Question

A study was conducted to determine the mean birth weight of a certain breed of kittens. Consider the birth weights of kittens to be normally distributed. A sample 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. What is the margin of error for a 90% confidence interval on the mean birth weight of all kittens of this breed.

Answer 1

45 kittens mean =3.56 standard deviation = 0.2 so youre looking for 90% of the bell curve... yeah?

Answer 2

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

= $\frac{0.2}{\sqrt{45} } \\=0.0298$

Margin of error = 1.68*std error

=0.0500