Essentials of Statistics for Business and Economics
Essentials of Statistics for Business and Economics
9th Edition
ISBN: 9780357118191
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams
Publisher: Cengage Learning US
bartleby

Videos

Question
Book Icon
Chapter 7, Problem 55SE

a.

To determine

Find the standard error for each of the three firms given a sample of size 50 using the finite population correction factor.

a.

Expert Solution
Check Mark

Answer to Problem 55SE

The standard error for N=2,000 is 20.11.

The standard error for N=5,000 is 20.26.

The standard error for N=10,000 is 20.31.

Explanation of Solution

Calculation:

The given information is that sample of 50 items from its inventory is taken. Firm A’s inventory contains 2,000 items, firm B’s inventory contains 5,000 items and firm C’s inventory contains 10,000 items. The population standard deviation for the cost of item’s in each firm’s inventory is σ=144.

Sampling distribution of x¯:

The probability distribution all possible values of the sample mean x¯ is termed as the sampling distribution of x¯.

  • The expected value of x¯ is, E(x¯)=μ.
  • The standard deviation of x¯ is

    For finite population, σx¯=NnN1(σn)

    For infinite population, σx¯=σn

  • When the population size is infinite or finite and nN0.05 then the standard deviation of x¯ is σx¯=σn.

For N=2,000:

The standard deviation of x¯ is σx¯=NnN1(σn)

Substitute σ as 144, N as 2,000 and n as 50 in the formula,

σx¯=NnN1(σn)=2,000502,0001(14450)=1,9501,999(1447.0711)=0.9877(20.3646)

     =20.1141

Thus, the standard deviation is 20.11.

For N=5,000:

The standard deviation of x¯ is σx¯=NnN1(σn)

Substitute σ as 144, N as 5,000 and n as 50 in the formula,

σx¯=NnN1(σn)=5,000505,0001(14450)=4,9504,999(1447.0711)=0.9951(20.3646)

     =20.2648

Thus, the standard deviation is 20.26.

For N=10,000:

The standard deviation of x¯ is σx¯=NnN1(σn)

Substitute σ as 144, N as 10,000 and n as 50 in the formula,

σx¯=NnN1(σn)=10,0005010,0001(14450)=9,9509,999(1447.0711)=0.9975(20.3646)

     =20.3137

Thus, the standard deviation is 20.31.

b.

To determine

Find the probability that each firm the sample mean x¯ will be within ±25 of the population mean μ.

b.

Expert Solution
Check Mark

Answer to Problem 55SE

The probability that the sample mean will be within ±25 of the population mean μ for N=2,000 is 0.785.

The probability that the sample mean will be within ±25 of the population mean μ for N=5,000 is 0.7814.

The probability that the sample mean will be within ±25 of the population mean μ for N=10,000 is 0.7814.

Explanation of Solution

Calculation:

For N=2,000:

The probability that the sample mean x¯ will be within ±25 of the population mean μ is P(x¯μ±25)

P(x¯μ±25)=P(25σx¯x¯μσx¯25σx¯)=P(2520.11z2520.11)=P(1.24z1.24)=P(z1.24)P(z1.24)

From Table 1: Cumulative probabilities for the standard normal distribution,

For z=1.24:

  • Locate the value 1.2 in the first column.
  • Locate the value 0.04 in the first row corresponding to the value 1.2 in the first column.
  • The intersecting value of row and column is 0.8925.

For z=1.24:

  • Locate the value –1.2 in the first column.
  • Locate the value 0.04 in the first row corresponding to the value –1.2 in the first column.
  • The intersecting value of row and column is 0.1075.

P(x¯μ±25)=P(z1.24)P(z1.24)=0.89250.1075=0.785

Thus, the probability that the sample mean will be within ±25 of the population mean μ is 0.785.

For N=5,000:

The probability that the sample mean x¯ will be within ±25 of the population mean μ is P(x¯μ±25)

P(x¯μ±25)=P(25σx¯x¯μσx¯25σx¯)=P(2520.26z2520.26)=P(1.23z1.23)=P(z1.23)P(z1.23)

From Table 1: Cumulative probabilities for the standard normal distribution,

For z=1.23:

  • Locate the value 1.2 in the first column.
  • Locate the value 0.03 in the first row corresponding to the value 1.2 in the first column.
  • The intersecting value of row and column is 0.8907.

For z=1.23:

  • Locate the value –1.2 in the first column.
  • Locate the value 0.03 in the first row corresponding to the value –1.2 in the first column.
  • The intersecting value of row and column is 0.1093.

P(x¯μ±25)=P(z1.23)P(z1.23)=0.89070.1093=0.7814

Thus, the probability that the sample mean will be within ±25 of the population mean μ is 0.7814.

For N=10,000:

The probability that the sample mean x¯ will be within ±25 of the population mean μ is P(x¯μ±25).

P(x¯μ±25)=P(25σx¯x¯μσx¯25σx¯)=P(2520.26z2520.26)=P(1.23z1.23)=P(z1.23)P(z1.23)

From Table 1: Cumulative probabilities for the standard normal distribution,

For z=1.23:

  • Locate the value 1.2 in the first column.
  • Locate the value 0.03 in the first row corresponding to the value 1.2 in the first column.
  • The intersecting value of row and column is 0.8907.

For z=1.23:

  • Locate the value –1.2 in the first column.
  • Locate the value 0.03 in the first row corresponding to the value –1.2 in the first column.
  • The intersecting value of row and column is 0.1093.

P(x¯μ±25)=P(z1.23)P(z1.23)=0.89070.1093=0.7814

Thus, the probability that the sample mean will be within ±25 of the population mean μ is 0.7814.

Here, all the probabilities are approximately same therefore for all three firms the sample of size 50 works.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
0|0|0|0 - Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were performed on Xt to obtain Yt? seasonal difference of order 2 simple difference of order 5 seasonal difference of order 1 seasonal difference of order 5 simple difference of order 2
Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.
Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO

Chapter 7 Solutions

Essentials of Statistics for Business and Economics

Ch. 7.3 - The following data are from a simple random...Ch. 7.3 - Prob. 12ECh. 7.3 - 13. A sample of 5 months of sales data provided...Ch. 7.3 - 14. Morningstar publishes ratings data on 1208...Ch. 7.3 - Rating Wines. According to Wine-Searcher, wine...Ch. 7.3 - AARP Survey. In a sample of 426 U.S. adults age 50...Ch. 7.3 - Attitudes Toward Automation. The Pew American...Ch. 7.5 - 18. A population has a mean of 200 and a standard...Ch. 7.5 - A population has a mean of 200 and a standard...Ch. 7.5 - 20. Assume the population standard deviation is σ...Ch. 7.5 - 21. Suppose a random sample of size 50 is selected...Ch. 7.5 - Sampling Distribution for Electronic Associates,...Ch. 7.5 - Finding Probabilities for Electronic Associates,...Ch. 7.5 - Barrons reported that the average number of weeks...Ch. 7.5 - SAT Scores. In May 2018, The College Board...Ch. 7.5 - Federal Income Tax Returns. The Wall Street...Ch. 7.5 - College Graduate-Level Wages. The Economic Policy...Ch. 7.5 - State Rainfalls. The state of California has a...Ch. 7.5 - Income Tax Return Preparation Fees. The CPA...Ch. 7.5 - To estimate the mean age for a population of 4000...Ch. 7.6 - A sample of size 100 is selected from a population...Ch. 7.6 - 32. A population proportion is .40. A sample of...Ch. 7.6 - Prob. 33ECh. 7.6 - 34. The population proportion is .30. What is the...Ch. 7.6 - Orders from First-Time Customers. The president of...Ch. 7.6 - The Wall Street Journal reported that the age at...Ch. 7.6 - Food Waste. In 2017, the Restaurant Hospitality...Ch. 7.6 - Unnecessary Medical Care. According to Readers...Ch. 7.6 - Better Business Bureau Complaints. In 2016 the...Ch. 7.6 - 40. The Grocery Manufacturers of America reported...Ch. 7.6 - 41. The Food Marketing Institute shows that 17% of...Ch. 7.9 - A population has a mean of 400 and a standard...Ch. 7.9 - Assume the population standard deviation is = 25....Ch. 7.9 - Prob. 44ECh. 7.9 - Assume that the population proportion is .44....Ch. 7.9 - Vacation Hours Earned by Blue-Collar and Service...Ch. 7.9 - MPG for New Cars. The New York Times reported that...Ch. 7.9 - Repeat Purchases. The president of Colossus.com,...Ch. 7.9 - Landline Telephone Service. According to the U.S....Ch. 7 - Shadow Stocks. Jack Lawler, a financial analyst,...Ch. 7 - Personal Health Expenditures. Data made available...Ch. 7 - 44. Foot Locker uses sales per square foot as a...Ch. 7 - Airline Fares. The mean airfare for flights...Ch. 7 - After deducting grants based on need, the average...Ch. 7 - Prob. 55SECh. 7 - Survey Research Results. A researcher reports...Ch. 7 - Production Quality Control. A production process...Ch. 7 - Australians and Smoking. Reuters reports that 15...Ch. 7 - 51. A market research firm conducts telephone...Ch. 7 - Advertisers contract with Internet service...Ch. 7 - The proportion of individuals insured by the...Ch. 7 - Lori Jeffrey is a successful sales representative...Ch. 7 - Life of Compact Fluorescent Lights. In 2018, the...Ch. 7 - Typical Home Internet Usage. According to USC...Ch. 7 - Undeliverable Mail Pieces. Of the 155 billion...Ch. 7 - U.S. Drivers and Speeding. ABC News reports that...Ch. 7 - Managerial Report Prepare a managerial report that...
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License