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Aug
14
comment Exact interpretration of p-value and significance of test
It is always an issue of plausibility; you can never reach absolute certain, but it is more plausible/likely , given the sample data and the distribution, that the mean is actually 100.
Aug
14
comment Exact interpretration of p-value and significance of test
But the number of observations is factored in the p-value already.
Aug
14
comment Exact interpretration of p-value and significance of test
Phrasing is confusing. I would say we conclude/assume the mean is 100, but we cannot be 100% sure that this is the case.
Aug
14
answered If a property holds for arbitrary compact set in a metric space, does it also holds for the metric space?
Aug
14
comment Exact interpretration of p-value and significance of test
Right, then you do not reject. But you do not conclude the mean is 100; the sample data may have been either unrepresentative or made up of outliers. You can only conclude that under assumptions (which you cannot guarantee) of data being representative (not outliers ) and randomly-selected, the claim is true.
Aug
14
revised Exact interpretration of p-value and significance of test
added 381 characters in body
Aug
14
comment Exact interpretration of p-value and significance of test
Sorry, I misread ;let me add something.
Aug
14
comment Exact interpretration of p-value and significance of test
Please see the bottom paragraph I just wrote and let me know if it answers your question.
Aug
14
revised Exact interpretration of p-value and significance of test
added 381 characters in body
Aug
14
answered Exact interpretration of p-value and significance of test
Aug
14
comment $f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.
Are you assuming there is a local homeomorphism in order to use this argument?
Aug
14
comment $f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.
O.K, so we agree ;).
Aug
14
comment $f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.
Sorry to nitpick, but then it is (pairwise) disjoint, not distinct.
Aug
14
comment $f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.
Why not? You can form an open interval (a,b) for every pair of Reals. Maybe you mean an open subset of Reals with uncountably-many components?
Aug
14
answered $f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.
Aug
14
comment Is $B = A^2 + A - 6I$ invertible when $A^2 + 2A = 3I$?
@StefanSmith: Don't count on it. Unfortunately some people here apparently cannot be bothered to offer constructive criticism, and they prefer to punish instead, with anonymous down-votes.
Aug
13
reviewed Approve Function with continuous inverse is continuous?