Question

The city of Omaha decided to build a new parking garage in hopes of luring more people to the Old Market district.The city will pay for the parking garage through parking fees. For a random sample of 42 weekdays, daily fees collected averaged $126 with a standard deviation of $16. The consultant who advised the city on the project estimated the potential parking revenues to be $135 per day, though the city believes that estimate is too high. Perform a hypothesis test (by hand) using α=0.05 to check the advisor's claim that the revenue would be $135 per day. Include the following:
1. Write your hypotheses in both symbols and words.
2. Test an appropriate hypothesis "by hand" using an alpha level of 0.05. Assume conditions are met (so don't check them). Be sure to include the correct probability notation and use the editor for all work.
3. Support your test with the corresponding confidence interval, again "by hand". Use the editor to show all steps.
4. Write your 3-part final conclusion using the test results and interval results. Include a sentence that states whether the test and interval results agree.

Answer 1

Answer:

Given:

Mean, u = 135

Sample size, n = 42

Sample mean, x' = 126

Standard deviation = 16

Significance level = 0.05

1) The null and alternative hypotheses:

H0 : u = 135 (the revenue per day is $135)

H1 : u < 135 (the revenue per day is less than $135)

2) In this case, we have a left tailed test.

Let's use the formula:

$= \frac{x' - u}{s/ \sqrt{n}} = \frac{126 - 135}{16 / \sqrt{42}} = -3.645$

t = - 3.645

Critical value: at a significance level of 0.05, left tailed test,

tcritical = -1.683

We reject null hypothesis, H0, since t calculated, -3.645 is less than tcritical, -1.683.

3) Given, a = 0.05, df = 41, left tailed test,

t-value = 2.02

For the corresponding confidence interval, we have:

$CI = (x' - \frac{t * \sigma}{\sqrt{n}}, x' + \frac{t * \sigma}{\sqrt{n}})$

$CI = (126 - \frac{2.02 * 16}{\sqrt{42}}, 126 + \frac{2.02 * 16}{\sqrt{42}})$

CI = (121.014, 130.986)

4) Conclusion:

We reject the estimated potential revenue advised by the consultant, H0, as it is too high, because the t-calculated falls in rejection region(i.e, less than critical value), also the upper limit of the confidence interval is less than $135.