Question

In: Statistics and Probability

Suppose that an accounting firm does a study to determine the time needed to complete one...

Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 175 people. The sample mean is 23.4 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 Xbar

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 175 customers.

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 175 customers.

Part (c) Which distribution should you use for this problem? (Round your answers to two decimal places.) X ~ , Explain your choice. The Student's t-distribution should be used because the sample mean is smaller than 30. The Student's t-distribution should be used because the sample standard deviation is given. The standard normal distribution should be used because the mean is given. The standard normal distribution should be used because the population standard deviation is known.

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?

    a)It should increase the number of people surveyed.

  • b)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?

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

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

  • c) 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?

    a)The number of people surveyed would decrease because more accurate information requires a smaller sample.

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

  • c) There would be no change.

Solutions

Expert Solution

a)

x̅ = 23.4

σ = 6.4

n = 175

(b)

X is the time needed to complete one person's tax forms, and Xbar is the mean time needed to complete tax forms from a sample of 175 customers.

--

(c)

Standard deviation = 6.4/SQRT(175) = 0.4838

Distribution of x̅ ~ N(23.4, 0.48)

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

(i)

90% Confidence interval :

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) = 1.645

Lower Bound = x̅ - z_c*σ/√n = 23.4 - 1.645 * 6.4/√175 = 22.60

Upper Bound = x̅ + z_c*σ/√n = 23.4 + 1.645 * 6.4/√175 = 24.20

22.60 < µ < 24.20

(iii) Margin of error , E = z-crit*σ/√n = 1.6449*6.4/√175 = 0.80

(e) Answer: b) It should decrease the number of people surveyed.    

(f) Answer: b)The level of confidence would be larger because we have collected a smaller sample, obtaining less accurate information.  

(g) Answer: b)The number of people surveyed would increase because more accurate information requires a larger sample.


Related Solutions

Suppose that an accounting firm does a study to determine the time needed to complete one...
Suppose that an accounting firm does a study to determine the time needed to complete one persons tax form. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. Construct a 90% confidence interval for the population mean time to complete the tax forms.
suppose that an accounting firm does a study to determine the time needed to complete one...
suppose that an accounting firm does a study to determine the time needed to complete one persons tax forms. it randomly surveys 100 people. the sample mean is 23.6 hours. there is a known standard deviation of 7.0 hours. the population distribution is assumed to be normal. a) if the firm wished to increase it level of confidence and keep the error bound the same by taking another survey, what changes should it make? b) if the firm did another...
Suppose that an accounting firm does a study to determine the time needed to complete one...
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....
An accounting firm does a study to determine the time needed to complete one person's tax...
An accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 40 people. The sample mean is 17.4 hours and the sample standard deviation is 6.2 hours. Based on this data, construct a 95% confidence interval for the time to complete one person's answer. Round to nearest hundredth of an hour
The U.S. Census Bureau conducts a study to determine the time needed to complete the short...
The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Identify the following. (Enter exact numbers as integers, fractions or decimals.) a. x-bar = b. σ = c. n =
An automobile dealer conducted a test to determine if the time in minutes needed to complete...
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow. Analyzer Computerized Electronic Car Compact 50 41 Intermediate 55 44 Full-sized 63 47 Use α = 0.05 to test for any...
An automobile dealer conducted a test to determine if the time in minutes needed to complete...
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow. Analyzer Computerized Electronic Car Compact 49 41 Intermediate 56 45 Full-sized 63 46 Use α = 0.05 to test for any...
An automobile dealer conducted a test to determine if the time in minutes needed to complete...
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow. Analyzer Computerized Electronic Car Compact 52 43 Intermediate 55 43 Full-sized 61 46 Use α = 0.05 to test for any...
An automobile dealer conducted a test to determine if the time in minutes needed to complete...
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow. Analyzer Computerized Electronic Car Compact 50 42 Intermediate 57 45 Full-sized 61 45 Use α = 0.05 to test for any...
An automobile dealer conducted a test to determine whether the time needed to complete a minor...
An automobile dealer conducted a test to determine whether the time needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data (time in minutes) obtained follow. Car Compact Intermediate Full Size Analyzer Computerized 50 55 63 Electronic 42 44 46 The following regression model can be...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT