Question

A very busy and fancy restaurant records the wait time for each customer that comes in. Below are 12 customers’ wait times (in minutes):

51        60        59        72        80        83        54        66        61        81        66        62

1. Assuming that the true mean wait time is 70 minutes with a variance of 9 minutes, what is the probability that a customer will wait between 65 to 80 minutes?

2. What is the standard deviation of the sampling distribution of x̄, the average wait time?

3. In practice we do not know the population mean wait time. Assuming this, construct and interpret a 90% confidence interval for μ, the true mean wait time. Assume that σ = 20 minutes.

4.In practice, we do not know the population standard deviation. Assuming this, construct and interpret a 95% confidence interval for μ, the true mean wait time. Assume that σ is unknown.

5.The manager wants to know if there is evidence that the true mean wait time is greater than 1 hour. What are the null and alternative hypotheses to test the claim? Will this be a one-sided or two-sided test?

Answer 1

1)

µ =    70                              
σ =    3                              
we need to calculate probability for ,                                  
P (   65   < X <   80   )                  
=P( (65-70)/3 < (X-µ)/σ < (80-70)/3 )                                  
                                  
P (    -1.667   < Z <    3.333   )                   
= P ( Z <    3.333   ) - P ( Z <   -1.667   )

=    0.9996   -    0.0478  

=    0.9518

......

2)

	X	(X - X̄)²
total sum	795	1240.25
n	12	12

sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1240.25/11)   =       10.618

......

3)

Level of Significance ,    α =    0.05
population std dev ,    σ =    20.0000
Sample Size ,   n =    12
Sample Mean,    x̅ = ΣX/n =    66.2500
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   20.0000   / √   12   =   5.773503
margin of error, E=Z*SE =   1.9600   *   5.77350   =   11.315857
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    66.25   -   11.315857   =   54.934143
Interval Upper Limit = x̅ + E =    66.25   -   11.315857   =   77.565857
95%   confidence interval is (   54.93   < µ <   77.57   )

..........

4)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   10.6184
Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   11          
't value='   tα/2=   2.2010   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.6184   / √   12   =   3.065262
margin of error , E=t*SE =   2.2010   *   3.06526   =   6.746597
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    66.25   -   6.746597   =   59.503403
Interval Upper Limit = x̅ + E =    66.25   -   6.746597   =   72.996597
95%   confidence interval is (   59.50   < µ <   73.00   )

.......

5)

( 1 hour = 60minutes)

Ho :   µ =   60   
Ha :   µ >   60      

one sided test

(Right tail test)  

...................

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