The manager of a fast-food restaurant wants to estimate the average service time

Question

The manager of a fast-food restaurant wants to estimate the average service time needed to complete a customer's order. He records service times for 50 randomly selected customers and obtains a mean of 181.0 seconds and a standard deviation of 82.1 seconds.

1. Find a 95% confidence interval for the mean service time at the restaurant.
2. Why is it reasonable to use a standard normal distribution to find the confidence interval?
3. By finding the endpoints of the confidence interval, what are we prediction about service times (measured in seconds) at this restaurant?
4. Suppose the manager decides to use a larger sample to find a more accurate confidence interval. How large a sample would he need to estimate the mean service time to within 10 seconds with 95% confdience. (Use the std deviation of the sample as an estimate for the population).

Explanation / Answer

1.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    181          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    82.1          
n = sample size =    50          
              
Thus,              
Margin of Error E =    22.7565408          
Lower bound =    158.2434592          
Upper bound =    203.7565408          
              
Thus, the confidence interval is              
              
(   158.2434592   ,   203.7565408   ) [ANSWER]

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b)

Because the sample size is big, greater than 30, so the sampling distribution of the mean is approximately normal by central limit theorem.

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c)

We are 95% confident that the true mean service time is within that interval.

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d)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    82.1  
E = margin of error =    10  
      
Thus,      
      
n =    258.9300745  
      
Rounding up,      
      
n =    259   [ANSWER]

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