A small group of 15 hummingbirds has been under study The av

A small group of 15 hummingbirds has been under study. The average weight for these birds if found to be 3.15g. Based on the previous studies, we can assume that the weights have a normal probability distribution with a standard deviation of 0.33g. Obtain 95% confidence interval for the average weights of these hummingbirds in the study region. Interpret your results.

Solution

sample size,n = 15
sample mean,xbar = 3.15
standard deviation,sigma = 0.33
alpha,a = 1-0.95 = 0.05

Za/2 = Z0.025 = 1.96

confidence interval:
xbar +/- [Za/2 * (sigma/sqrt(n))]
3.15 +/- [1.96 * (0.33/sqrt(15))]
3.15 +/- 0.167
= ( 2.983 , 3.317 )

We are 95% confident that true mean weights of hummingbirds in the study region lies in the interval ( 2.983 , 3.317 ).