Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central district. The city plans to pay for the structure through parking fees. During a two-week period (14 days), daily fees collected average $126 with standard deviation $15. We want to construct a confidence interval for the true mean daily fees collected at this parking garage.

(a) To construct the confidence interval, should you use the normal distribution orat distribution?
(b) Construct a 90% confidence interval.
(c) 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 could have been correct? Why or why not?

c) For this case since confidence interval include tha value of 130 so then we don't have enough evidence to conclude that the claim by the consultant is incorrect.

Step-by-step explanation:

Part a

For this case we need to use a t distribution since we know the information about a sample and we don't know the population deviation.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

[tex]\bar X=126[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=15 represent the sample standard deviation

n=14 represent the sample size

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=14-1=13[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,13)".And we see that [tex]t_{\alpha/2}=1.77[/tex]

Now we have everything in order to replace into formula (1):

[tex]126-1.77\frac{15}{\sqrt{14}}=118.904[/tex]

[tex]126+1.77\frac{15}{\sqrt{14}}=133.096[/tex]

So on this case the 90% confidence interval would be given by (118.904;133.096)

Part c

For this case since confidence interval include tha value of 130 so then we don't have enough evidence to conclude that the claim by the consultant is incorrect.