Question

We wish to estimate the mean serum indirect bilirubin level of 4-day-old infants. The mean for a sample of 16 infants was found to be 5.98 mg/100 cc. Assume that bilirubin levels in 4-day-old infants are approximately normally distributed with a standard deviation of 3.5 mg/100 cc. Construct the 99 percent confidence interval for the population mean.

Answer 1

Solution :

Given that,

= 5.98

s = 3.5

n = 16

Degrees of freedom = df = n - 1 = 16- 1 = 15

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,15 = 2.947

Margin of error = E = t_/2,df * (s /n)   

= 2.947 * (3.5/ 16) = 2.5789

The 99% confidence interval estimate of the population mean is,

- E < < + E

5.98 - 2.5789 < < 5.98 + 2.5789

3.4011 < < 8.5589

(3.4011 , 8.5589)