Answer:
0.0500
Step-by-step explanation:
Given that a ample of 45 kittens was randomly selected from all kittens of this breed at a large veterinary hospital. The birth weight of each kitten in the sample was recorded. The sample mean was 3.56 ounces, and the sample standard deviation was 0.2 ounces.
For 90%
df = 45-1 =44
t critical value = 1.68
Std error = std dev/sq rt n
= [tex]\frac{0.2}{\sqrt{45} } \\=0.0298[/tex]
Margin of error = 1.68*std error
=0.0500
