F value in ANOVA table?

F value in ANOVA table? Topic: Rejecting null hypothesis of anova tests
June 16, 2019 / By Donough
Question: I have an ANOVA table where I have F value = 0.00 and P<0.0001. How do I interpret these results? what does F value tell me in this case? Thanks note: DF=1
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Best Answers: F value in ANOVA table?

Brannon Brannon | 10 days ago
A low P value indicates you can reject your null hypothesis, not the opposite as another answerer stated. That is arbitrarily set (generally .05 or .01). No matter what your P value is you cannot "accept" your null hypothesis. You only have two choices, reject your null or not reject it. Rejecting it can support your alternative hypothesis, but you cannot say you can accept it, only support. Not rejecting the null obviously means you cannot support your alternative, but you cannot accept a null hypothesis through statistics. The F value is basically what plays into determining the p value. A low F indicates any variation in the data can be accounted for essentially by chance. A large F value indicates the variation cannot entirely be accounted for by chance. This is a rather basic and fundamental part of statistics, so I would suggest looking through your lecture notes and textbook to see how an ANOVA test is performed. Simply doing that should have answered all of your questions before needing to ask here.
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Brannon Originally Answered: A database contains a table named tb00_Categories. This table has two fields: CategoryCode (int) and Cat?
This is not a question asking for helpso that you can complete possibly homework, or a project at school/college. It is an instruction to do something for you without even a please!

Adonijah Adonijah
For the best answers, search on this site https://shorturl.im/axnCS I don´t know what your null hypothesis is, and what your alpha is. Therefore my help will be limited. Ok, you can interpret the ANOVA results in two different ways. 1. If you use the p value, then you compare that with your alpha or confidence level For example, if you are working with a confidence level of 95% you alpha is 0.05. Therefore, if your p value is less than 0.05 then you reject your null hypothesis. In your case is a lot less than 0.05 so you reject your null hypothesis. 2. If you use the F value, you have to see that your results also show another value that says F critical. So basically you compare both values. Here the criteria used is if your F value is less than F critical then you reject your null hypothesis. I don´t know what your null hypothesis is, but usually when you work with an ANOVA your hypothesis should look like this: the means of the populations are equal. And your alternate hypothesis is: at least one of the means in your populations is not equal to the others. If you prove that you have to reject your null hypothesis, than you have to do a post-hoc test to prove which mean is the one that´s different. But I don´t know if you have or need to go that far. I hope my explanation is clear. If not let me know and I will elaborate a little more. What software did you use to compute your results?? Source: I´m a biologist
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Storm Storm
You seem to be missing some details from your ANOVA table, but your values are telling of another problem. 1. The F value is calculated generally by dividing MS source / MS error or this can be thought of as the variance due to effect / variance due to error. An F that is equal to 1 would suggest that both the variance from effect and the variance from the error are equal or the same... This is very common and is usually associate with a very high probability that your result is due to chance... but in your case the F value is less than 1 suggesting that the variance due to error is higher than the variance due to source. Since the F probability distribution has two tails it is also very unlikely to have very low F values due to chance. This is why you have such a low P value... the problem is this F value makes your result uninterpretable because as I mentioned you have the opposite of an effect... you may have some other issue with your data set.
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Quinta Quinta
OK. Your F value is so less which is good. Your F value has p<0.0001 which indicate that your null hypothesis can not be rejected. In other words the probability of being wrong of your decision (on accepting null hypothesis )is less than 0.0001. The 1-0.0001=0.9999 is the probability of your decision be reliable.
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Melantha Melantha
guys help me please.. my computed value was 0.913.. can you guys tell me what is my tabular value at alpha .05? I really need the answer as soon as possible guys.. thank you so much.. hoping for your answer....
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Melantha Originally Answered: Anova or Kruskal-Wallis? 10 points?
REGION cannot be scale. You have to change it to Nominal. (Correction) I just realize the fatality rate has already been provided. You only need Table 1083: Traffic Fatalities by State. Your homework is still to assign which REGION a state should belong to. You have a total of 51 cases. You should run exploratory analysis to see if your three groups (regions) have normal distributions. However, since your smallest case still have a sample size of 43, I would simply assume it is ok to run ANOVA.

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