Відео

Thank you!🙏

So nice of you

Hello friend, that is a very good question! Yes, the lower value of AD shows better fit. Choosing Weibull is still safe as Exponential is only a special case of Weibull with a shape parameter of 1.

Thank you for asking an excellent question! Yes, Minitab can analyse data of an accelerated life test with two variables.

@@uhemant1 thank you for the response 😊

Thank you! I am glad to know you are finding the videos useful! I am not aware of the software mentioned by you. Apologize.

My pleasure!

A standby pump would be in parallel with the primary pump with a switch. The reliability of the switch also needs to be considered.

You need to separate the failure modes for taking improvement actions. For growth calculations, all failure modes should be considered.

@@instituteofqualityandrelia7902 Thank you so much for the quick answer. I'm studying the warranty. So I need consider all failure modes. Rght?

Thank you!🙏

I am not sure why Minitab does this often. However, Exponential is only a special case of Weibull Distribution with a shape parameter of 1. So if you select Weibull, Exponential will get included, if it fits.

Glad it was helpful!

Hello Fernando, Thanks for your interest. There are two gensets. So the time is time column shows data when any of the genset fails. Hope this is clear.

Yes I got it. When performing this analysis using Minitab the results are different. Also, how this Eta and Beta relates to shape and scale parameters informed when using Minitab?

Glad you think so! Keep watching!🎉

Thank you!🙏

Appreciate your feedback! I suggest that you create the xl worksheet for better learning and understanding!

Thank you!🙏

Most welcome!

No. 63% is correct. For Weibull distribution, by time equal to scale parameter, 63.2% parts are expected to fail. You may like to watch my video on Weibull distribution. Here is the link: ua-cam.com/video/dxwlYZ-8_GM/v-deo.html

Glad to hear that it was helpful! Appreciate your feedback!

I have used Minitab software. There are other software such as Weibull++.

It is basically for test time in multiples of lower test mtbf.

These are calculated based on probabilities as per assumed distribution.

Thank you for your interest. 1/i is delta hazard where i is the reverse rank. The reverse rank represents items at risk just prior to t. Thus proportion of items failing out of SURVIVING population during the interval is by definition delta of hazard rate. You may like to watch my video on hazard rate. Here is the link: ua-cam.com/video/0a-BF036Di4/v-deo.html

Thank you for your suggestion. Kindly clarify your expectations.

It is explained in the video. You can also refer to the paper referenced.

Glad you think so!

Thx. But I haven't understood your question.

Glad you enjoyed it

So nice of you

You’re welcome 😊

Thank you! Please watch my video on how to estimate beta and eta parameters using probability plotting. Here is the link: ua-cam.com/video/dsuLVS2yQ4U/v-deo.htmlsi=i7saq3ej5S-wkYKk

@@instituteofqualityandrelia7902 Thank you very much sir..

Watch my other video: ua-cam.com/video/dsuLVS2yQ4U/v-deo.htmlsi=i7saq3ej5S-wkYKk

@@instituteofqualityandrelia7902 Sir can you tell me how we can decide which plots are good and which one is to choose or very helpful for further analysis after conducting the ALT How to decide plots according to what??

I am glad you liked it!🎉🙏

Beta value is estimated from the data of past failures.

This is possible when you do not have any degrees of freedom for the error term.

@@instituteofqualityandrelia7902 because of this in minitab I didn't get the values of p test , f test and SE Coefficient So how I can resolve this??

@@instituteofqualityandrelia7902 basically I'm facing this problem when I use the minitab software and I have created 2 level 2 factor full factorial design in this, when I checked for the significance then I didn't get any P values ...

Yes. This is called a saturated design. If you replicate the design, your problem will be solved as then you will get enough degrees of freedom for error. With two replicates, there will be 8 runs. Total degrees of freedom will be 7 and there will now be 3 degrees of freedom for error. Hope this helps.

@@instituteofqualityandrelia7902 Thanks sir . .👍🏻

ioqr@world-class-quality.com

Thank you! The left censored data can be analysed using similar commands in arbitratily censored data.

This is only because in XL, the notation is different. In most literature, the shape parameter is denoted as beta. In XL, the shape parameter is denoted as alpha.

Hello there! This is a property of Weibull Distribution. You may like to watch my videos on Weibull Distribution. Links shown below: Weibull Distribution Part-1 ua-cam.com/video/dxwlYZ-8_GM/v-deo.html Weibull Distribution Part-2 ua-cam.com/video/WbG5Z7irUu4/v-deo.html

You are welcome! I am really glad to know!

Standard Deviation can be for sample or population. In this particular case, standard deviation is calculated based on 250 data points and therefore can be practically considered as population standard deviation, If control charts are use and process is stable, the standard deviation can be calculated using the control chart, but then it is considered as 'within' subgroup variation and represents the random portion of the variation.

It depends on what the blocking factor is. If p-value shows that it is significant, then we need to consider the block in the model and prediction. There is no question of accepting or not accepting!

Thank you for your interest! You can watch my other videos on the sample size. Links are given below: ua-cam.com/video/Aa8FFh4GxCw/v-deo.htmlsi=2blPOcNFGSKgW9zt ua-cam.com/video/XjKY8ELjupQ/v-deo.html Let me know if you still have questions!

Glad you enjoyed it!

If we take cumulative frequency value as claculated, it will mean that no tube will fail below 26.8 and all tubes will fail by 34.9! It is not appropriate to draw this conclusion based on a sample of 10! Thus this formula, knows as Modified Kapaln Meier Method, makes a correction based on sample size n. As na becomes larger, the correction becomes smaller! A more populalar method of such a correction is called median rank and is given by (j-0.3)/(n+0.4). Applicability of median rank is wider as many distributions, such as Weibull, are skewed. You can watch my video on Weibull probability plotting: ua-cam.com/video/dsuLVS2yQ4U/v-deo.html. Hope this helps.

@@instituteofqualityandrelia7902 that's super detailed response. Got it sir. Thank you for sharing the resource... and thank you for taking out your time to reply.

Welcome! Keep watching my technical videos!

Hello Lokesh, thank you for your interest and observation. Unfortunately that is a typo error. It should be same as (Xuc-13)/3. I apologise for the error. I am adding a note in the video description. UA-cam videos cannot be edited for such errors. Thanks you for pointing out.