Question

#10. A random sample of size 120 was obtained from a distribution presumed to be well behaved but of unknown form. The sample had a mean of 78.3 and s.d. of 8.1. An interval was reported as (77.56, 79.039) If this were to be a confidence interval, what would be the level of confidence?

Answer 1

Here Margin of Error is $E=\frac{79.039-77.56}{2}=0.7395$

So we will find z from formula of E

$E=z*\frac{\sigma }{\sqrt{n}}$

So $z=\frac{E}{\frac{\sigma }{\sqrt{n}}}=\frac{0.7395}{\frac{8.1}{\sqrt{120}}}=1$

So CI level is 68.27% as P(-1<z<1)=0.6827