Question

Question 15 Write down the null and alternative hypothesis to test for first-order autocorrelation. Question 16 Calculate the Durbin-Watson test statistic of autocorrelation? Use two decimal places and show clear steps for calculating the test statistic in Excel. Answers without supporting Excel calculation will be marked as Zero. Answer: 

          Question 15
Write down the null and alternative hypothesis to test for first-order autocorrelation.
Question 16
Calculate the Durbin-Watson test statistic of autocorrelation? Use two decimal places and show clear steps for calculating the test statistic in Excel. Answers without supporting Excel calculation will be marked as Zero.
Answer:
Show moreā€¦
Question 15 Write down the null and alternative hypothesis to test for first-order autocorrelation. Question 16 Calculate the Durbin-Watson test statistic of autocorrelation? Use two decimal places and show clear steps for calculating the test statistic in Excel. Answers without supporting Excel calculation will be marked as Zero. Answer:

Added by Johnny S.

Close

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
The null hypothesis for the Durbin-Watson test is H0: Ļ = 0, and the alternative hypothesis is Ha: Ļ ā‰  0. To calculate the Durbin-Watson test statistic in Excel, follow these steps: 1. Calculate the first-order differences of the data. 2. Square the first-order differences. 3. Sum the squared first-order differences. 4. Divide the sum of squared first-order differences by the sum of squared data. 5. The result is the Durbin-Watson test statistic. Round to two decimal places.
Close icon
Play audio
Feedback
Powered by NumerAI
David Collins Ivan Kochetkov
Danielle Fairburn verified

Akash M and 70 other subject Intro Stats / AP Statistics educators are ready to help you.

Ask a new question
*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs
*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.
*

Recommended Videos

-
perform-a-test-for-the-autocorrelation-in-the-residuals-of-the-estimated-model-with-detailed-steps-hypothesis-test-statistics-decision-rule-decision-and-conclusion-eviews-outputs-of-the-test-55097

Akash M.
to-perform-a-test-of-the-null-and-alternative-hypotheses-shown-below-random-samples-were-selected-from-the-two-normally-distributed-populations-with-equal-variances-the-data-are-shown-below-14073

To perform a test of the null and alternative hypotheses shown below, random samples were selected from the two normally distributed populations with equal variances. The data are shown below. Test the null hypothesis using an alpha level equal to 0.10. Sample from Population 1: 38,28,28,39,39,33,29,37,43,38 Sample from Population 2: 45,53,37,47,44,38,43,46,46,41 H0: ?1 ? ?2 = 0 HA: ?1 ā€“ ? ? 0 Determine the rejection region for the test statistic t. Select the correct choice below and fill in the answer box(es) to complete your choice. A. t < ____ or t > ____ B. t > ___ C. t < ___ Calculate the value of the test statistic. t = ___ Since the test statistic ____ (is not, or is) in the rejection region, _____ (reject, or do not reject) the null hypothesis. There is ____ (sufficient, or insufficient) evidence to conclude that the two population means are different. show formulas in excel

James K.
the-null-and-alternate-hypotheses-are-h0-1-2-h1-1-2-a-random-sample-of-24-items-from-the-first-population-showed-a-mean-of-104-and-a-standard-deviation-of-10-a-sample-of-14-items-for-the-sec-05651

The null and alternate hypotheses are: H0: Ī¼1 ā‰¤ Ī¼2 H1: Ī¼1 > Ī¼2 A random sample of 24 items from the first population showed a mean of 104 and a standard deviation of 10. A sample of 14 items for the second population showed a mean of 99 and a standard deviation of 6. Use the 0.05 significant level. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) State the decision rule for 0.050 significance level. (Round your answer to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 3 decimal places.) What is your decision regarding the null hypothesis? Use the 0.05 significance level.

Adi S.
*

Recommended Textbooks

-
Elementary Statistics a Step by Step Approach

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition
achievement 1,404 solutions
The Practice of Statistics for AP

The Practice of Statistics for AP

Daren S. Starnes, Daniel S. Yates, David S. Moore 4th Edition
achievement 1,450 solutions
Introductory Statistics

Introductory Statistics

Barbara Illowsky, Susan Dean 1st Edition
achievement 1,270 solutions
*

Transcript

-
00:01 First state the hypothesis null hypothesis for the test of correlation in the residuals is that there is no autocorrelation present in the residuals the alternate hypothesis is that the autocorrelation present in the residuals second is test statistics the test statistics used to test for autocorrelation is the q statistics the q statistics calculated as the sum of the squared autocorrelations up to a certain log third is decision rule to determine whether to reject or fail the to reject the null hypothesis will compare the calculated q statistic to the critical value from the chi -square distribution so the critical value is determined based on the desired level of significance and degrees of freedom which is equal to the number of lags being tested for this decision if the calculated q statistics is greater than the critical value we reject the null hypothesis and conclude that there is evidence of auto correlation in the residuals if the calculated q statistics is less than or equal to the critical value we fail to reject the null hypothesis and conclude that there is no evidence of autocorrelation the residual first step is conclusion.
01:45 So based on the provided e views outputs we can see that the q statistics for each log is significantly greater than 0 with values of 0 .000 so this indicates that there is evidence of autocorrelation in the residuals therefore we reject the null hypothesis and conclude that the residuals do not follow a white noise process...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade

Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever