Question

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. For a random sample of

43 ​weekdays, daily fees collected averaged ​$128​, with standard deviation of ​1717.

Complete parts a through e below.

​

a) Find a 99​% confidence interval for the mean daily income this parking garage will generate.

The 99​% confidence interval for the mean daily income is ​($_________________, $___________​).

​(Round to two decimal places as​ needed.)

​

b) Explain in context what this confidence interval means.

Choose the correct answer below.

1. There is 99​% confidence that the daily income for a weekday falls in the interval.
2. There is 99​% confidence that the interval contains the mean daily income.
3. There is 99​% confidence that the daily income for all weekdays falls in the interval.
4. There is 99​% confidence that the mean daily income will always fall in the interval.

​

c) Explain what 99​% confidence means in this context.

Choose the correct answer below.

1. 99​% of all samples of size 43 produce intervals that contain the mean daily income.
2. 99​% of all samples of size 43 have a mean daily income that is in the interval.
3. 99​% of all weekdays sampled have daily incomes that fall in the interval.
4. 99​% of all weekdays have daily incomes that fall in the interval.

​

e) The consultant who advised the city on this project predicted that parking revenues would average ​$133 per day. Based on your confidence​ interval, what do you think of the​ consultant's prediction? ​ Why?

Since the 99​% confidence interval (Contains or Does Not Contain) the predicted​ average, the​ consultant's prediction is (Not Plausible or Plausible)

Answer 1

sample std dev ,    s =    17.0000
Sample Size ,   n =    43
Sample Mean,    x̅ =   128.0000

a)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   42          
't value='   tα/2=   2.6981   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   17.000   / √   43   =   2.5925
margin of error , E=t*SE =   2.6981   *   2.592   =   6.995
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    128.00   -   6.995   =   121.0053
Interval Upper Limit = x̅ + E =    128.00   -   6.995   =   134.9947
99%   confidence interval is (   121.01   < µ <   134.99   )

b)

There is 99​% confidence that the interval contains the mean daily income.

c)

99​% of all samples of size 43 produce intervals that contain the mean daily income.

d)

Ho :   µ =   133
Ha :   µ ╪   133

Since the 99​% confidence interval (Contains ) the predicted​ average, the​ consultant's prediction is ( Plausible)