Question

A marine biologist wanted to construct a t interval to estimate the mean weight of marine otters using 98% confidence. They took a random sample of n = 8 marine otters to measure their weights. These weights were roughly symmetric with a mean of 2 = 4.5 kg and a standard deviation of sx = 1.1 kg.What critical value t* should they use?

Answer 1

CI 98 % = ( 3.332 ; 5.668 )

t(c) = 2.9979

Explanation:

Confidence Interval CI 98 % then α = 2 % α = 0,02 α/2 = 0.01

Sample information:

Sample size n = 8

sample mean x = 4.5

standard deviation of sample s = 1.1 kg

degree of freedom df = n - 1 df = 8 - 1 df = 7

With α/2 0.01 and df = 7 from t-studente table we find t(c)

t(c) = 2.9979

CI 98 % = ( x ± t(c) * s/√n )

CI 98 % = ( 4.5 ± 2.9979 * 1.1/ √8 )

CI 98 % = ( 4.5 ± 3.2976/2.8228 )

CI 98 % = ( 4.5 ± 1.168 )

CI 98 % = ( 3.332 ; 5.668 )