Rejection Region
This is two tailed test, for α = 0.1
Critical value of z are -1.645 and 1.645.
Hence reject H0 if z < -1.645 or z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (12700 - 12600)/(1450/sqrt(44))
z = 0.46

P-value Approach
P-value = 0.6455
As P-value >= 0.1, fail to reject null hypothesis.