A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim.
***I am getting stuck on finding the sample proportion.
p= #successes/ n
This is how I tried to find p-value:
Step I
H sub naught: p is less than or equal too .28
H sub alternative p is greater than .28
Step II
Signifance level= .05
Step III
Calculate standard error for Z statistic manaully and w/Megastat.
Manuall I tried .32*130= 41.6 for p
inputting 41.6 as the sample proportion into the Z equation my results were:
1051.39 referenced this too the z table, z=.27 (.5 -.27=.23)
Megastat returned z= 1.02
Using Megastat results z=1.02 Z table area=.3461
.5-.3461= .1539 (manually able to compute using z=1.02 from Megastat)
***Here in lies my confusion, using megastat it gives the return
p value as .1549
z=1.02
the choices given for p value are:
0.1539
0.3078
0.3461
0.1635
??? Why the difference of p value .1549 and the answer provided .1539.
Trying to to figure out why this is
I know t test you can find rejection region but not p value
the z test I ought to be able to find rejection region and p value
Thank you. Jesse.
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