Answer:
Give
null hypothesis: H0 = 5.5
alternative hypothesis: Ha different 5.5
standard error = 1 / (31) 1/2= 0.1796
we deduce that:
t stat = (5.2-5.5) /0.1796 = -1.67033
p-value = 0.105
the null hypothesis cannot be rejected because the value of p is greater