In: Statistics and Probability
Arandomsampleofthebirthweightsof24babieshasameanof3103 grams and a standard deviation of 696 grams. Construct a 95% confidence interval estimate of the mean birth weight for all such babies.
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)
= 293.9
The 95% confidence interval mean is,
- E < 
 < 
 + E
3103 - 293.9 < 
 < 3103+ 293.9
2809.1 < 
 < 3396.9