Let [tex]\mu[/tex] be the population mean.
As per given , we have
Alternative hypothesis : [tex]\mu>55000[/tex] , since it is right -tailed so the test is a right tailed test.
[tex]\overline{x}=56500[/tex]
s= 3750 , since population standard deviation is unknown so we use t-test.
n=61
Degree of freedom : df= n-1=60
Test statistic : [tex]t=\dfrac{x-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]\\\\=\dfrac{56500-55000}{\dfrac{3750}{\sqrt{61}}}\\\\=3.12409987036\approx3.12[/tex]
P-value (For right tailed test) =0.0014 [using right-tailed p-value table for t-test]