Show all work Question 3 a A sample of size n100 produced a

Show all work.

Question 3.

(a). A sample of size n=100 produced a sample mean Xbar=16. Assuming the population standard deviation =3

(i). Compute a 95% confidence interval for the population mean µ. (5 points)

(ii). How large should a sample be to estimate the population mean µ with a margin of error not exceeding 0.5? (5 points)

(b). The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assuming the standard deviation of this assembly time is 3.6 minutes.

(i). After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 92% confidence interval for the mean assembly time. (5 points)

(ii). How many workers should be involved in this study in order to have the mean assembly time estimated up to 15 seconds with 92% confidence. (5 points)

(C). A Bank charges an overdraft fee if a customer does not have enough money in a checking account to cover a check or a debit card withdrawal. A random sample of 12 banks resulted in the following data regarding overdraft fees.

$25.00 $36.00 $30.00 $20. $15.00 $20.00 $30.00 $35.00 $10.00 $36.00 $25.00 $20.00.

(i)Find the sample mean. (2 points).

(ii).Find the sample standard deviation. (6 points).

(iii). Construct a 90% confidence interval for the population mean overdraft fee at banks and provide an interpretation of your interval. (3 points).

(iv). Construct a 95% confidence interval for the population mean overdraft fee at banks and provide an interpretation of your interval. (3 points).

(v). Construct a 99% confidence interval for the population mean overdraft fee at banks and provide an interpretation of your interval. (3 points).

(vi).By comparing your results from b(i), b(ii), and b(iii), which confidence level gives the best precision?, and why do you say so? (3 points).

Solution

a)

i)

Note that              
              
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 =    16          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3          
n = sample size =    100          
              
Thus,              
              
Lower bound =    15.4120108          
Upper bound =    16.5879892          
              
Thus, the confidence interval is              
              
(   15.4120108   ,   16.5879892   ) [ANSWER]

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ii)

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 =    3  
E = margin of error =    0.5  
      
Thus,      
      
n =    138.2925175  
      
Rounding up,      
      
n =    139   [ANSWER]

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