Re: [R] One-sample test for p

2008-10-16 Thread Greg Snow
--- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > project.org] On Behalf Of rr400 > Sent: Thursday, October 16, 2008 6:38 AM > To: r-help@r-project.org > Subject: Re: [R] One-sample test for p > > > Thanks for your responses. I know it is bad form asking about these > thi

Re: [R] One-sample test for p

2008-10-16 Thread rr400
Thanks for your responses. I know it is bad form asking about these things but i was really having trouble getting my head around it, and i wanted to make sure that the cause wasn't due to the commands i was entering into R. At least now i know it's a conceptual error i am making rather than a tec

Re: [R] One-sample test for p

2008-10-16 Thread Peter Dalgaard
Duncan Murdoch wrote: 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 a

Re: [R] One-sample test for p

2008-10-16 Thread Duncan Murdoch
On 10/16/2008 7:35 AM, Peter Dalgaard wrote: 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 i

Re: [R] One-sample test for p

2008-10-16 Thread Peter Dalgaard
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*

[R] One-sample test for p

2008-10-16 Thread rr400
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) >