In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a 90% confidence interval for the proportion of adults in the whole population who use email.
605/1000 is our "mean" as i recall
have to review a little to make sure what to do next :)
z*sqrt(pq/n)
the zscore for 90% is ...
\[\frac{605}{1000}\pm z*\sqrt{\frac{605}{1000}\frac{395}{1000}\frac{1}{1000}}\]
Is it OK if you can find out some more background info on this matter? Like, maybe a site to explain this more?
http://www.stat.wmich.edu/s160/book/node47.html this is what refreshed it for me
Thanks for it... So, from the last step you wrote, how do you continue?
you determine the zscore for 90% and plug it into the rest to determine your interval
So, how do you determine the zscore? With a table... or?
table, yes; or ti83
I don't have a graphic calc... Would you use the Modulus Function table... or?
oy this thing is slooowwww this morning
this measures the z from the mean itself; so we want .4500 from the mean so that the total area from the mean is 90%
.4500 in the feild is there in the middle; it rests neatly between 1.64 and 1.65 1.645 should be the zscore then
\[.605\pm 1.645\sqrt{.605*.305*.001}\]
the ;ower and upper limits of that stuff define the interval
Hmm.. Ok... I don't have this table in my statistics book. Do you think they'll give it me in the exam?
lol, i doubt it. the table should be on the back cover
or in a leaflet; its called a ztable
Hmmm... I hope they do! ztable? What's the difference between a ztable and a modulus function table?
depends on what a modulus function table is :)
I mean a normal distribution table...?
But... the z table you first gave looks different than the normal distribution table?
there are a few ways that it is written up depending on how the authors like to display it the main difference is in what they measure from.
in my book; they measure from the mean itself; in other books they measure from a tail
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