In the following tables *n* is the sample size and *k* is the number of independent variables. See Autocorrelation for details.

**Alpha = .01**

**Alpha = .05**

Everything you need to do real statistical analysis using Excel

In the following tables *n* is the sample size and *k* is the number of independent variables. See Autocorrelation for details.

**Alpha = .01**

**Alpha = .05**

hi charles

can you tell me n=304

Eric,

This depends on the number of independent variables and the alpha value. See the following for more information

https://web.stanford.edu/~clint/bench/dwcrit.htm

Charles

Hy Charles,

thats very great articles.

but I want to ask you something.

My teacher give me some homework, and there’s n >200. how if i want to know the value from n = 240?

thanks before.

Linda,

The value is probably fairly similar to that given for n = 200, but you can also look at the following webpages:

https://stats.stackexchange.com/questions/160830/durbin-watson-critical-values-for-large-sample-sizes

http://www.inderscienceonline.com/doi/abs/10.1504/IJCEE.2016.073370

Charles

Hi Charles,

I have data of 1.176 households over 13 weeks in my dataset, which makes 25.288 lines of data. I have 22 explanatory variables in my model + an intercept. My DW statistic is 1.511.

How can I test whether autocorrelation is an issue?

Thanks,

Mark

Mark,

Yes, this is much larger than the largest tables. See the following webpages:

https://stats.stackexchange.com/questions/160830/durbin-watson-critical-values-for-large-sample-sizes

http://www.inderscienceonline.com/doi/abs/10.1504/IJCEE.2016.073370

Charles

Hello, can you help me to get dl and du for 1346 samples, 3 variables, and alpha 0.5? Thank you.

Diva,

Here are the values for 1300 and 1350. You will need to interpolate to get the value for 1346 (although obviously it will be pretty close to the value for 1350). I assume that alpha is .05 (and not .5) and that 3 variables means in addition to the constant term.

n = 1300: dl = 1.90420 and du = 1.91345

n = 1350: dl = 1.90607 and du = 1.91498

You can also use the Real Statistics functions DLowerCrit and DUpperCrit.

Charles

Charles,

I have a time sequence of 48 numbers (quarterly crime stats in fact). DW test statistic calculates as 1.632. The alpha=5% lower and upper bounds for k=1 n=48 (interpolated) are 1.492 & 1.577. As d is > than du there is failure to reject the null of no auto-correlation. When checking against the alpha=1% bounds the interpolated lower and upper bounds are 1.310 & 1.392. d is now even farther from the upper critical value. However as the 1% test is a more rigorous test, I would have intuitively thought that the 1% upper bound would move towards the value of 2 (no auto-correlation) rather than away from it. Why is the 1% test less demanding than the 5% one for a DW test of a hypothesis?

Joe

Joe,

When alpha = 1% it should be easier to retain the null hypothesis than when alpha = 5%. This is indeed the case.

When alpha = 1% it should be harder to reject the null hypothesis than when alpha = 5%. This is indeed the case.

Charles

what is the dl and du for 258 observations with 40 independent variables at 5%, noting that the 32 of these 40 idependent variable are dummy variables used for the fixed effect model.

Sawsan,

I have not seen any tables of critical values with more than 20 independent variables. I will look into creating an approximation for such critical values.

Charles

can you help me to explain how to interpolate for n=205?

i have read your link about it, but i still don’t understand

Merin,

You won’t be able to interpolate for values of n larger than 200 since the table provided on the website only goes up to n = 200.

The following are the table values for n = 210 and alpha = .05. You can now interpolate between n = 200 and n = 210. In fact, the linear interpolations are simply the average of the values for n = 200 and n = 210 for any given value of k

n k dL dU

210 1 1.76445 1.78358

210 2 1.75483 1.79326

210 3 1.74513 1.80305

210 4 1.73537 1.81295

210 5 1.72554 1.82294

210 6 1.71563 1.83305

210 7 1.70566 1.84325

210 8 1.69561 1.85355

210 9 1.6855 1.86394

210 10 1.67532 1.87445

210 11 1.66508 1.88505

210 12 1.65478 1.89574

210 13 1.64441 1.90653

210 14 1.63398 1.91742

210 15 1.62348 1.92839

210 16 1.61293 1.93947

210 17 1.60232 1.95063

210 18 1.59165 1.96188

210 19 1.58094 1.97323

210 20 1.57015 1.98467

Charles

ok thank you so much .. that helpfull

How can I find DL and DU for n=48 with 4 independent variable

Zhyan,

This depends on the alpha value, but if alpha = .05, then the table has values for n=45 and n=50 and you will need to interpolate between these values to get the answer you are looking for. See the following webpage for details abouit interpolation:

Interpolation

Charles

Hi Charles,

How if I want to know the value for n = 665?

Fiona,

See the following webpage>

http://web.stanford.edu/~clint/bench/dw05d.htm

You will need to interpolate between the values in the table.

Charles

Hy charles, can I know how the table of durbin watson for n=204

reply my coment please 🙂

Ade,

Please see the following website, which contains values for 200 and 210. You will need to interpolate to get the value at 204.

http://web.stanford.edu/~clint/bench/dw05c.htm

Charles

please i have TTF- TIME TO FAILURES values as follows; 28,52,42,8,14,13,47,38,25,12,50,42. How do i know my n and k so to check on DW table for du and dl.

Thanks

Joel,

See the following webpage

http://www.real-statistics.com/multiple-regression/autocorrelation/

Charles

Please am still waiting for the answer to my question.

thanks

If a (one period) auto-correlation adjustment is made, does the adjustment count as an additional k (independent) variable in evaluating DW limits. Thanks

Brad,

What sort of (one period) auto-correlation adjustment are your referring to? Are you referring to differencing?

Charles

If I want to know for n=41, how can I know value

Aditya,

You need to interpolate the values in the table between the entries for 40 and the entries for 45. This is explained on the webpage

Interpolation

Alternatively, you can use the Real Statistics functions DLowerCrit and DUpperCrit

Charles

Thank you, Charles!

Great job.

I appreciate you spending time to have this important table available.

Warmest regards,

Francisco

Hi Charles,

thank you for all the info on this website!

I was wondering where you got those tables from? I noticed that in Durbin & Watson’s paper from 1951 (Testing for serial correlation), they only go up to an N = 100 (and I am particularly interested in higher N’s). Is there another paper that I have been missing?

Thank you!

Olga

Hi Olga,

One source is

https://www3.nd.edu/~wevans1/econ30331/Durbin_Watson_tables.pdf

Charles

Hi Charles,

thanks for the quick reply! When I click on the link, it required authorization. Do you know if there’s another source I could click on? Thanks again!

Olga

Olga, no I don’t.

Charles

Thanks for this!

Hi Charles,

Thank you for your excellent website.

Can I confirm that the k in durbin watson table excludes intercept.

Thanks

Hi Cheng,

The tables are for the case where there is an intercept, but k does not include the intercept. Thus if k = 2, there are 2 independent variables plus an intercept.

Charles

Dear Dr. Zaiontz. I was wondering, how to interpret bounds for Durbin-Watson test (dU, dL) if dU is greater than 2.0. I know that it is an extreme case, but just want to know…

For example, for alpha=0.01 if n=13 and k=8 than dU=3.182, dL=0.090. Then 4-dU=0.818, 4-dL=3.91… Does it mean that between 0.090 and 3.190 test is inconlusive? It is quite strange situation, I know that there is plenty of tests better than Durbin-Watson one, but it is nice to know such details.

Love your website (thank you!) and your nice Slavic surname.

Greetings from Central Europe.

Good to hear someone from Central Europe and thanks for your kind words about the website.

The bounds for the Durbin-Watson test are indeed strange. If the test produces a value between dL and dU, i.e. between .818 and 3.10 in this case, then yes the test is inconclusive.

Charles