FAQ: Hypothesis Testing - ANOVA

This community-built FAQ covers the “ANOVA” exercise from the lesson “Hypothesis Testing”.

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Data Science

FAQs on the exercise ANOVA

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I don’t get it:
In the explanation: “The null hypothesis, in this case, is that all three populations have the same mean … If we reject this null hypothesis (if we get a p-value less than 0.05), we can say that we are reasonably confident that a pair of datasets is significantly different.”
But in the exercise:
With store_b the means are : 58.349636084 65.6262871356 62.3611731859
and p-value is 0.000153411660078 ie we can reject the null hypothesis (see above) and the samples are different.
With store_b_new the means are: 58.349636084 148.354940186 62.3611731859
and p-value is 8.49989098083e-215 ie we cannot reject the null hypothesis (see above) and the samples are basically the same.
Surely that is the wrong way round?


No, it’s correct.

The null hypothesis in this case is “There is no significant difference in sales between the stores.”

Rejecting the null hypothesis (p-value < 0.05) would mean there IS a significant difference between the at least one store.

The new sales numbers for Store B easily pass the eye test and you’d expect to reject the null hypothesis. And that’s exactly what happened in the ANOVA test (p-value = 8.49989098083e-215). You would say that there is a 99.999999…% chance that a store is significant.


I found myself still confused as to how the p-value was less than 0.05 until I learned what the “e-” means. This wasn’t taught anywhere in the Data Science path prior to this exercise so I am posting it here in case it is new to anyone else.

Basically, the “e-” format in this case tells you that the p-value is 8.49989098083 times 10^-215, so it is 0 point followed by 214 zeroes and then 849…

In the project that comes after the end of this module, there is a p-value of 2.74631179866e-10. So, this should be read as the p-value equaling 0.000000000274631178966.


Why ANOVA’s prediction is much precise than Two Sample T-Tests?

My first suggestion:
With both methods we check: if at least one pair of samples in, let us say in 3 samples set, has significant differencies.
-Two Sample T-Test checks it via comparisson pairs like this: a_b & a_c & b_c (all the experiments depend on each other)
-ANOVA checks it via comparisson pairs like this: a_b | a_c | b_c (all the experiments are isolated)

My second suggestion:
Both methods check different things:
-Two Sample T-Test checks which of 3 pairs has significant differences (it is more specific conclusion, but much demanding to accuracy)
-ANOVA checks if there are significant differencies between at least one pair (it is more general conclusion, since we don’t know the specific pair, but it is less demanding to accuracy)

I found this very helpful.
I do not remember this e-format as well. I was getting curious when it asked if it is less than 0.05

Can you please tell how the tests are dependent (2 sample) and independent( anova).