Question

A simple random sample of 28 recent birth records at the local hospital was taken. In the sample, the average birth weight was 119.6 ounces and the sample standard deviation was 6.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean µ. Construct a 98% confidence interval for µ, the average birthweight of newborns.

(Show work please)

Answer 1

Solution :

Given that,

= 119.6

s =6.5

n =28

Degrees of freedom = df = n - 1 =28 - 1 = 27

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t _/2,df = t_0.01,27 =2.473 ( using student t table)

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

= 2.473 * (6.5 / 28)

E=3.0378

The 98% confidence interval estimate of the population mean is,

- E < < + E

119.6 -3.0378   < < 119.6+ 3.0378

116.5622 < < 122.6378

(116.5622 , 122.6378 )