Loose Leaf for Statistical Techniques in Business and Economics
Loose Leaf for Statistical Techniques in Business and Economics
17th Edition
ISBN: 9781260152647
Author: Douglas A. Lind
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 64DA

a.

To determine

Check whether people can conclude that less than 40% of the districts buses are old.

Find the p-value.

a.

Expert Solution
Check Mark

Answer to Problem 64DA

Yes, people cannot conclude that less than 40% of the districts buses are old.

The p-value is 0.181.

Explanation of Solution

Calculation:

In this case, the test is to check whether less than 40% of the districts buses are old.

Let π represents hypothesized population proportion of old buses in districts.

H0:π0.40H1:π<0.40

From the data, it can be observed that the number of old buses is 28 out of 80.

The level of significance is 0.01.

Therefore, the value of z score using the table B.3: Areas under the normal curve is –2.33.

Decision rule:

Reject the null hypothesis if z < –2.326.

The sample size n is 80, p is 0.40 and the number of old buses (x) is 28.

Step-by-step procedure to find the test statistic using MINITAB software:

  • Choose Stat > Basic Statistics > 1 Proportion.
  • Choose Summarized data.
  • In Number of events, enter 28. In Number of trials, enter 80.
  • Enter Hypothesized proportion as 0.40.
  • Check Options, enter Confidence level as 99.0.
  • Choose greater than in alternative.
  • Select Method as Normal approximation.
  • Click OK in all dialogue boxes.

Output is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  1

Thus, the value of the test statistic is –0.91 and the p-value is 0.181.

In this case, the critical value is –2.33 and the test statistic is –0.91.

Here, the test statistic –0.91 is greater than the critical value –2.33.

That is, –0.91 > –2.33.

Therefore, do not reject the null hypothesis.

Therefore, people cannot conclude that less than 40% of the districts buses are old.

b.

To determine

Find the median maintenance cost and the median age of the buses.

Check whether the age of the bus is related to the amount of the maintenance cost.

b.

Expert Solution
Check Mark

Answer to Problem 64DA

The median maintenance cost and the median age of the buses are $4,179 and 7.00, respectively.

Yes, the age of the bus is related to the amount of the maintenance cost.

Explanation of Solution

Calculation:

In this case, the test is to check whether the age of the bus is related to the amount of the maintenance cost.

H0: There is no relationship between age of the bus and maintenance cost.

H1: There is a relationship between age of the bus and maintenance cost.

Step-by-step procedure to find the median for age of the bus and maintenance cost using MINITAB software:

  • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • In Variables enter the columns Age and Maintenance cost.
  • Choose option statistics, and select Median.
  • Click OK.

Output using MINITAB software is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  2

From the output, the median maintenance cost and the median age of the buses are $4,179 and 7.00, respectively.

Using the given conditions, the contingency table is obtained as follows:

High MaintenanceAge
Lower halfTop halfTotal
No33740
Yes93140
Total423880

The number of degrees of freedom is obtained as follows:

Degrees of freedom=(Rows1)(Columns1)=(21)(21)=1

Therefore, the number of degrees of freedom is 1.

Step-by-step procedure to find the critical value using MINITAB software:

  • Choose Graph > Probability Distribution Plot > View Probability > OK.
  • From Distribution, choose ‘Chi-Square’ distribution.
  • Enter Degrees of freedom is 1.
  • Click the Shaded Area tab.
  • Choose Probability and Right Tail for the region of the curve to shade.
  • Enter the data value as 0.01.
  • Click OK.

Output using MINITAB software is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  3

From the output, the critical value of chi-square is 3.841.

The general decision rule is reject the null hypothesis if χ2> critical value.

Therefore, the decision rule is reject the null hypothesis if χ2>3.84.

Test statistic:

Step-by-step procedure to find the test statistic using MINITAB software:

  • Choose Stat > Tables > Chi-Square Test for Association.
  • Choose Summarized data in a two-way table.
  • In Columns containing the table, enter the columns of Lower half and Top half.
  • In Rows under Labels for the table, enter the column of High Maintenance.
  • Click OK.

Output is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  4

From the output, the test statistic is 28.872.

The critical value is 3.84.

Here, the test statistic is greater than the critical value.

That is, 28.872 > 3.84.

Thus, reject the null hypothesis.

Therefore, there is sufficient evidence to conclude that age of the bus is related to the amount of the maintenance cost.

c.

To determine

Check whether there is a relationship between the maintenance cost and the manufacturer of the bus.

c.

Expert Solution
Check Mark

Answer to Problem 64DA

No, there is no relationship between the maintenance cost and the manufacturer of the bus.

Explanation of Solution

Calculation:

In this case, the test is to check whether there is a relationship between the maintenance cost and the manufacturer of the bus.

H0: There is no relationship between age of the maintenance cost and manufacturer of the bus.

H1: There is a relationship between age of the maintenance cost and manufacturer of the bus.

Using the given conditions, the contingency table for the maintenance cost and manufacturer of the bus is obtained as follows:

High MaintenanceManufacturer of the bus
BluebirdKeiserThompsonTotal
No2513240
Yes2212640
Total4725880

The number of degrees of freedom is obtained as follows:

Degrees of freedom=(Rows1)(Columns1)=(21)(31)=2

Therefore, the number of degrees of freedom is 2.

Step-by-step procedure to find the critical value using MINITAB software:

  • Choose Graph > Probability Distribution Plot > View Probability > OK.
  • From Distribution, choose ‘Chi-Square’ distribution.
  • Enter Degrees of freedom is 1.
  • Click the Shaded Area tab.
  • Choose Probability and Right Tail for the region of the curve to shade.
  • Enter the data value as 0.05.
  • Click OK.

Output using MINITAB software is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  5

From the output, the critical value of chi-square is 5.991.

The general decision rule is reject the null hypothesis if χ2> critical value.

Therefore, the decision rule is reject the null hypothesis if χ2>5.991.

Test statistic:

Step-by-step procedure to find the test statistic using MINITAB software:

  • Choose Stat > Tables > Chi-Square Test for Association.
  • Choose Summarized data in a two-way table.
  • In Columns containing the table, enter the columns of Lower half and Top half.
  • In Rows under Labels for the table, enter the column of High Maintenance.
  • Click OK.

Output is obtained as follows:

Loose Leaf for Statistical Techniques in Business and Economics, Chapter 15, Problem 64DA , additional homework tip  6

From the output, the test statistic is 2.231.

The critical value is 5.991.

Here, the test statistic is less than the critical value.

That is, 2.231 > 5.991.

Thus, do not reject the null hypothesis.

Therefore, there is no sufficient evidence to conclude that there is a relationship between the maintenance cost and the manufacturer of the bus.

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
Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a.  Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.)   I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…
Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 0. What level of type II error would you recommend here? = Round your answer to four decimal places (e.g. 98.7654). Use α = 0.05. β = 0.0594 What sample size would be required? Assume the sample sizes are to be…
Consider the hypothesis test Ho: 0 s² = = 4.5; s² = 2.3. Use a = 0.01. = σ against H₁: 6 > σ2. Suppose that the sample sizes are n₁ = 20 and 2 = 8, and that (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = 1.96 The critical value is f = 6.18 Conclusion: fail to reject the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/2/622 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) 035

Chapter 15 Solutions

Loose Leaf for Statistical Techniques in Business and Economics

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
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
The Shape of Data: Distributions: Crash Course Statistics #7; Author: CrashCourse;https://www.youtube.com/watch?v=bPFNxD3Yg6U;License: Standard YouTube License, CC-BY
Shape, Center, and Spread - Module 20.2 (Part 1); Author: Mrmathblog;https://www.youtube.com/watch?v=COaid7O_Gag;License: Standard YouTube License, CC-BY
Shape, Center and Spread; Author: Emily Murdock;https://www.youtube.com/watch?v=_YyW0DSCzpM;License: Standard Youtube License