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Mathematics 19 Online
OpenStudy (allison.ortiz):

does anyone know how to do hypothesis test for population proportion binomial distribution?

jimthompson5910 (jim_thompson5910):

did you have a problem you wanted to go over?

OpenStudy (allison.ortiz):

Ho: p=.065 Hi: p< .065 p=.062 n=60

OpenStudy (nincompoop):

go on

OpenStudy (allison.ortiz):

That is the problem. I know how to solve hypothesis test proportions when the Hi is > but I don't know how to solve it when the Ho : is <

OpenStudy (nincompoop):

okay what is it that you know about p-values?

OpenStudy (nincompoop):

and tail tests

OpenStudy (nincompoop):

for binomial probability distribution this is the familiar formula \(\sf \large P(x) = _nC_xp^{x}(1-p)^{n-x}\) but this will not really help you if you don't have a good grasp of the concept.

OpenStudy (nincompoop):

and perhaps from your previous probability/stat lessons you've encountered p-values \(\hat{p} = \dfrac{x}{n}\)

OpenStudy (agent0smith):

"That is the problem. I know how to solve hypothesis test proportions when the Hi is > but I don't know how to solve it when the Ho : is <" It doesn't make a difference, at all. You're still looking for the p-value to be in the critical region, and if it is, you reject the null hypothesis.

OpenStudy (nincompoop):

testing claims really

OpenStudy (allison.ortiz):

OpenStudy (allison.ortiz):

the problem is that the requirements are not satisfied np(1-p) >10 n=60 p=0.062 60*0.062(1-.062) =3.489 > 10

OpenStudy (agent0smith):

Your numbers don't match at all with the document you posted. And if the requirements aren't met, that just means you can't use the normal approximation. You just have to use the actual binomial formula.

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