Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List)
Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List)
6th Edition
ISBN: 9781337115186
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
Publisher: Cengage Learning
bartleby

Videos

Question
Chapter 7, Problem 47SE

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 47SE

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 47SE

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!

Chapter 7 Solutions

Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List)

Ch. 7.3 - 15. The National Football League (NFL) polls fans...Ch. 7.3 - AARP Survey. In a sample of 426 U.S. adults age 50...Ch. 7.3 - 17. One of the questions in the Pew Internet &...Ch. 7.3 - In this section we showed how a simple random...Ch. 7.5 - 19. A population has a mean of 200 and a standard...Ch. 7.5 - Prob. 16ECh. 7.5 - 21. Suppose a random sample of size 50 is selected...Ch. 7.5 - Refer to the EAI sampling problem. Suppose a...Ch. 7.5 - In the EAI sampling problem (see Figure 7.8), we...Ch. 7.5 - 24. Barron’s reported that the average number of...Ch. 7.5 - The average full-sized front-loading Energy Star...Ch. 7.5 - Federal Income Tax Returns. The Wall Street...Ch. 7.5 - 27. The Economic Policy Institute periodically...Ch. 7.5 - State Rainfalls. The state of California has a...Ch. 7.5 - The mean preparation fee H&R Block charged retail...Ch. 7.5 - 30. To estimate the mean age for a population of...Ch. 7.6 - 31. A sample of size 100 is selected from a...Ch. 7.6 - 32. A population proportion is .40. A sample of...Ch. 7.6 - Prob. 29ECh. 7.6 - 34. The population proportion is .30. What is the...Ch. 7.6 - Prob. 31ECh. 7.6 - 36. The Wall Street Journal reported that the age...Ch. 7.6 - Seventy two percent of American adults have read a...Ch. 7.6 - Unnecessary Medical Care. According to Reader’s...Ch. 7.6 - Thirty-six percent of all Americans drink bottled...Ch. 7.6 - Prob. 36ECh. 7.6 - 41. The Food Marketing Institute shows that 17% of...Ch. 7.8 - Martina Levitt, director of marketing for the...Ch. 7.8 - Consider the Spontanversation sampling problem for...Ch. 7.8 - The Wall Street Journal reported that 37% of all...Ch. 7.8 - The vice president of sales for Blasterman...Ch. 7 - Shadow Stocks. Jack Lawler, a financial analyst,...Ch. 7 - 43. The latest available data showed health...Ch. 7 - 44. Foot Locker uses sales per square foot as a...Ch. 7 - 45. Allegiant Airlines charges a mean base fare of...Ch. 7 - 46. After deducting grants based on need, the...Ch. 7 - Prob. 47SECh. 7 - Prob. 48SECh. 7 - Production Quality Control. A production process...Ch. 7 - Prob. 50SECh. 7 - Prob. 51SECh. 7 - Prob. 52SECh. 7 - Prob. 53SECh. 7 - 54. Lori Jeffrey is a successful sales...Ch. 7 - Managerial Report Prepare a managerial report that...
Knowledge Booster
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.
Recommended textbooks for you
  • Glencoe Algebra 1, Student Edition, 9780079039897...
    Algebra
    ISBN:9780079039897
    Author:Carter
    Publisher:McGraw Hill
  • Glencoe Algebra 1, Student Edition, 9780079039897...
    Algebra
    ISBN:9780079039897
    Author:Carter
    Publisher:McGraw Hill
    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