Question

A Parking Enforcement Officer in Adelaide is conducting an analysis on the amount of time left on parking meters in the CBD. A quick survey of 14 metered parking spaces where cars had just left provided the following times (in minutes)

14	18	2	12	4	13	17
20	31	15	31	9	25	27

(a) What assumptions must you make to obtain a confidence interval for the mean amount of time left for all vacant meters?

Calculated Mean: 16.2

Calculated Sample Standard Deviation =

(b) Construct a 95% confidence interval estimate for the mean amount of time left for all vacant meters.

(c) Interpret the confidence interval obtained in (b).

(d) The Policy Manager is not totally convinced with the confidence interval estimated in (b), because it is too wide. They would like to be 99% sure that the sample mean is within 3 minutes of the true mean. What sample size is required to meet this requirement? Assume the population standard deviation is equal to the one identified for the sample of the 14 parking spaces.

Answer 1

a)

Assumption is normality of data.

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   9
Sample Size ,   n =    14
Sample Mean,    x̅ = ΣX/n =    17

b)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   13          
't value='   tα/2=   2.1604   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   9.1062   / √   14   =   2.4337
margin of error , E=t*SE =   2.1604   *   2.43374   =   5.2578
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    17.00   -   5.257770   =   11.7422
Interval Upper Limit = x̅ + E =    17.00   -   5.257770   =   22.2578
95%   confidence interval is (   11.74   < µ <   22.26   )

c)

We are 95% confident that true mean will lie in above interval

d)

Standard Deviation ,   σ =    9                  
sampling error ,    E =   3                  
Confidence Level ,   CL=   99%                  
                          
alpha =   1-CL =   1%                  
Z value =    Zα/2 =    2.576   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   2.576   *   9   /   3   ) ² =   59.714
                          
                          
So,Sample Size needed=       60                  

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