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.