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

1) Call x the sample mean = 3.56

2) Call s the sample standard deviation = 0.2

3) Given that the variable is normally distributed and the sample is large, you determine the interval of confidence from:

x +/- Z(0.5) s/√n

Wehre Z(0.5) is the value of the probabilities over 5% (90% of confidence mean to subtract 10%, which is 5% for each side (tails) of the normal distribuition) and is taken from tables.

Z(0.5) = 0.3085

Then the inteval is

x +/- 0.385 *s /√n = 3.56 +/- 0.385 * 0.2/√45

3.56 +/- 0.011 = ( 3.549, 3.571). This is the answer.