The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 570 peas with 111 of them having yellow pods. Find the value of the test statistic.
hi
Hey there
hi
don`t understand the question!!!
Nope
what do you not understand about the question?
How to put the equation into my TI83 to obtain the value
Of the test statistic
looks like we have 1 sample, and are tring to evaluate a proportion ...
what does 2nd, vars, tests (maybe calc) get for us?
ok I have it up but how would I input the information?
tell me what the inputs are ... are what youd assume they should be
po = ? n = ? x = ?
po=0.25 n=600 x= no sure about
not 600 I meant 570
po is good, now from the information we have only 2 numbers left to work 570 peas with 111 of them having yellow pods
Yes. So, then to put that into an equation is what I am having difficulty with.
there is no equation, you simply type them into the calculator po = .25 n=570 x=111 the last thing to do is to determine the type of tails we want to test.
you can work it as an equation, but then why would we have used the function to start with?
Oh ok... So how would I input it into the ti83?
i already covered that ... with links and pictures and everything.
Ok, I will try it out.
once you get to the function and fill in those 3 parts, theres one last part to cover. let me know when you are there.
ok, I have it calculated
and what is your value?
also, what did you choose for the last portion? \(\ne\)p, <p, or >p
All I see is: z=-3.047 p=.0011 p^=.1947 n=570
your test stat is z :)
I just clicked on calculate
thats fine, we werent asked anything that would alter the z value.
OMG thank you thank you!!!
another way to have done is: z = (111/570 - .25)/sqrt(.25(1-.25)/570) which is -3.047 http://www.wolframalpha.com/input/?i=%28111%2F570+-+.25%29%2Fsqrt%28.25%281-.25%29%2F570%29
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