The Minitab and TI results are (modulo different levels of rounding)
just what you'd get from
doing the problem ``by hand'' in the good old-fashioned way. :-)
The Excel result appears to be the same with an excessive level of rounding.
The ``by-hand'' procedure uses the plug-in method to get an
approximation to the standard
error of p.hat. The prop.test() function uses a more sophisticated
approach which involves
solving a quadratic equation to determine the endpoints of the
confidence interval. This
more sophisticated solution is a pain in the pohutukawa ( :-) ) to
calculate by hand, but if you've
got a computer to do the nasty arithmetic for you, then why not?
The formula for the confidence interval endpoints using the more
sophisticated method can
be found e.g. in Jay L. Devore, Probability and Statistics for
Engineering and the Sciences,
Thomson --- Brooks/Cole, 6th ed., 2004, page 295. equation (7.10). The
much simpler old-fashioned
formula is given on the same page as equation (7.11).
Presumably these formulae can be found in the Newcombe reference cited
in the help for
prop.test(); I haven't checked.
HTH
cheers,
Rolf Turner
On 18/07/11 04:27, Jack Sofsky wrote:
I have just joined this list (and just started using R), so please
excuse any etiquette breaches as I do not yet have a feel for how the
list operates.
I am in the process of teaching myself statistics using R as my
utility as my ultimate goals cannot be satisfied by Excel or any of
the plug-ins I could afford.
I am currently looking at chap12 page 552 of Weiss's Introductory
Statistics 9th edition. Example 12.5 demonstrates using "Technology"
to obtain a One-Proportion z-Interval.
n=202
x=1010
confidence interval = .95.
Answer given by Minitab
0.175331, .224669
Answer given by TI-83/84
.17533, .22467
Answer given by Weiss's Excel Plug-in
0.175 < p < 0.225
Here is what I got with R
prop.test(202,1010,correct="FALSE")
1-sample proportions test without continuity correction
data: 202 out of 1010, null probability 0.5
X-squared = 363.6, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.1764885 0.2257849
sample estimates:
p
0.2
I'm also getting slight differences in the answers for exercises and
find this disconcerting.
Why are these differences present (or am I doing something wrong)?
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