How to test the hypothesis : "16% of vegetarians are vegans" with 19 out of 200 respondents being vegan?
Answer
1. Here H0 must be an equality so H0 is : p=0.16 and H1 is : \(p \neq .16\)
2. We use a Two Tailed test SOOOO =>
3. Z score/T score with formula : \(Z={(\hat{p}-p)\over{\sqrt{p\times(1-p)\over{n}}}}\) needs to be > Zscore of .975 OR < Zscore of .025 (Here \(Z_\alpha = \pm1.96\))
4. Here Z = 1.158 so we cannot reject H0
5. Altenatively p value for Z is 0.13 so > 0.05 so we can not rejet H0
NB : as H0 is an equality we are interested in areas to the RIGHT of a positive Z (in EXCEL 1-norm.dist) and to the LEFT of a negative Z (in EXCEL normdist)
Question
How to test the hypothesis : "16% of vegetarians are vegans" with 19 out of 200 respondents being vegan?
Answer
?
Question
How to test the hypothesis : "16% of vegetarians are vegans" with 19 out of 200 respondents being vegan?
Answer
1. Here H0 must be an equality so H0 is : p=0.16 and H1 is : \(p \neq .16\)
2. We use a Two Tailed test SOOOO =>
3. Z score/T score with formula : \(Z={(\hat{p}-p)\over{\sqrt{p\times(1-p)\over{n}}}}\) needs to be > Zscore of .975 OR < Zscore of .025 (Here \(Z_\alpha = \pm1.96\))
4. Here Z = 1.158 so we cannot reject H0
5. Altenatively p value for Z is 0.13 so > 0.05 so we can not rejet H0
NB : as H0 is an equality we are interested in areas to the RIGHT of a positive Z (in EXCEL 1-norm.dist) and to the LEFT of a negative Z (in EXCEL normdist)
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owner: naxplast06 - (no access) - Deviant S. - The Practically Cheating Statistics Handbook (2010, CreateSpace Independent Publishing Platform).pdf, p130
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