Question

Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 150 people. The sample mean is 23.1 hours. There is a known population standard deviation of 6.4 hours. The population distribution is assumed to be normal.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)

  Find the following. (Enter exact numbers as integers, fractions, or decimals.)(i)    

  x =

  
  
  (ii)    

  σ =

  
  
  (iii)    

  n =

• Part (b)

  In words, define the random variables X and X

  X is the number of tax forms that an accounting firm completes, and X is the mean number of tax forms that an accounting firm completes.X is the number of tax forms that an accounting firm completes, and X is the mean number of tax forms that an accounting firm completes.    X is the time needed to complete one person's tax forms, and X is the mean time needed to complete tax forms from a sample of 150 customers.X is the time needed to complete one person's tax forms, and X is the mean time needed to complete tax forms from a sample of 150 customers.

• Part (c)

  Which distribution should you use for this problem? (Round your answers to two decimal places.)

  X ~

    ? Exp N B H U   ,  
  
  Explain your choice.

  The standard normal distribution should be used because the mean is given.The Student's t-distribution should be used because the sample mean is smaller than 30.    The standard normal distribution should be used because the population standard deviation is known.The Student's t-distribution should be used because the sample standard deviation is given.

• Part (d)

  Construct a 90% confidence interval for the population mean time to complete the tax forms.(i) State the confidence interval. (Round your answers to two decimal places.)   ,  
  
  (ii) Sketch the graph. (Round your answers to two decimal places.) (iii) Calculate the error bound. (Round your answer to two decimal places.)

• Part (e)

  If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what change should it make?

  It should increase the number of people surveyed.It should decrease the number of people surveyed.    

• Part (f)

  If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?

  The level of confidence would be larger because we have collected a smaller sample, obtaining less accurate information.The level of confidence would be smaller because we have collected a smaller sample, obtaining less accurate information.    There would be no change.

• Part (g)

  Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?

  The number of people surveyed would decrease because more accurate information requires a smaller sample.The number of people surveyed would increase because more accurate information requires a larger sample.    There would be no change.

Answer 1

a)

x̅ =   23.1000
σ =    6.4000
n =    150

b)

X is the time needed to complete one person's tax forms, and X is the mean time needed to complete tax forms from a sample of 150 customers.

c) X~ N(23.1, 6.4/√150)

The standard normal distribution should be used because the population standard deviation is known

d)

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.64   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   6.4000   / √   150   =   0.5226
margin of error, E=Z*SE =   1.6449   *   0.5226   =   0.8595
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    23.10   -   0.859531   =   22.2405
Interval Upper Limit = x̅ + E =    23.10   -   0.859531   =   23.9595
90%   confidence interval is (   22.24   < µ <   23.96   )

error bound =margin of error, E=Z*SE =   1.6449   *   0.5226   =   0.86

e) It should decrease the number of people surveyed.

f)

The level of confidence would be larger because we have collected a smaller sample, obtaining less accurate information

g)

The number of people surveyed would increase because more accurate information requires a larger sample.