Parking Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. Dur-ing a two-month period (44 weekdays), daily fees collected av-eraged $126, with a standard deviation of $15. a) What assumptions must you make in order to use these statistics for inference? b) Write a 90% confidence interval for the mean daily income this parking garage will generate. c) Interpret this confidence interval in context. d) Explain what "90% confidence" means in this context. e) The consultant who advised the city on this project predicted that parking revenues would average $130 per day. Based on your confidence interval, do you think the consultant was correct? Why?

The parameter to estimate is the population mean (μ) of the daily income of the parking garage. Since the population mean is a parameter of the normal distribution you need your study variable to be normally distributed to study it. If your study variable hasn't the required distribution, since you have a big enough sample, apply the Central Limit Theorem to approximate the sample mean distribution to normal.

So the assumption you need to make is that X[bar]≈N(μ;δ²/n)

b.

The formula for the Confidence Interval is:

X[bar]±[tex]Z_{1-\alpha /2}[/tex]*(S/√n)

I don't have the value of the population standard deviation, so I'll use the approximation with the sample standard deviation.

Sample

n=44

X[bar]= $126

S= $15

[tex]Z_{1-\alpha /2} = Z_{0.90} =1.64[/tex]

X[bar]±[tex]Z_{1-\alpha /2}[/tex]*(S/√n)

[126±1.64*(15/√44)]

[122.29;129.71]

c.

With a confidence level of 90%, you can expect the interval [122.29;129.71] will contain the true mean of the daily income of the parking garage.

d.

90% of confidence means that if you were to take 100 samples and calculate a confidence interval for the population mean of the daily income of the parking garages, you'd expect 90 of those intervals to contain the true value of the parameter.

e.

If I were to make a hypothesis test to see if the consultant was right, using the complemental level of significance, since the interval doesn't contain the supposed value of $130, I would reject the null hypothesis and conclude that he wasn't right.

H₀: μ = 130

H₁: μ ≠ 130

α: 0.10

[122.29;129.71]

Interval doesn't include 130 ⇒ to Reject the null hypothesis.

I hope this helps!