Answer to Question #293820 in Statistics and Probability for Emu

Question

We wish to estimate the mean serum indirect bilirubin level of 4 days old infants.the mean for the sample of 16 infants was found to be 5.98mg/100cc.assume the bilirubin levels in 4 days old infant are approximately normally distributed with a standard deviation of 3.5 mg/100cc.calculate the 90,95 and 99% confidence interval and compare the intervals?

Answer 1

$90\%CI=(5.98-1.64\frac{3.5}{\sqrt{16}},5.98+1.64\frac{3.5}{\sqrt{16}})=(-0.60,12.56).$

$95\%CI=(5.98-1.96\frac{3.5}{\sqrt{16}},5.98+1.96\frac{3.5}{\sqrt{16}})=(-1.86,13.82).$

$99\%CI=(5.98-2.58\frac{3.5}{\sqrt{16}},5.98+2.58\frac{3.5}{\sqrt{16}})=(-4.32,16.28).$

The 99% confidence interval is the widest.

The 90% confidence interval is the narrowest.