Question

Arandomsampleofthebirthweightsof24babieshasameanof3103 grams and a standard deviation of 696 grams. Construct a 95% confidence interval estimate of the mean birth weight for all such babies

Answer 1

Solution :

Given that,

=3103

s = 696

n =24

Degrees of freedom = df = n - 1 =24 - 1 = 23

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,23 =2.069 ( using student t table)

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

= 2.069 * ( 696/ 24)

= 293.9437

The 95% confidence interval estimate of the population mean is,

- E < < + E

3103 -293.9437 < < 3103+ 293.9437

2807.0563 < < 3396.9437

( 2807.0563 ,3396.9437 )