Question 1192839
I am not sure what the "32" means so will ignore it.
Mean is 5.33 and s is 0.63
95% t-half interval for a sample of 6 has t (0..975, df=5)=t-table value is2.571*s/sqrt(n)
=2.571*0.63/sqrt(6)
=0.66
the interval is (4.67, 5.99)  units mg%, Since 4.8 is contained in the interval, there is insufficient evidence to state the patient is hyperphosphatemic at the 0.05 level.