In: Statistics and Probability
Suppose a certain species bird has an average weight of x=3.85 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with 0=0.29 grams. For a small group of 17 birds, find a 75% confidence interval for the average weights of these birds.
Solution :
Given that,
= 3.85
= 0.29
n = 17
At 75% confidence level the z is ,
= 1 - 75% = 1 - 0.75 = 0.25
/ 2 = 0.25 / 2 = 0.125
Z/2 = Z0.255 = 1.15
Margin of error = E = Z/2* ( /n)
= 1.15* (0.29 / 17)
= 0.08
At 99% confidence interval estimate of the population mean is,
- E < < + E
3.85-0.08 < < 3.85+0.08
3.77< < 3.93