rr400 wrote:
Hi, i am doing a statistics course and am having trouble with an exercise
where i need to determine whether my success rate at something is higher
than 80%.
I was successful in 29 out of 60 trials, so these were the commands i
entered into R:
n=60
p.hat=29/n
p.0=0.8
se.0=sqrt(p.0*(1-p.0)/n)
z=(p.hat-p.0)/se.0
print(z)
Which returned:
[1] -6.132224
1-pnorm(z)
Which returned
[1] 1
My problem is that i am meant to state a null and alternative hypothesis
which at the moment i have stated as p>0.8 (null) and p≤0.8 (alternative).
As things stand, though, a p-value of 1 suggests i should reject the null
hypothesis which can't be right since i am obviously successful less than
80% of the time.
I am not sure where i am getting muddled. Any advice would be greatly
appreciated. Thanks!
This isn't really about R, and maybe it is homework, but now that we got
you in the appropriate frame of mind:
(a) p values should look at "this or more unfavourable" events. You have
arranged things so that that translates to -6.13 or _lower_. I.e. you're
looking at the wrong tail.
(b) Make sure you get your accept/reject logic right. You _reject_ the
null when data would be _un_likely if the null hypothesis were true.
(c) You might also want to play with binom.test and prop.test
--
O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.