APPLIED STAT.IN BUS.+ECONOMICS
APPLIED STAT.IN BUS.+ECONOMICS
6th Edition
ISBN: 9781259957598
Author: DOANE
Publisher: RENT MCG
bartleby

Videos

Question
Book Icon
Chapter 16, Problem 36CE
To determine

State the hypotheses, test statistic and the two-tailed p-value.

Make a decision for the research question “whether there is significant correlation between gasoline price and carbon dioxide emission?” and identify whether the research question is close to the obtained decision.

Identify the issues of sample size.

Identify whether non-normality is concerned or not.

Expert Solution & Answer
Check Mark

Answer to Problem 36CE

The hypotheses for the test are given below:

Null hypothesis:

H0:ρs0

The rank correlation between gasoline price and carbon dioxide emission is zero.

Alternate Hypothesis:

H1:ρs>0

The rank correlation between gasoline price and carbon dioxide is greater than zero.

The test statistic and the p-value are –0.588 and 0.355 respectively.

There is a support evidence to conclude that there exists a significant correlation between gasoline price and carbon dioxide emission at 5% level of significance.

The decision is close since the rejection of null hypothesis is close to the research question “whether there is significant correlation between gasoline price and carbon dioxide emission?”

The sample size is not an issue.

The gasoline price follows normality but the carbon dioxide emission does not follow normality.

Explanation of Solution

Calculation:

The given information is that, the data shows gasoline price and carbon emissions for selected nations. The level of significance is 0.05.

The hypotheses for the test are given below:

Null hypothesis:

H0:ρs0

The rank correlation between gasoline price and carbon dioxide emission is zero.

Alternate Hypothesis:

H1:ρs>0

The rank correlation between gasoline price and carbon dioxide emission is greater than zero.

Software procedure:

Step-by-step procedure to find the Spearman rank correlation using MINITAB is given below:

  • Choose MegaStat > Nonparametric Tests > Spearman Coefficient of Rank Correlation.
  • In Input range, select the cells A1:B32.
  • Unselect Output ranked data and Correct for ties.
  • Click OK.

Output obtained from MegaStat is given below:

APPLIED STAT.IN BUS.+ECONOMICS, Chapter 16, Problem 36CE , additional homework tip  1

Decision Rule:

Reject the null hypothesis (H0) when the p-value is lesser than or equal to the level of significance α, otherwise do not reject the null hypothesis (H0).

Conclusion:

The absolute value of the test statistic is 0.588 and the critical value for the desired level of significance α is 0.355.

The test statistic is greater than the critical value.

That is, 0.588(=test statistic)>0.355(=critical value).

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to conclude that there exists a significant correlation between gasoline price and carbon dioxide at 5% level of significance.

Histogram for gasoline price:

Step-by-step procedure to construct a histogram using MINITAB is given below:

  • Choose Basic Statistics > Graphical Summary.
  • Choose Simple, and then click OK.
  • In variables, enter the column of Gasoline price.
  • In Confidence level, enter 95.0.
  • Click OK.

Output obtained from MINITAB is given below:

APPLIED STAT.IN BUS.+ECONOMICS, Chapter 16, Problem 36CE , additional homework tip  2

Interpretation:

The histogram appears to be left skewed since the tail is elongated towards the left than to the right side. Hence, the test of normality is recommended.

Testing the normality using Anderson Darling test:

Null hypothesis:

H0: The given data follows normal distribution.

Alternate Hypothesis:

H1: The given data do not follow normal distribution.

Decision Rule:

Reject the null hypothesis (H0) when the p-value is lesser than or equal to the level of significance α, otherwise do not reject the null hypothesis (H0).

Conclusion:

The p-value for the A-D test is 0.169 and the level of significance is 0.05.

The p-value for the A-D test is greater than the level of significance.

That is, 0.169(=p-value)>0.05(=α)

Thus, the null hypothesis is rejected.

Hence, there is a support of evidence to assume that gasoline follows normal distribution at 5% level of significance.

Histogram for carbon dioxide emission:

Step-by-step procedure to construct a histogram using MINITAB is given below:

  • Choose Basic Statistics > Graphical Summary.
  • Choose Simple, and then click OK.
  • In variables, enter the column of carbon dioxide emission.
  • In Confidence level, enter 95.0.
  • Click OK.

Output obtained from MINITAB is given below:

APPLIED STAT.IN BUS.+ECONOMICS, Chapter 16, Problem 36CE , additional homework tip  3

Interpretation:

The histogram appears to be right skewed since the tail is elongated towards the right than to the left side. Hence, the test of normality is recommended.

Testing the normality using Anderson Darling test:

Null hypothesis:

H0: The given data follows normal distribution.

Alternate Hypothesis:

H1: The given data do not follow normal distribution.

Decision Rule:

Reject the null hypothesis (H0) when the p-value is lesser than or equal to the level of significance α, otherwise do not reject the null hypothesis (H0).

Conclusion:

The p-value for the A-D test is lesser than 0.005 and the level of significance is 0.05.

The p-value for the A-D test is lesser than the level of significance.

That is, <0.005(=p-value)<0.05(=α)

Thus, the null hypothesis is rejected.

Hence, there is no support of evidence to assume that the carbon dioxide emission follows normal distribution at 5% level of significance.

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
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
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
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
College Algebra
Algebra
ISBN:9781938168383
Author:Jay Abramson
Publisher:OpenStax
Text book image
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell
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