## 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 stru

Question

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?

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3 months 2022-02-06T17:23:21+00:00 1 Answer 0 views 0

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.

b) $$126-1.77\frac{15}{\sqrt{14}}=118.904$$

$$126+1.77\frac{15}{\sqrt{14}}=133.096$$

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

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.

$$\bar X=126$$ represent the sample mean

$$\mu$$ 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:

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

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

$$df=n-1=14-1=13$$

Since the Confidence is 0.90 or 90%, the value of $$\alpha=0.1$$ and $$\alpha/2 =0.05$$, 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 $$t_{\alpha/2}=1.77$$

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

$$126-1.77\frac{15}{\sqrt{14}}=118.904$$

$$126+1.77\frac{15}{\sqrt{14}}=133.096$$

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.